a) \(3\sqrt{3}=\sqrt{27}>\sqrt{12}\)
b) \(3\sqrt{5}=\sqrt{45}>\sqrt{27}\)
c) \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{51}{9}}< \sqrt{\dfrac{54}{9}}=6=\sqrt{\dfrac{150}{25}}=\dfrac{1}{5}\sqrt{150}\)
d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{6}{4}}=\sqrt{\dfrac{3}{2}}< \sqrt{\dfrac{36}{2}}=6\sqrt{\dfrac{1}{2}}\)
LG a
12√48−2√75−√33√11+5√1131248−275−3311+5113;
Phương pháp giải:
+ Cách đổi hỗn số ra phân số: abc=a.c+bcabc=a.c+bc.
+ Sử dụng quy tắc đưa thừa số ra ngoài dấu căn:
√A2.B=A√BA2.B=AB, nếu A≥0, B≥0A≥0, B≥0.
√A2.B=−A√BA2.B=−AB, nếu A<0, B≥0A<0, B≥0.
+ √ab=√a√bab=ab, với a≥0, b>0a≥0, b>0.
+ √a.√b=√aba.b=ab, với a, b≥0a, b≥0.
+ A√B=A√BBAB=ABB, với B>0B>0.
Lời giải chi tiết:
Ta có:
12√48−2√75−√33√11+5√1131248−275−3311+5113
=12√16.3−2√25.3−√3.11√11+5√1.3+13=1216.3−225.3−3.1111+51.3+13
=12√42.3−2√52.3−√3.√11√11+5√43=1242.3−252.3−3.1111+543
=12.4√3−2.5√3−√3+5√4√3=12.43−2.53−3+543
=42√3−10√3−√3+5√4.√3√3.√3=423−103−3+54.33.3
=2√3−10√3−√3+52√33=23−103−3+5233
=2√3−10√3−√3+10√33=23−103−3+1033
=(2−10−1+103)√3=(2−10−1+103)3
=−173√3=−1733.
LG b
√150+√1,6.√60+4,5.√223−√6;150+1,6.60+4,5.223−6;
Phương pháp giải:
+ Cách đổi hỗn số ra phân số: abc=a.c+bcabc=a.c+bc.
+ Sử dụng quy tắc đưa thừa số ra ngoài dấu căn:
√A2.B=A√BA2.B=AB, nếu A≥0, B≥0A≥0, B≥0.
√A2.B=−A√BA2.B=−AB, nếu A<0, B≥0A<0, B≥0.
+ √ab=√a√bab=ab, với a≥0, b>0a≥0, b>0.
+ √a.√b=√aba.b=ab, với a, b≥0a, b≥0.
+ A√B=A√BBAB=ABB, với B>0B>0.
Lời giải chi tiết:
Ta có:
√150+√1,6.√60+4,5.√223−√6150+1,6.60+4,5.223−6
=√25.6+√1,6.60+4,5.√2.3+23−√6=25.6+1,6.60+4,5.2.3+23−6
=√52.6+√1,6.(6.10)+4,5√83−√6=52.6+1,6.(6.10)+4,583−6
=5√6+√(1,6.10).6+4,5√8√3−√6=56+(1,6.10).6+4,583−6
=5√6+√16.6+4,5√8.√33−√6=56+16.6+4,58.33−6
=5√6+√42.6+4,5√8.33−√6=56+42.6+4,58.33−6
=5√6+4√6+4,5.√4.2.33−√6=56+46+4,5.4.2.33−6
=5√6+4√6+4,5.√22.63−√6=56+46+4,5.22.63−6
=5√6+4√6+4,5.2√63−√6=56+46+4,5.263−6
=5√6+4√6+9√63−√6=56+46+963−6
=5√6+4√6+3√6−√6=56+46+36−6
=(5+4+3−1)√6=11√6.=(5+4+3−1)6=116.
Cách 2: Ta biến đổi từng hạng tử rồi thay vào biểu thức ban đầu:
+ √150=√25.6=5√6150=25.6=56
+ √1,6.60=√1,6.(10.6)=√(1,6.10).6=√16.61,6.60=1,6.(10.6)=(1,6.10).6=16.6
=4√6=46
+ 4,5.√223=4,5.√2.3+23=4,5.√83=4,5√8.334,5.223=4,5.2.3+23=4,5.83=4,58.33
=4,5.√4.2.33=4,5.2.√63=9.√63=3√6.=4,5.4.2.33=4,5.2.63=9.63=36.
Do đó:
√150+√1,6.√60+4,5.√223−√6150+1,6.60+4,5.223−6
=5√6+4√6+3√6−√6=56+46+36−6
=(5+4+3−1)√6=11√6=(5+4+3−1)6=116
LG c
(√28−2√3+√7)√7+√84;(28−23+7)7+84;
Phương pháp giải:
+ Cách đổi hỗn số ra phân số: abc=a.c+bcabc=a.c+bc.
+ Hằng đẳng thức số 1: (a+b)2=a2+2ab+b2(a+b)2=a2+2ab+b2.
+ Sử dụng quy tắc đưa thừa số ra ngoài dấu căn:
√A2.B=A√BA2.B=AB, nếu A≥0, B≥0A≥0, B≥0.
√A2.B=−A√BA2.B=−AB, nếu A<0, B≥0A<0, B≥0.
+ √ab=√a√bab=ab, với a≥0, b>0a≥0, b>0.
+ √a.√b=√aba.b=ab, với a, b≥0a, b≥0.
+ A√B=A√BBAB=ABB, với B>0B>0.
Lời giải chi tiết:
Ta có:
=(√28−2√3+√7)√7+√84=(28−23+7)7+84
=(√4.7−2√3+√7)√7+√4.21=(4.7−23+7)7+4.21
=(√22.7−2√3+√7)√7+√22.21=(22.7−23+7)7+22.21
=(2√7−2√3+√7)√7+2√21=(27−23+7)7+221
=2√7.√7−2√3.√7+√7.√7+2√21=27.7−23.7+7.7+221
=2.(√7)2−2√3.7+(√7)2+2√21=2.(7)2−23.7+(7)2+221
=2.7−2√21+7+2√21=2.7−221+7+221
=14−2√21+7+2√21=14−221+7+221
=14+7=21=14+7=21.
LG d
(√6+√5)2−√120.(6+5)2−120.
Phương pháp giải:
+ Cách đổi hỗn số ra phân số: abc=a.c+bcabc=a.c+bc.
+ Hằng đẳng thức số 1: (a+b)2=a2+2ab+b2(a+b)2=a2+2ab+b2.
+ Sử dụng quy tắc đưa thừa số ra ngoài dấu căn:
√A2.B=A√BA2.B=AB, nếu A≥0, B≥0A≥0, B≥0.
√A2.B=−A√BA2.B=−AB, nếu A<0, B≥0A<0, B≥0.
+ √a.√b=√aba.b=ab, với a, b≥0a, b≥0.
Lời giải chi tiết:
Ta có:
(√6+√5)2−√120(6+5)2−120
=(√6)2+2.√6.√5+(√5)2−√4.30=(6)2+2.6.5+(5)2−4.30
=6+2√6.5+5−2√30=6+26.5+5−230
=6+2√30+5−2√30=6+5=11.=6+230+5−230=6+5=11.
a) Ta có:
\(2\sqrt{3}=\sqrt{2^2.3}=\sqrt{12}.\)
Mà \(\sqrt{12}< \sqrt{13}\)
Nên \(2\sqrt{3}< \sqrt{13}\)
+ Ta có:
2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)
=2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5
=2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).
+ Ta có:
3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)
=3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7
=3(√10−√7)3=√10−√7=3(10−7)3=10−7.
+ Ta có:
1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)
=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y
+ Ta có:
2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)
=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.
\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)
\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
a) a) Biến đổi vế trái thành 32√6+23√6−42√6326+236−426 và làm tiếp.
b) Biến đổi vế trái thành (√6x+13√6x+√6x):√6x(6x+136x+6x):6x và làm tiếp
a) VT=\left(\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}VT=(8−223−6−3216)⋅61
=\left(\dfrac{\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{3}-\sqrt{6}}{\sqrt{2^{2} \cdot 2}-2}-\dfrac{\sqrt{6^{2} .6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}=(22⋅2−22⋅2⋅3−6−362.6)⋅
+ Ta có:
3√3+1=3(√3−1)(√3+1)(√3−1)=3√3−3.1(√3)2−1233+1=3(3−1)(3+1)(3−1)=33−3.1(3)2−12
=3√3−33−1=3√3−32=33−33−1=33−32.
+ Ta có:
2√3−1=2(√
\(\frac{3}{\sqrt{3}+1}=\frac{3\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{3\sqrt{3}-3}{3-1}=\frac{3\sqrt{3}-3}{2}\)
\(\frac{2}{\sqrt{3}-1}=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{2\left(\sqrt{3}+1\right)}{3-1}=\sqrt{3}-1\)
\(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3}=\left(2+\sqrt{3}\right)^2=4+4\sqrt{3}+3=7+4\sqrt{3}\)
\(\frac{b}{3+\sqrt{b}}=\frac{b\left(3-\sqrt{b}\right)}{\left(3+\sqrt{b}\right)\left(3-\sqrt{b}\right)}=\frac{b\left(3-\sqrt{b}\right)}{9-b}\)
\(\frac{p}{2\sqrt{p}-1}=\frac{p\left(2\sqrt{p}+1\right)}{\left(2\sqrt{p}-1\right)\left(2\sqrt{b}+1\right)}=\frac{p\left(2\sqrt{b}+1\right)}{4p-1}\)
a, \(\sqrt{\left(2x-1\right)^2}=3\Leftrightarrow\left|2x-1\right|=3\)
Với \(x\ge\frac{1}{2}\)pt có dạng : \(2x-1=3\Leftrightarrow x=2\)( tm )
Với \(x< \frac{1}{2}\)pt có dạng : \(-2x+1=3\Leftrightarrow x=-1\)( tm )
Vậy tập nghiệm của pt là S = { -1 ; 2 }
b, \(\frac{5}{3}\sqrt{15x}-\sqrt{15x}-2=\frac{1}{3}\sqrt{15x}\)ĐK : \(x\ge0\)
\(\Leftrightarrow\frac{2}{3}\sqrt{15x}-2=\frac{1}{3}\sqrt{15x}\Leftrightarrow\frac{1}{3}\sqrt{15x}=2\)
\(\Leftrightarrow\sqrt{15x}=6\)bình phương 2 vế : \(\Leftrightarrow15x=36\Leftrightarrow x=\frac{36}{15}=\frac{12}{5}\)( tm )
Vậy tập nghiệm của pt là S = { 12/5 }
a, \(3\sqrt{3}\) >\(2\sqrt{3}\) =>\(3\sqrt{3}\) >\(\sqrt{12}\)
b,có \(3\sqrt{5}=\sqrt{45}\) <\(\sqrt{49}=7\) =>7 >\(3\sqrt{5}\)
c,\(\sqrt{\dfrac{51}{9}}\) <\(\sqrt{6}\) => \(\dfrac{1}{3}\sqrt{51}\) <\(\dfrac{1}{5}\sqrt{150}\)
d.\(\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
a) 3\(\sqrt{3}\)=\(\sqrt{27}\)>\(\sqrt{12}\)
c) \(\frac{1}{3}\)\(\sqrt{51}\)=\(\sqrt{\frac{51}{9}}\)<\(\frac{1}{5}\)\(\sqrt{150}\)=\(\sqrt{\frac{150}{25}}\)=\(\sqrt{6}\)
b) 3\(\sqrt{5}\)=\(\sqrt{45}\)< 7=\(\sqrt{49}\)
d) \(\frac{1}{2}\sqrt{6}\)=\(\sqrt{\frac{6}{4}}\)=\(\sqrt{\frac{3}{2}}\)< 6\(\sqrt{\frac{1}{2}}\)=\(\sqrt{\frac{36}{2}}\)=\(\sqrt{18}\)
a) Ta có: 3√3=√32.3=√9.3=√2733=32.3=9.3=27
Vì √27>√1227>12 nên 3√3>√1233>12
Vậy 3√3>√1233>12.
b) Ta có: 3√5=√32.5=√4535=32.5=45
7=√72=√497=72=49
Vì √49>√4549>45 nên 7>3√57>35
Vậy
Đúng(0)
a) \(3\sqrt{3}=\sqrt{9}.\sqrt{3}=\sqrt{27}>\sqrt{12}\)
b) \(3\sqrt{5}=\sqrt{9}.\sqrt{5}=\sqrt{45}< \sqrt{49}=7\)
c) \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{1}{9}}.\sqrt{51}=\sqrt{\dfrac{51}{9}}=\sqrt{\dfrac{17}{3}}< \sqrt{6}=\dfrac{1}{5}\sqrt{150}\)
d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{3}{2}}< \sqrt{18}=6\sqrt{\dfrac{1}{2}}\)
\(\dfrac{1}{2}\sqrt{6}=\dfrac{\sqrt{6}}{2}< 6\sqrt{\dfrac{1}{2}}=3\sqrt{2}\)\(3\sqrt{3}=\sqrt{27}>\sqrt{12}\)\(7< 3\sqrt{5}=45\)\(\dfrac{1}{3}\sqrt{51}=\dfrac{\sqrt{51}}{3}< \dfrac{1}{5}\sqrt{150}=\sqrt{6}\)
a) 3\(\sqrt{3}\)>\(\sqrt{12}\)
b)7>3\(\sqrt{5}\)
C)\(\dfrac{1}{3}\sqrt{51}\)<\(\dfrac{1}{5}\sqrt{150}\)
d)\(\dfrac{1}{2}\sqrt{6}\)<\(6\sqrt{\dfrac{1}{2}}\)
\(\sqrt{5}\)
a) Ta có: 3 \sqrt{3}=\sqrt{3^2.3}=\sqrt{9.3}=\sqrt{27}33=32.3=9.3=27
Vì \sqrt{27}>\sqrt{12}27>12 nên 3 \sqrt{3}>\sqrt{12}33>12
Vậy 3 \sqrt{3}>\sqrt{12}33>12.
b) Ta có: 3 \sqrt{5}=\sqrt{3^{2}. 5}=\sqrt{45}35=32. 5=45
7=\sqrt{7^{2}}=\sqrt{49}7=72=49
<...a lớn hơn
b lớn hơn
c nhỏ hơn
d nhỏ hơn
a) ta có: 3\(\sqrt{3}\)=\(\sqrt{3^2.3}\)= \(\sqrt{9.4}\) =\(\sqrt{27}\) vì \(\sqrt{27}\)>\(\sqrt{12}\)nên 3\(\sqrt{3}>\sqrt{12}\) b) tao có: 3\(\sqrt{5}\)= \(\sqrt{3^2.5}=\sqrt{9.5}=\sqrt{45}\) ; 7=\(\sqrt{7^2}=\sqrt{49}\) vì \(\sqrt{49}>\sqrt{45}nên\) 7> 3\(\sqrt{5}\) c)ta có: \(\dfrac{1}{3}\sqrt{51}=\sqrt{\left(\dfrac{1}{3}\right)^2.51}=\dfrac{51}{9}\); \(\dfrac{1}{5}\sqrt{50}=\sqrt{\left(\dfrac{1}{5}\right)^2.150}=\sqrt{\dfrac{150}{25}}=\sqrt{6}=\sqrt{\dfrac{69}{9}}\)
a)\(3\sqrt{3}=\sqrt{3^2.3}=\sqrt{9.3}=\sqrt{27}>\sqrt{12}\)
d)\(\dfrac{1}{2}\sqrt{6}=\sqrt{\left(\dfrac{1}{2}\right)^2.6}=\sqrt{\dfrac{3}{2}}=\sqrt{3.\dfrac{1}{2}}=\sqrt{3}.\sqrt{\dfrac{1}{2}}< 6\sqrt{\dfrac{1}{2}}\)
c)\(\dfrac{1}{3}\sqrt{51}=\sqrt{\left(\dfrac{1}{3}\right)^2.51}=\sqrt{\dfrac{51}{9}};\\ \dfrac{1}{5}\sqrt{150}=\sqrt{\left(\dfrac{1}{5}\right)^2.150}=\sqrt{\dfrac{150}{25}}=\sqrt{6}\Rightarrow\dfrac{1}{3}\sqrt{51}< \dfrac{1}{5}\sqrt{150}\)
b)\(3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45};7=\sqrt{7^2}=\sqrt{49}=>\sqrt{49}>\sqrt{45}\)
a) \(3\sqrt{3}=\sqrt{3^2.3}=\sqrt{27}\) vậy \(\sqrt{27}>\sqrt{12}=>3\sqrt{3}>\sqrt{12}\) b) \(3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45}\) ; \(7=\sqrt{7^2}=\sqrt{49}\) vậy \(\sqrt{45}< \sqrt{49}=>7>3\sqrt{5}\) c)\(\dfrac{1}{3}\sqrt{51}=\sqrt{\left(\dfrac{1}{3}\right)^2.51}=\sqrt{\dfrac{51}{9}}\) ; \(\dfrac{1}{5}\sqrt{150}=\sqrt{\left(\dfrac{1}{5}\right)^2.150}=\sqrt{6}\) vậy \(\dfrac{1}{3}\sqrt{51}< \dfrac{1}{5}\sqrt{150}\) d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\left(\dfrac{1}{2}\right)^2.6}=\sqrt{\dfrac{6}{4}}=\sqrt{\dfrac{3}{2}}=\sqrt{3.\dfrac{1}{2}}\) vậy \(\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
1/2 căn 6 < 6 căn 1/2
a)có \(\sqrt{12}=\sqrt{4\cdot3}=2\sqrt{3}\)
VÌ \(3\sqrt{3}>2\sqrt{3}\) \(\Rightarrow3\sqrt{3}>\sqrt{12}\)
b)có \(7=\sqrt{7^2}=\sqrt{49}\)
\(3\sqrt{5}=\sqrt{3^2\cdot5}=\sqrt{45}\)
Vì \(\sqrt{49}>\sqrt{45}\)
\(\Rightarrow7>3\sqrt{5}\)
c) có \(\sqrt{\left(\dfrac{1}{3}\right)^2\cdot51}=\sqrt{\dfrac{17}{3}}\)
\(\dfrac{1}{5}\cdot\sqrt{150}=\sqrt{\left(\dfrac{1}{5}\right)^2\cdot150}=\sqrt{6}=\sqrt{\dfrac{18}{3}}\)
Vì \(\sqrt{\dfrac{17}{3}}< \sqrt{\dfrac{18}{3}}\)
\(\Rightarrow\dfrac{1}{3}\sqrt{51}>\dfrac{1}{5}\sqrt{150}\)
d) có\(\dfrac{1}{2}\cdot\sqrt{6}=\sqrt{\left(\dfrac{1}{2}\right)^2\cdot6}=\sqrt{\dfrac{3}{2}}=\sqrt{1,5}\)
\(6\sqrt{\dfrac{1}{2}}=\sqrt{6^2\cdot\dfrac{1}{2}}=\sqrt{18}\)
Vì \(\sqrt{1,5}< \sqrt{18}\)
\(\Rightarrow\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
b)\(7>3\sqrt{5}\)
a)\(3\sqrt{3}>\sqrt{12}\)
c)\(\dfrac{1}{3}\sqrt{51}< \dfrac{1}{5}\sqrt{150}\)
d)\(\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
a) Ta có: 3\(\sqrt{3}\)=\(\sqrt{3^2.3}\)=\(\sqrt{9.3}\)=\(\sqrt{27}\)
Vì \(\sqrt{27}\)>\(\sqrt{12}\)
Nên \(3\sqrt{3}\)>\(\sqrt{12}\)
Vậy \(3\sqrt{3}\)>\(\sqrt{12}\)
b) Ta có: \(3\sqrt{5}\)=\(\sqrt{3^2.5}\)=\(\sqrt{45}\)
7=\(\sqrt{7^2}\)=\(\sqrt{49}\)
Vì \(\sqrt{45}\)<\(\sqrt{49}\)
Nên 7>\(3\sqrt{3}\)
Vậy 7>\(3\sqrt{3}\)
c) Ta có: \(\dfrac{1}{3}\sqrt{51}\)=\(\sqrt{(\dfrac{1}{3})^2.51}\)=\(\sqrt{\dfrac{51}{9}}\)
\(\dfrac{1}{5}\sqrt{150}\)=\(\sqrt{(\dfrac{1}{5})^2.150}\)=\(\sqrt{\dfrac{150}{25}}=\sqrt{6}=\sqrt{\dfrac{6.9}{9}}=\sqrt{\dfrac{54}{9}}\)
Vì \(\sqrt{\dfrac{54}{9}}>\sqrt{\dfrac{51}{9}}\)
Nên \(\dfrac{1}{3}\sqrt{51}< \dfrac{1}{5}\sqrt{150}\)
Vậy \(\dfrac{1}{3}\sqrt{51}< \dfrac{1}{5}\sqrt{150}\)
d) Ta có: \(\dfrac{1}{2}\sqrt{6}\)=\(\sqrt{(\dfrac{1}{2})^2.6}\)=\(\sqrt{\dfrac{6}{4}}\)=\(\sqrt{\dfrac{3}{2}}=\sqrt{3.\dfrac{1}{2}}=\sqrt{3}.\sqrt{\dfrac{1}{2}}\)
Vì \(\sqrt{3}\).\(\sqrt{\dfrac{1}{2}}< 6\sqrt{\dfrac{1}{2}}\)
Nên \(\dfrac{1}{2}.\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
Vậy \(\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
a = \(\sqrt{ }\)27 > \(\sqrt{ }\)12 ⇒3\(\sqrt{ }\)3 > \(\sqrt{ }\)12 b = \(\sqrt{ }\)7^2 &\(\sqrt{ }\)45 = \(\sqrt{ }\)49 > \(\sqrt{ }\)45 ⇒7 > 3\(\sqrt{ }\)5 c = \(\sqrt{ }\)(1/3)^2 .51 & \(\sqrt{ }\)(1/5)^2. 150 = \(\sqrt{ }\)17/3 < \(\sqrt{ }\)18/3 ⇒ 1/3 \(\sqrt{ }\)51 < 1/5\(\sqrt{ }\)150 d = \(\sqrt{ }\)(1/2)^2.6 & \(\sqrt{ }\)36.1/2 = \(\sqrt{ }\)1,5 < \(\sqrt{ }\)18 ⇒ 1/2\(\sqrt{ }\)6<6\(\sqrt{ }\)1/2
a) Có 3√3=√32.3=√9.3=√27
Vì √27>√12 nên 3√3>√12
=> 3√3>√12
b) Ta có 3√5=√32.5=√45
7=√72=√49
Vì √49> √45 nên 7>3√5
=> 7>3√5
c) Ta có: 1351=(13)2.51=519\dfrac{1}{3} \sqrt{51}=\sqrt{\left(\dfrac{1}{3}\right)^{2} .51}=\sqrt{\dfrac{51}{9}}315
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a) Ta có: 3 \sqrt{3}=\sqrt{3^2.3}=\sqrt{9.3}=\sqrt{27}33=32.3=9.3=27
Vì \sqrt{27}>\sqrt{12}27>12 nên 3 \sqrt{3}>\sqrt{12}33>12
Vậy 3 \sqrt{3}>\sqrt{12}33>12.
b) Ta có: 3 \sqrt{5}=\sqrt{3^{2}. 5}=\sqrt{45}35=32. 5=45
7=\sqrt{7^{2}}=\sqrt{49}7=72=49
a) có \(3\sqrt{3}=\sqrt{3^2\cdot3}=\sqrt{9\cdot3}=\sqrt{27}\) mà \(\sqrt{27}>\sqrt{12}\) vậy \(3\sqrt{3}>\sqrt{12}\) b) có\(3\sqrt{5}=\sqrt{3^2\cdot5}=\sqrt{45}\) \(7=\sqrt{7^2}=\sqrt{49}\) mà\(\sqrt{49}>\sqrt{45}\) vậy 7>\(3\sqrt{5}\) c) có \(\dfrac{1}{3}\sqrt{51}=\sqrt{\left(\dfrac{1}{3}\right)^2\cdot51}=\sqrt{\dfrac{51}{9}}\) \(\dfrac{1}{5}\sqrt{150}=\sqrt{\left(\dfrac{1}{5}\right)^2\cdot150}=\sqrt{\dfrac{150}{25}}=\sqrt{6}\) mà \(\dfrac{1}{5}\sqrt{150}=\sqrt{6}=\sqrt{\dfrac{6\cdot9}{9}}=\sqrt{\dfrac{54}{9}}>\sqrt{\dfrac{51}{9}}\) vậy\(\dfrac{1}{3}\sqrt{51}< \dfrac{1}{5}\sqrt{150}\) d)có \(\dfrac{1}{2}\sqrt{6}=\sqrt{\left(\dfrac{1}{2}\right)^2\cdot6}=\sqrt{\dfrac{6}{4}}=\sqrt{\dfrac{3}{2}}=\sqrt{3\cdot\dfrac{1}{2}}=\sqrt{3}\sqrt{\dfrac{1}{2}}< 6\sqrt{\dfrac{1}{2}}\) vậy \(\dfrac{1}{2}\cdot\sqrt{6}\) <\(6\sqrt{\dfrac{1}{2}}\)
a) 3\(\sqrt{3}\) > \(\sqrt{12}\)
b) 7 > 3\(\sqrt{5}\)
c) \(\dfrac{1}{3}\)\(\sqrt{51}\) < \(\dfrac{1}{5}\)\(\sqrt{150}\)
d) \(\dfrac{1}{2}\)\(\sqrt{6}\) < \(6\sqrt{\dfrac{1}{2}}\)
a) Ta có: 3√3=√32.3=√9.3=√2733=32.3=9.3=27
Vì √27>√1227>12 nên 3√3>√1233>12
Vậy 3√3>√1233>12.
b) Ta có: 3√5=√32.5=√4535=32.5=45
7=√72=√497=72=49
Vì √49>√4549>45 nên 7>3√57>35
Vậy
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