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25 tháng 7 2023

\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\dfrac{3}{7}}\)

\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=\sqrt{5}\)

\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}=-\dfrac{2\sqrt{6}}{6}\)

`(sqrt 15 - sqrt 6)/(sqrt 35 - sqrt 14)`

`= (sqrt 3 . (sqrt 5 - sqrt 2))/(sqrt 7. (sqrt 5 - sqrt 2))`

`= sqrt3/sqrt 7`

`@ (sqrt 15 - sqrt 5)/(sqrt 3 - 1)`

`= (sqrt 5(sqrt 3 - 1))/(sqrt 3 - 1)`

`= sqrt5`

`@ (2 sqrt 8 - sqrt 12)/(sqrt18 - sqrt 48)`

`= (2(sqrt 8 - sqrt 3)/(sqrt 6(sqrt 3 - sqrt 8))`

`= (-2)/(sqrt 6) = (-2 sqrt 6)/6`

1 tháng 7 2016

câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :

\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)

\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)      

\(=3-\sqrt{6}+2\sqrt{6}-3\)   ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )

\(=\sqrt{6}\)

 

18 tháng 6 2017

sai ngay từ đầu limdim

4 tháng 7 2019

Câu e mình chịu, bạn 😔😔

18 tháng 9 2019

a)\(\frac{3\sqrt{6}-\sqrt{2}}{1-3\sqrt{3}}=\frac{3\sqrt{3}.\sqrt{2}-\sqrt{2}}{1-3\sqrt{3}}=\frac{\sqrt{2}.\left(3\sqrt{3}-1\right)}{-\left(3\sqrt{3}-1\right)}=-\sqrt{2}\)

b)\(\frac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}=\frac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}=\frac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2.\left(\sqrt{2}-\sqrt{3}\right)}=\frac{\sqrt{5}}{2}\)

c)\(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}\)

d)\(\frac{5\sqrt{6}-6\sqrt{5}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{5^2.6}-\sqrt{6^2.5}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{30}.\sqrt{5}-\sqrt{30}.\sqrt{6}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{30}.\left(\sqrt{5}-\sqrt{6}\right)}{\sqrt{5}-\sqrt{6}}=\sqrt{30}\)

e)\(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{6}}=\frac{\sqrt{2^2.3}-\sqrt{3^2.2}}{\sqrt{6}}=\frac{\sqrt{6}.\sqrt{2}-\sqrt{6}.\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{6}.\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}}=\sqrt{2}-\sqrt{3}\)

f)\(\frac{6\sqrt{2}-4}{\sqrt{2}}=\frac{6\sqrt{2}-\sqrt{16}}{\sqrt{2}}=\frac{6\sqrt{2}-\sqrt{2}.2\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}.\left(6-2\sqrt{2}\right)}{\sqrt{2}}=6-2\sqrt{2}\)

g)\(\frac{6-5\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{36}-5\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}.2\sqrt{3}-5\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}.\left(2\sqrt{3}-5\right)}{\sqrt{3}}=2\sqrt{3}-5\)

19 tháng 9 2019

Cảm ơn bạn nha

25 tháng 6 2023

\(a,\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\\ =\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{\sqrt{2}.\sqrt{4}-\sqrt{3}.\sqrt{4}}\\ =\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2^2}}\\ =\dfrac{\sqrt{5}}{2}\)

\(b,\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\\ =\dfrac{\sqrt{5}.\sqrt{3}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}\)

\(c,\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\dfrac{\sqrt{5}}{\sqrt{2}}\)

25 tháng 6 2023

\(a,=\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}\\ =\dfrac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{2}\)

\(b,=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}=\dfrac{\sqrt{21}}{7}\)

\(c,=\dfrac{\sqrt{5}.\sqrt{5}+\sqrt{5}}{\sqrt{2}.\sqrt{5}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{10}}{2}\)

Bài 1: Rút gọn biểu thức1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\)              2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\)      4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\) ...
Đọc tiếp

Bài 1: Rút gọn biểu thức

1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\)              2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)

3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\)      4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)

5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\)   6) \(\left(3\sqrt{20}-\sqrt{125}-15\sqrt{\frac{1}{5}}\right).\sqrt{5}\)

7) \(\left(6\sqrt{128}-\frac{3}{5}\sqrt{50}+7\sqrt{8}\right):3\sqrt{2}\)  8) \(\left(2\sqrt{48}-\frac{3}{2}\sqrt{\frac{4}{3}}+\sqrt{27}\right).2\sqrt{3}\)

9) \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}\)    10) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)

11) \(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\)      12) \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

13) \(\sqrt{15-6\sqrt{6}}\)    14) \(\sqrt{8-2\sqrt{15}}\)    15) \(\sqrt[3]{-2}.\sqrt[3]{32}+\sqrt{2}.\sqrt{32}\)

 

1
26 tháng 11 2017

Giúp mình :<

6 tháng 8 2017

\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)

\(=\sqrt{121}.\sqrt{2}.\sqrt{2}.\sqrt{13}.\sqrt{13}.\sqrt{10}.\sqrt{0,9}-\left(2-1\right)\)

\(=11.2.13.\sqrt{9}-1=286.3-1=857\)

6 tháng 8 2017

\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)

\(=\frac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\left(\sqrt{3}-\sqrt{2}\right)}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{4\left(\sqrt{3}-1\right)}+\frac{\sqrt{\left(2\sqrt{3}-\sqrt{5}\right)^2}}{4}\)

\(=\frac{\sqrt{3}}{2}-\frac{\sqrt{5}}{4}+\frac{2\sqrt{3}-\sqrt{5}}{4}\)

\(=\sqrt{3}-\frac{\sqrt{5}}{4}\)