Question

[Trân Trân's]

Giải giúp mik vs ạ! Cảm ơn mn

Answer 1

[Nguyễn Lê Phước Thịnh]

a) Ta có: \(S=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)

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

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

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

b) Thay \(x=4-2\sqrt{3}\) vào S, ta được:

\(S=\dfrac{1-\left(4-2\sqrt{3}\right)\left(\sqrt{3}-1\right)}{4-2\sqrt{3}}\)

\(=\dfrac{1-\left(4\sqrt{3}-4-6+2\sqrt{3}\right)}{4-2\sqrt{3}}\)

\(=\dfrac{1-2\sqrt{3}+10}{4-2\sqrt{3}}=\dfrac{9-2\sqrt{3}}{4-2\sqrt{3}}\)

\(=\dfrac{\left(9-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}{4}\)

\(=\dfrac{36+18\sqrt{3}-8\sqrt{3}-12}{4}\)

\(=\dfrac{24+10\sqrt{3}}{4}=\dfrac{12+5\sqrt{3}}{2}\)

Answer 2

[LanAnk]

a)\(S=\left(\dfrac{x+1+\sqrt{x}}{x+1}\right):\left(\dfrac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right)\) \(đk:x\ne\pm1\)

\(S=\dfrac{x+1+\sqrt{x}}{x+1}.\dfrac{\left(\sqrt{x}-1\right)\left(x+1\right)}{\left(\sqrt{x}-1\right)^2}\)

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

b)\(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\left(TMĐK\right)\)

\(\sqrt{x}=\sqrt{3}-1\)

Từ đó ta có :

\(S=\dfrac{4-2\sqrt{3}+1+\sqrt{3}-1}{\sqrt{3}-1-1}\)

\(S=-5-2\sqrt{3}\)