Question

[locloc]

làm 3-5 câu nhé có thể lm hết cũng được :))

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(A=\left(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right):\dfrac{2}{\sqrt{5}-\sqrt{3}}\)

\(=\left(\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right):\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}\)

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

a:

\(A=\left(\dfrac{\sqrt{14}-\sqrt{7}}{2\sqrt{2}-2}+\dfrac{\sqrt{15}-\sqrt{5}}{2\sqrt{3}-2}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\dfrac{\sqrt{7}+\sqrt{5}}{2}\right)\cdot\dfrac{\sqrt{7}-\sqrt{5}}{1}=\dfrac{7-5}{2}=1\)

a: 

\(A=\left(a-b\right)\cdot\sqrt{\dfrac{ab}{\left(a-b\right)^2}}\)

\(=\left(a-b\right)\cdot\dfrac{\sqrt{ab}}{\left|a-b\right|}\)

a<b<0

=>a-b<0

=>\(A=\left(a-b\right)\cdot\dfrac{\sqrt{ab}}{-\left(a-b\right)}=-\sqrt{ab}\)

a:

\(A=\sqrt{\dfrac{9+12a+4a^2}{b^2}}\)

\(=\sqrt{\dfrac{\left(2a+3\right)^2}{b^2}}=\left|\dfrac{2a+3}{b}\right|\)

a>=-3/2

=>2a+3>=0

b<0

=>\(\dfrac{2a+3}{b}< =0\)

=>\(A=\dfrac{-\left(2a+3\right)}{b}\)

b:

\(A=\left(\dfrac{1}{3-2\sqrt{2}}-\dfrac{6}{2+\sqrt{2}}\right)\left(3+5\sqrt{2}\right)\)

\(=\left(\dfrac{3+2\sqrt{2}}{1}-\dfrac{6\left(2-\sqrt{2}\right)}{2}\right)\left(3+5\sqrt{2}\right)\)

\(=\left(3+2\sqrt{2}-3\left(2-\sqrt{2}\right)\right)\cdot\left(3+5\sqrt{2}\right)\)

\(=\left(5\sqrt{2}-3\right)\left(5\sqrt{2}+3\right)\)

=50-9

=41

b:

\(A=\left(\dfrac{1}{\sqrt{5}-2}-\dfrac{59}{3\sqrt{7}-2}\right)\left(3\sqrt{7}+\sqrt{5}\right)\)

\(=\left(\dfrac{\sqrt{5}+2}{5-4}-\dfrac{59\left(3\sqrt{7}+2\right)}{63-4}\right)\left(3\sqrt{7}+\sqrt{5}\right)\)

\(=\left(\sqrt{5}+2-3\sqrt{7}-2\right)\left(3\sqrt{7}+\sqrt{5}\right)\)

\(=\left(\sqrt{5}-3\sqrt{7}\right)\left(\sqrt{5}+3\sqrt{7}\right)\)

=5-63

=-58

b:

\(A=\dfrac{x}{\sqrt{x}-1}+\dfrac{2x-\sqrt{x}}{\sqrt{x}-x}\)

\(=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\sqrt{x}-1\)