a) \(\sqrt{16}\cdot\sqrt{25}+\sqrt{196}:\sqrt{49}\)
\(=\sqrt{16\cdot25}+\sqrt{196:49}\)
\(=20+2=22\)
b) \(36:\sqrt{2\cdot3^2\cdot18}-\sqrt{169}\)
\(=36:\sqrt{324}-\sqrt{169}\)
\(=36:18-13=2-13=-11\)
c) \(\sqrt{\sqrt{81}}\)
\(=\sqrt{9}=3\)
d) \(\sqrt{3^2+4^2}\)
\(=\sqrt{9+16}=\sqrt{25}=5\)
a) \(\sqrt{16}.\sqrt{25}+\sqrt{196}\div\sqrt{49}\)
\(=4.5+14:7\)
\(=20+2=22\)
b) \(36:\sqrt{2.3^2.18}-\sqrt{169}\)
\(=36:18-13=-11\)
c) \(\sqrt{\sqrt{81}}=\sqrt{9}=3\)
d) \(\sqrt{3^2+4^2}=\sqrt{25}=5\)
Bài làm:
Bài 1:
a)\(\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)
= 4.5 + 14 : 7
= 20 + 2
= 22
b)\(36:\sqrt{2.3^2.18}-\sqrt{169}\)
= 36 : 18 - 14
= 2 - 14
= - 12
c)\(\sqrt{\sqrt{81}}\) = \(\sqrt{9}\) = 3
d)\(\sqrt{3^2+4^2}\)
= \(\sqrt{9+16}\)
= \(\sqrt{25}\)
= 5
a) (\(\sqrt{3}\)-1)2=3-2\(\sqrt{3}\)+1= 4-2\(\sqrt{3}\) (ĐPCM)
b) \(\sqrt{4-2\sqrt{3}}\)=\(\sqrt{3}\)-1 >0
Bình phương 2 vế, ta có:
4-2\(\sqrt{3}\)=3-2\(\sqrt{3}\)+1= 4-2\(\sqrt{3}\) (ĐPCM)
a) \(\left(\sqrt{3}-1\right)^2\)=\(\left(\sqrt{3}\right)^2\)- 2\(\sqrt{3}\) +1= 3- 2\(\sqrt{3}\) +1=4-2\(\sqrt{3}\)
b) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\) = \(\sqrt{\left(\sqrt{3}-1\right)^2}\) - \(\sqrt{3}\)= \(|\sqrt{3}-1|\)-\(\sqrt{3}\)=\(\sqrt{3}\)-1-\(\sqrt{3}\)=-1
a, Ta có \(\sqrt{25-16}=\sqrt{9}=3\)
\(\sqrt{25}-\sqrt{16}=5-4=1\)
Do 3 > 1 nên \(\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)
a) căn 25 - 16 > căn 25 - căn 16
b)Với a>b>0a>b>0 nên \sqrt{a},\sqrt{b},\sqrt{a-b}a,b,− đều xác định
Để so sánh \sqrt{a}-\sqrt{b}a−b và \sqrt{a-b}− ta quy về so sánh \sqrt{a}a và \sqrt{a-b}+\sqrt{b}−+b.
+) (\sqrt{a})^2=a(a)2=a.
+) (\sqrt{a-b}+\sqrt{b})^2=(\sqrt{a-b})^2+2\sqrt{a-b}.\sqrt{b}+(\sqrt{b})^2=a-b+b+2\sqrt{a-b}.\sqrt{b}=a+2\sqrt{a-b}.\sqrt{b}(−+b)2=(−)2+2−.b+(b)2=a−b+b+2−.b=a+2−
a) Ta có:
+)√25+9=√34+)25+9=34.
+)√25+√9=√52+√32=5+3+)25+9=52+32=5+3
=8=√82=√64=8=82=64.
Vì 34<6434<64 nên √34<√6434<64
Vậy √25+9<√25+√925+9<25+9
b) Với a>0,b>0a>0,b>0, ta có
+)(√a+b)2=a+b+)(a+b)2=a+b.
+)(√a+√b)2=(√a)2+2√a.√b+(√b)2+)(a+b)2=(a)2+2a.b+(b)2
=a+2√ab+b=a+2ab+b
=(a+b)+2√ab=(a+b)+2ab.
Vì a>0, b>0a>0, b>0 nên √ab>0⇔2√ab>0ab>0⇔2ab>0
⇔(a+b)+2√ab>a+b⇔(a+b)+2ab>a+b
⇔(√a+√b)2>(√a+b)2⇔(a+b)2>(a+b)2
⇔√a+√b>√a+b⇔a+b>a+b (đpcm)
a, Ta có : \(\sqrt{25+9}=\sqrt{34}\)
\(\sqrt{25}+\sqrt{9}=5+3=8=\sqrt{64}\)
mà 34 < 64 hay \(\sqrt{25+9}< \sqrt{25}+\sqrt{9}\)
b, \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)
bình phương 2 vế ta được : \(a+b< a+2\sqrt{ab}+b\)
\(\Leftrightarrow2\sqrt{ab}>0\)vì \(a;b>0\)nên đẳng thức này luôn đúng )
Vậy ta có đpcm
a) Điều kiện: x≥0x≥0
√16x=816x=8⇔(√16x)2=82⇔(16x)2=82 ⇔16x=64⇔16x=64 ⇔x=6416⇔x=4⇔x=6416⇔x=4 (thỏa mãn điều kiện)
Vậy x=4x=4.
Cách khác:
√16x=8⇔√16.√x=8⇔4√x=8⇔√x=2⇔x=22⇔x=416x=8⇔16.x=8⇔4x=8⇔x=2⇔x=22⇔x=4
b) Điều kiện: 4x≥0⇔x≥04x≥0⇔x≥0
√4x=√54x=5 ⇔(√4x)2=(√5)2⇔4x=5⇔x=54⇔(4x)2=(5)2⇔4x=5⇔x=54 (thỏa mãn điều kiện)
Vậy x=54x=54.
c) Điều kiện: 9(x−1)≥0⇔x−1≥0⇔x≥19(x−1)≥0⇔x−1≥0⇔x≥1
√9(x−1)=219(x−1)=21⇔3√x−1=21⇔3x−1=21⇔√x−1=7⇔x−1=7 ⇔x−1=49⇔x=50⇔x−1=49⇔x=50 (thỏa mãn điều kiện)
Vậy x=50x=50.
Cách khác:
√9(x−1)=21⇔9(x−1)=212⇔9(x−1)=441⇔x−1=49⇔x=509(x−1)=21⇔9(x−1)=212⇔9(x−1)=441⇔x−1=49⇔x=50
d) Điều kiện: x∈Rx∈R (vì 4.(1−x)2≥04.(1−x)2≥0 với mọi x)x)
√4(1−x)2−6=04(1−x)2−6=0⇔2√(1−x)2=6⇔2(1−x)2=6 ⇔|1−x|=3⇔|1−x|=3 ⇔[1−x=31−x=−3⇔[1−x=31−x=−3 ⇔[x=−2x=4⇔[x=−2x=4
Vậy x=−2;x=4.
a, \(\sqrt{16x}=8\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
b, \(\sqrt{4x}=\sqrt{5}\)ĐK : x \(\ge0\)
bình phương 2 vế ta được : \(4x=5\Leftrightarrow x=\frac{5}{4}\)
c, \(\sqrt{9\left(x-1\right)}=21\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\)
bình phương 2 vế ta được : \(x-1=49\Leftrightarrow x=50\)
d, \(\sqrt{4\left(1-x\right)^2}-6=0\Leftrightarrow2\left|1-x\right|=6\Leftrightarrow\left|1-x\right|=3\)
TH1 : \(1-x=3\Leftrightarrow x=-2\)
TH2 : \(1-x=-3\Leftrightarrow x=4\)
(do xy > 0 (gt) nên đưa thừa số xy vào trong căn để khử mẫu)
#Học tốt!!!
\(ab\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{ab}\)
\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)
\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)
\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)
\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)
a) \(\sqrt{16}\).\(\sqrt{25}\)+\(\sqrt{196}\):\(\sqrt{49}\)
=4.5+14/7
=20+2
=22
a) \(\sqrt{16}\).\(\sqrt{25}\) + \(\sqrt{196}\) : \(\sqrt{49}\) = 4.5+14:9=22
b) 36:\(\sqrt{2.3^2.18}\) - \(\sqrt{169}\)= 36 : \(\)18 - 13 = -11
c) \(\sqrt{\sqrt{81}}\) = 3
d) \(\sqrt{3^2+4^2}\)= \(\sqrt{25}\)=5
a, 4.5+ 14:7= 20+2= 22
b, 36: √2.9.18 - 13 = 36:18 -13 =-11
c, √9 = 3
d, √9+16 = √25 = 5
a) \(\sqrt{16}\times\sqrt{25}+\sqrt{196}\div\sqrt{49}=4\times5+14\div7=20+2=22\)
b) \(36\div\sqrt{2\times3^2\times18}-\sqrt{169}=36\div18-13=2-13=-11\)
c) \(\sqrt{\sqrt{81}}=3\)
d) \(\sqrt{3^2+4^2}=5\)
a.22
b.-11
c.3
d.5
a. \(\sqrt{16}\) . \(\sqrt{25}\) + \(\sqrt{196}\) : \(\sqrt{49}\)
= \(\sqrt{4^2}\) . \(\sqrt{5^2}\) + \(\sqrt{14^2}\) : \(\sqrt{7^2}\)
= 4.5 + 14:7 = 20+2=22
b. \(\sqrt{36}\) : \(\sqrt{2.3^2.18}\) - \(\sqrt{169}\)
= 36 : \(\sqrt{3^2.36}\) - \(\sqrt{13^2}\)
= 36 : 18 - 13 = 2-13 = -11
c.\(\sqrt{\sqrt{81}}\) = \(\sqrt{\sqrt{9^2}}\) = \(\sqrt{9}\) = \(\sqrt{3^2}\) = 3
d. \(\sqrt{3^2+4^2}\) = \(\sqrt{25}\) = \(\sqrt{5^2}\) = 5
a) √16.√25+√196:√4916.25+196:49
=√42.√52+√142:√72=42.52+142:72
=4.5+14:7=20+2=22=4.5+14:7=20+2=22
b) √36:√2.32.18−√16936:2.32.18−169
=36:√32.36−√132=36:32.36−132
=36:√32.62...
a) = 4.5 + 14 : 7 = 20 + 2 = 22
b) 36 : 18 - 13 = 2 - 13 = - 11
c) \(\sqrt{9}\) = 3
d) \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5
a) √16.√25+√196:√4916.25+196:49
=√42.√52+√142:√72=42.52+142:72
=4.5+14:7=20+2=22=4.5+14:7=20+2=22.
b) √36:√2.32.18−√16936:2.32.18−169
=36:√32.36−√132=36:32.36−132
=36:√32.62
Đúng(0)
a) √16.√25+√196:√4916.25+196:49
=√42.√52+√142:√72=42.52+142:72
=4.5+14:7=20+2=22=4.5+14:7=20+2=22.
b) √36:√2.32.18−√16936:2.32.18−169
=36:√32.36−√132=36:32.36−132
=36:√32.6
Đúng(0)
a) \(\sqrt{4^2}.\sqrt{5^2}+\sqrt{14^2}:\sqrt{7^2}\) = \(|4|.|5|+|14|:|7|\) = 4.5+14:7 = 20+2 = 22
b) 36:\(\sqrt{2.9.2.9}-\sqrt{169}\) = 36:\(\sqrt{4}.\sqrt{9}.\sqrt{9}-\sqrt{169}\) = 2.3.3-13 = 18-3 = 15
c)\(\sqrt{\sqrt{9.9}}\) = \(\sqrt{\sqrt{9}.\sqrt{9}}\) = \(\sqrt{3.3}\) = \(\sqrt{9}\) = 3
d)\(\sqrt{9+16}=\sqrt{25}=5\)
a) Ta có: √16.√25+√196:√4916.25+196:49
=√42.√52+√142:√72=42.52+142:72
=|4|.|5|+|14|:|7|=|4|.|5|+|14|:|7|
=4.5+14:7=4.5+14:7
=20+2=22=20+2=22.
b) Ta có:
36:√2.32.18−√16936:2.32.18−169
=36:√(2.32)
a) Ta có: √16.√25+√196:√4916.25+196:49
=√42.√52+√142:√72=42.52+142:72
=|4|.|5|+|14|:|7|=|4|.|5|+|14|:|7|
=4.5+14:7=4.5+14:7
=20+2=22=20+2=22
b) Ta có:
36:√2.32.18−√16936:2.32.18−169
=36:√(2.32)
a) \(\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\) = \(\sqrt{4^2}.\sqrt{5^2}+\sqrt{14^2}:\sqrt{7^2}\) = \(\left|4\right|.\left|5\right|+\left|14\right|.\left|7\right|\) = 4.5 + 14.7 =20 + 2 = 22
b)
\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}16.25+196:49
=\sqrt{4^2}.\sqrt{5^2}+\sqrt{14^2}:\sqrt{7^2}=42.52+142:72
= 4.5 + 14:7 = 20 + 2 = 22=4.5+14:7=20+2=22.
b) \sqrt{36}:\sqrt{2.3^2.18}-\sqrt{169}36:2.32.18−169
=36:\sqrt{3^2.36}-\sqrt{13^2}=36:32.36
Đúng(0)
a) \(\sqrt{16}.\sqrt{25}\) + \(\sqrt{169}:\sqrt{49}\)
= \(\sqrt{4^2}\).\(\sqrt{5^2}\) + \(\sqrt{14^2}:\sqrt{7^2}\)
= 4.5 + 14 : 7
= 20 + 2 = 22
b) \(\sqrt{36}:\sqrt{2.3^2.18}-\sqrt{169}\)
= 36 : \(\sqrt{3^2.36}-\sqrt{13^2}\)
= 36 : \(\sqrt{3^2.6^2}-13\)
= 36 : \(\sqrt{18^2}-13\)
= 36 : 18 - 13 = 2 - 13 = -11
c) \(\sqrt{\sqrt{81}}=\sqrt{\sqrt{9^2}}=\sqrt{9}=\sqrt{3^2}=3\)
d) \(\sqrt{3^2+4^2}=\sqrt{25}=\sqrt{5^2}=5\)
\(a,\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}=4.5+14:7=20+2=22\)
b,\(36:\sqrt{2.3^2.18}-\sqrt{169}=36:\sqrt{324}-13=36:18-13=2-13=-11\)
c, \(\sqrt{\sqrt{81}}=3\)
d, \(\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\)