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Giải:
#Học tốt!!!
a) 7 /24
b) 1, 08
c) 17/2
d) 15 /29
a) √1916.549.0,01=√(1.16+916).(5.9+49).11001916.549.0,01=(1.16+916).(5.9+49).1100
=√2516.√499.√1100=√25√16.√49√9.√1
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a) √1916.549.0,01=√(1.16+916).(5.9+49).11001916.549.0,01=(1.16+916).(5.9+49).1100
=√2516.√499.√1100=√25√16.√49√9.√1
Đúng(0)
a) √1916.549.0,01=√(1.16+916).(5.9+49).11001916.549.0,01=(1.16+916).(5.9+49).1100
=√2516.√499.√1100=√25√16.√49√9.√1
Đúng(0)
a) \(\sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01}=\sqrt{\dfrac{25}{16}.\dfrac{49}{9}.0,01}=\dfrac{5}{4}.\dfrac{7}{3}.0,1=\dfrac{7}{24}\)
b) \(\sqrt{1,44.1,21-1,44.0,4}=\sqrt{1,44.\left(1,21-0,4\right)}=\sqrt{1,44.0,81}=1,2.0,9=1,08\)
c) \(\sqrt{\dfrac{165^2-124^2}{164}}=\sqrt{\dfrac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\dfrac{41.289}{164}}=\sqrt{\dfrac{289}{4}}=\dfrac{17}{2}\)
d) \(\sqrt{\dfrac{149^2-76^2}{457^2-384^2}}=\sqrt{\dfrac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}=\sqrt{\dfrac{73.225}{73.841}}=\sqrt{\dfrac{225}{841}}=\dfrac{15}{29}\)
\(a=\dfrac{7}{24};b=1.08;\dfrac{17}{2};\dfrac{15}{29}\)
a) √1916.549.0,01=√(1.16+916).(5.9+49).11001916.549.0,01=(1.16+916).(5.9+49).1100
=√2516.√499.√1100=√25√16.√49√9.√1
Đúng(0)
a) √1916.549.0,01=√(1.16+916).(5.9+49).11001916.549.0,01=(1.16+916).(5.9+49).1100
=√2516.√499.√1100=√25√16.√49√9.√1
Đúng(0)
a\(\dfrac{7}{24}\)
b 1,08
c 17/2
d 15/29
a, \(\sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01}=\sqrt{\dfrac{25}{16}.\dfrac{49}{9}.\dfrac{1}{100}}=\sqrt{\dfrac{25}{16}}.\sqrt{\dfrac{49}{9}}.\sqrt{\dfrac{1}{100}}=\dfrac{5}{4}.\dfrac{7}{3}\dfrac{1}{10}=\dfrac{7}{24}\)
b,\(\sqrt{1,44.1,21-1,44.0,4}=\sqrt{1,44.\left(1,21-0,4\right)}=\sqrt{1,44.0,81}=\sqrt{144}.\sqrt{0,81}=12.0,9=10,8\)
c,
\(\sqrt{\dfrac{165^2-124^2}{164}}=\sqrt{\dfrac{\left(165-124\right).\left(165+124\right)}{164}}=\sqrt{\dfrac{41.289}{164}}=\sqrt{\dfrac{289}{4}}=\dfrac{\sqrt{289}}{\sqrt{4}}=\dfrac{17}{4}\)
d, \(\sqrt{\dfrac{149^2-76^2}{457^2-384^2}}=\sqrt{\dfrac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}=\sqrt{\dfrac{73.225}{73.841}}=\sqrt{\dfrac{225}{841}}=\dfrac{\sqrt{225}}{\sqrt{841}}=\dfrac{15}{29}\)
a) \sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01}=\sqrt{(\dfrac{1.16+9}{16}).(\dfrac{5.9+4}{9}).\dfrac{1}{100}}1169.594.0,01=(161.16+9).(95.9+4).1001
=\sqrt{\dfrac{25}{16}}.\sqrt{\dfrac{49}{9}}.\sqrt{\dfrac{1}{100}}=\dfrac{\sqrt{25}}{\sqrt{16}}.\dfrac{\sqrt{49}}{\sqrt{9}}.\dfrac{\sqrt{1}}{\sqrt{100}}=1625.949.1001=1625.949
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a, = \(\sqrt{\dfrac{25}{16}.\dfrac{49}{9}}.\dfrac{1}{100}\) = \(\sqrt{\dfrac{25}{16}}\). \(\sqrt{\dfrac{49}{9}}\) .\(\sqrt{\dfrac{1}{100}}\) = \(\dfrac{\sqrt{25}}{\sqrt{16}}\). \(\dfrac{\sqrt{49}}{\sqrt{9}}\).\(\dfrac{\sqrt{1}}{\sqrt{100}}\) = \(\dfrac{5}{4}\) . \(\dfrac{7}{3}\) .\(\dfrac{1}{10}\)= \(\dfrac{35}{120}\)=\(\dfrac{7}{24}\)
b, = \(\sqrt{1,44}.\left(1,21-0,4\right)\) = \(\sqrt{1,44}.0,8\) = \(\sqrt{\dfrac{144}{100}.\dfrac{81}{100}}\) = \(\sqrt{\dfrac{144}{100}}\).\(\sqrt{\dfrac{81}{100}}\)= \(\dfrac{\sqrt{144}}{\sqrt{100}}\).\(\dfrac{\sqrt{81}}{\sqrt{100}}\)
= \(\dfrac{12}{10}\). \(\dfrac{9}{10}\) =\(\dfrac{108}{100}\)=\(\dfrac{27}{25}\)
c, = \(\sqrt{\dfrac{\left(165-124\right).\left(165+124\right)}{164}}\)= \(\sqrt{\dfrac{41.289}{164}}\)= \(\sqrt{\dfrac{289}{4}}\)=\(\dfrac{\sqrt{289}}{\sqrt{4}}\)=\(\dfrac{17}{2}\)
d, = \(\sqrt{\dfrac{\left(149-76\right).\left(149+76\right)}{\left(457-384\right).\left(457+384\right)}}\)=\(\sqrt{\dfrac{73.225}{73.841}}\)=\(\sqrt{\dfrac{225}{841}}\)=\(\dfrac{\sqrt{225}}{\sqrt{841}}\)=\(\dfrac{15}{29}\)
a) √1916.549.0,01=√(1.16+916).(5.9+49).11001916.549.0,01=(1.16+916).(5.9+49).1100
=√2516.√499.√1100=√25√16.√49√9.√1
Đúng(0)
a) \sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01}=\sqrt{(\dfrac{1.16+9}{16}).(\dfrac{5.9+4}{9}).\dfrac{1}{100}}1169.594.0,01=(161.16+9).(95.9+4).1001
=\sqrt{\dfrac{25}{16}}.\sqrt{\dfrac{49}{9}}.\sqrt{\dfrac{1}{100}}=\dfrac{\sqrt{25}}{\sqrt{16}}.\dfrac{\sqrt{49}}{\sqrt{9}}.\dfrac{\sqrt{1}}{\sqrt{100}}=1625.949.1001=1625.949
.
a) = 7/24
b) = 27/25
c) = 17/2
d) = 15/29
a) √1916.549.0,01=√(1.16+916).(5.9+49).11001916.549.0,01=(1.16+916).(5.9+49).1100
=√2516.√499.√1100=√25√16.√49√9.√1
Đúng(0)
a,\(\dfrac{7}{24}\)
b,1,08
c,\(\dfrac{17}{2}\)
d,\(\dfrac{15}{29}\)
a) \sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01}=\sqrt{(\dfrac{1.16+9}{16}).(\dfrac{5.9+4}{9}).\dfrac{1}{100}}1169.594.0,01=(161.16+9).(95.9+4).1001
=\sqrt{\dfrac{25}{16}}.\sqrt{\dfrac{49}{9}}.\sqrt{\dfrac{1}{100}}=\dfrac{\sqrt{25}}{\sqrt{16}}.\dfrac{\sqrt{49}}{\sqrt{9}}.\dfrac{\sqrt{1}}{\sqrt{100}}=1625.949.1001=1625.949
a)\(\dfrac{7}{24}\)
b)\(1,08\)
c)\(\dfrac{17}{2}\)
d)\(\dfrac{15}{29}\)
a) \(\sqrt{1\dfrac{9}{16}\cdot5\dfrac{4}{9}\cdot0,01}\)
= \(\sqrt{\left(\dfrac{1\cdot16+9}{16}\right)\cdot\left(\dfrac{5\cdot9+4}{9}\right)\cdot\dfrac{1}{100}}\)
= \(\sqrt{\dfrac{25}{16}}\cdot\sqrt{\dfrac{49}{9}}\cdot\sqrt{\dfrac{1}{100}}\)
= \(\dfrac{\sqrt{25}}{\sqrt{16}}\cdot\dfrac{\sqrt{49}}{\sqrt{9}}\cdot\dfrac{\sqrt{1}}{\sqrt{100}}\)
= \(\dfrac{5}{4}\cdot\dfrac{7}{3}\cdot\dfrac{1}{10}\)
= \(\dfrac{7}{24}\)
b) \(\sqrt{1,44\cdot1,21-1,44\cdot0,4}\)
= \(\sqrt{1,44.\left(1,21-0,4\right)}\)
= \(\sqrt{1,44\cdot0,81}\)
= \(\sqrt{1,44}\cdot\sqrt{0,81}\)
= 1,2 . 0,9
= 1,08
c) \(\sqrt{\dfrac{165^2-124^2}{164}}\)
= \(\sqrt{\dfrac{\left(165-124\right)\left(165+124\right)}{164}}\)
= \(\sqrt{\dfrac{41\cdot289}{164}}\) = \(\sqrt{\dfrac{289}{4}}\) = \(\dfrac{\sqrt{289}}{\sqrt{4}}\) = \(\dfrac{17}{2}\)
d) \(\sqrt{\dfrac{149^2-76^2}{457^2-384^2}}\)
= \(\sqrt{\dfrac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}\)
\(\sqrt{\dfrac{73\cdot225}{73\cdot841}}\) = \(\sqrt{\dfrac{225}{841}}\) = \(\dfrac{\sqrt{225}}{\sqrt{841}}\) = \(\dfrac{15}{29}\)
\(\sqrt{\dfrac{1.16+9}{16}.\dfrac{5.9+4}{9}.\dfrac{1}{100}}\) =\(\sqrt{\dfrac{25}{16}}.\sqrt{\dfrac{49}{9}}.\sqrt{\dfrac{1}{100}}\) =\(\dfrac{5}{4}.\dfrac{7}{3}.\dfrac{1}{10}\) =\(\dfrac{7}{24}\) b)\(\sqrt{1,44\left(1,21-0,4\right)}=\sqrt{1,44}.0,81\)
a, \(\dfrac{7}{24}\) ;b= 1,08 ;c=\(\dfrac{17}{2}\) ;d=\(\dfrac{15}{29}\)
a.\(\sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01}\)
b.\(\sqrt{1,44.1,21-1,44.0,4}\)
c.\(\sqrt{\dfrac{165^2-124^2}{164}}\)
d.\(\sqrt{\dfrac{149^2-76^2}{457^2-384^2}}\)
a: \(=\sqrt{\dfrac{25}{16}\cdot\dfrac{49}{9}\cdot\dfrac{1}{100}}=\dfrac{5}{4}\cdot\dfrac{7}{3}\cdot\dfrac{1}{10}=\dfrac{35}{120}=\dfrac{7}{24}\)
b: \(=\sqrt{1.44\cdot0.81}=1.2\cdot0.9=1.08\)
c: \(=\sqrt{\dfrac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\dfrac{1}{4}\cdot289}=\dfrac{17}{2}\)
d: \(=\sqrt{\dfrac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}=\sqrt{\dfrac{225}{841}}=\dfrac{15}{29}\)
Tính:
a. \(\sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01};\) b. \(\sqrt{1,44.1,21-1,44.0,4};\)
c. \(\sqrt{\dfrac{165^2-124^2}{164}};\) d. \(\sqrt{\dfrac{149^2-76^2}{457^2-384^2}}.\)
a) HD: Đổi hỗn số và số thập phân thành phân số.
ĐS:
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b)
= 
=
= 
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=
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d) ĐS:
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Bài 29 (trang 19 SGK Toán 9 Tập 1)
Tính
a) $\dfrac{\sqrt{2}}{\sqrt{18}}$ ; b) $\dfrac{\sqrt{15}}{\sqrt{735}}$ ;
c) $\dfrac{\sqrt{12500}}{\sqrt{500}}$ ; d) $\dfrac{\sqrt{6^5}}{\sqrt{2^3.3^5}}$.
a, \(\frac{\sqrt{2}}{\sqrt{18}}=\sqrt{\frac{2}{18}}=\sqrt{\frac{1}{9}}=\frac{1}{3}\)
b, \(\frac{\sqrt{15}}{\sqrt{735}}=\sqrt{\frac{15}{735}}=\sqrt{\frac{1}{49}}=\frac{1}{7}\)
c, \(\frac{\sqrt{12500}}{\sqrt{500}}=\sqrt{\frac{12500}{500}}=\sqrt{\frac{125}{5}}=\sqrt{25}=5\)
d, \(\frac{\sqrt{6^5}}{\sqrt{2^3.3^5}}=\sqrt{\frac{6^5}{2^3.3^5}}=\sqrt{\frac{2^5.3^5}{2^3.3^5}}=\sqrt{2^2}=2\)
a) căn 2 / căn 18 = 1/3
b) căn 15/ căn 735 = 1/7
c) căn 12500 / căn 500 = 5
d) căn 6^5 / 2^3 * 3^5 = 2
Bài 28 (trang 18 SGK Toán 9 Tập 1)
Tính
a) $\sqrt{\dfrac{289}{25}}$ ; b) $\sqrt{2\dfrac{14}{25}}$ ;
c) $\sqrt{\dfrac{0,25}{9}}$ ; d) $\sqrt{\dfrac{8,1}{16}}$.
a, \(\sqrt{\frac{289}{25}}=\frac{\sqrt{289}}{\sqrt{25}}=\frac{17}{5}\)
b, \(\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\frac{8}{5}\)
c, \(\sqrt{\frac{0,25}{9}}=\frac{\sqrt{0,25}}{\sqrt{9}}=\frac{0,5}{3}=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)
d, \(\sqrt{\frac{8,1}{16}}\)đề có sai ko cô ?
a) căn 289 / 225 = 17/15
b) căn 64/ 25 = 8/5
c) căn 0,25 / 9 = 1/6
d) căn 8,1 / 1,6 = 9/4
Bài 1: Tính
a) \(\sqrt{1,44.1,21-1,44.0,4}\)
b) \(\dfrac{\sqrt{5}-2}{\sqrt{5}+2}+\sqrt{80}\)
c) \(\sqrt[3]{16}+\sqrt[3]{2}\left(\sqrt[3]{4}-\sqrt[3]{2}\right)\)
Bài 2: C/m
\(\dfrac{1}{\sqrt{a}-\sqrt{b}}:\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}=\dfrac{1}{a-b}\) với a,b>0, a khác 0
Bài 1:
a) \(\sqrt{1,44\cdot1,21-1,44\cdot0,4}\)
\(=\sqrt{1,44\cdot\left(1,21-0,4\right)}\)
\(=\sqrt{1,44\cdot0,81}\)
\(=\sqrt{1,44}\cdot\sqrt{0,81}\)
\(=1,2\cdot0,9\)
\(=1,08\)
b) \(\dfrac{\sqrt{5}-2}{\sqrt{5}+2}+\sqrt{80}\)
\(=\dfrac{\left(\sqrt{5}-2\right)^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+4\sqrt{5}\)
\(=\dfrac{5-4\sqrt{5}+4}{1}+4\sqrt{5}\)
\(=9-4\sqrt{5}+4\sqrt{5}\)
\(=9\)
c) \(\sqrt[3]{16}+\sqrt[3]{2}\left(\sqrt[3]{4}-\sqrt[3]{2}\right)\)
\(=\sqrt[3]{2^3\cdot2}+\sqrt[3]{2\cdot4}-\sqrt[3]{2\cdot2}\)
\(=2\sqrt[3]{2}+\sqrt[3]{8}-\sqrt[3]{4}\)
\(=2\sqrt[3]{2}+2-\sqrt[3]{4}\)
Bài 2: Ta có:
\(VT=\dfrac{1}{\sqrt{a}-\sqrt{b}}:\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}:\dfrac{\sqrt{ab}\cdot\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\cdot\dfrac{1}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{1}{a-b}=VP\left(dpcm\right)\)
Bài 50 (trang 30 SGK Toán 9 Tập 1)
Trục căn thức ở mẫu với giả thiết các biểu thức chữ đều có nghĩa
$\dfrac{5}{\sqrt{10}}$; $\dfrac{5}{2 \sqrt{5}}$ ; $\dfrac{1}{3 \sqrt{20}}$ ; $\dfrac{2 \sqrt{2}+2}{5 \sqrt{2}}$ ;$\dfrac{y+b.\sqrt{y}}{b.\sqrt{y}}$.
\(\frac{5}{\sqrt{10}}=\frac{5\sqrt{10}}{10}=\frac{\sqrt{10}}{2}\)
\(\frac{5}{2\sqrt{5}}=\frac{10\sqrt{5}}{20}=\frac{\sqrt{5}}{2}\)
\(\frac{1}{3\sqrt{20}}=\frac{3\sqrt{20}}{180}=\frac{\sqrt{20}}{60}=\frac{2\sqrt{5}}{60}=\frac{\sqrt{5}}{30}\)
\(\frac{2\sqrt{2}+2}{5\sqrt{2}}=\frac{10\sqrt{2}\left(\sqrt{2}+1\right)}{50}=\frac{20+10\sqrt{2}}{50}=\frac{10\left(2+\sqrt{2}\right)}{50}=\frac{2+\sqrt{2}}{5}\)
\(\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{y\left(\sqrt{y}+b\right)}{by}=\frac{\sqrt{y}+b}{b}\)
+ Ta có:
5√10=5.√10√10.√10=5√10(√10)2=5√1010510=5.1010.10=510(10)2=51010
=5.√105.2=5.105.2=√102=102.
+ Ta có:
52√5=5.√52√5.√5=
Bài 61 (trang 33 SGK Toán 9 Tập 1)
Chứng minh các đẳng thức sau:
a) $\dfrac{3}{2} \sqrt{6}+2 \sqrt{\dfrac{2}{3}}-4 \sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{6}$;
b) $\left(x \sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2 x}{3}}+\sqrt{6 x}\right): \sqrt{6 x}=2 \dfrac{1}{3} $ với $x>0$.
a) -17√3/3 b) 11√6
c) 21 d) 11
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
Bài 49 (trang 29 SGK Toán 9 Tập 1)
Khử mẫu của biểu thức lấy căn
$ab\sqrt{\dfrac{a}{b}}$ ; $\dfrac{a}{b} \sqrt{\dfrac{b}{a}}$ ; $\sqrt{\dfrac{1}{b}+\dfrac{1}{b^{2}}}$ ; $\sqrt{\dfrac{9 a^{3}}{36 b}}$ ; $3 xy \sqrt{\dfrac{2}{x y}}$.
(Giả thiết các biểu thức có nghĩa).
(do xy > 0 (gt) nên đưa thừa số xy vào trong căn để khử mẫu)
#Học tốt!!!
\(ab\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{ab}\)
\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)
\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)
\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)
\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)
Bài 52 (trang 30 SGK Toán 9 Tập 1)
Trục căn thức ở mẫu với giả thiết các biểu thức chữ đều có nghĩa
$\dfrac{2}{\sqrt{6}-\sqrt{5}}$ ; $\dfrac{3}{\sqrt{10}+\sqrt{7}}$ ; $\dfrac{1}{\sqrt{x}-\sqrt{y}}$ ; $\dfrac{2 a b}{\sqrt{a}-\sqrt{b}}$.
+ 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.
\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)
\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
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