Question

\(A=\left(\dfrac{3}{1+\sqrt{x}}-\dfrac{\sqrt{x}}{1-\sqrt{x}}-\dfrac{6\sqrt{x}-4}{x-1}\right):\dfrac{1-\sqrt{x}}{5+5\sqrt{x}}\)

Rút gonkj

Answer 1

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$

\(A=\left[\frac{3(\sqrt{x}-1)+\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-1)}-\frac{6\sqrt{x}-4}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\frac{5(\sqrt{x}+1)}{-(\sqrt{x}-1)}\)

\(=\frac{(3\sqrt{x}-3)+(x+\sqrt{x})-(6\sqrt{x}-4)}{(\sqrt{x}-1)(\sqrt{x}+1)}.\frac{5(\sqrt{x}+1)}{-(\sqrt{x}-1)}=\frac{x-2\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}.\frac{5(\sqrt{x}+1)}{-(\sqrt{x}-1)}=\frac{(\sqrt{x}-1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}.\frac{5(\sqrt{x}+1)}{-(\sqrt{x}-1)}=-5\)

Answer 2

Với `x >= 0,x \ne 1` có:

`A=(3/[1+\sqrt{x}]-\sqrt{x}/[1-\sqrt{x}]-[6\sqrt{x}-4]/[x-1]):[1-\sqrt{x}]/[5+5\sqrt{x}]`

`A=[3(1-\sqrt{x})-\sqrt{x}(1+\sqrt{x})+6\sqrt{x}-4]/[(1-\sqrt{x})(1+\sqrt{x})].[5(1+\sqrt{x})]/[1-\sqrt{x}]`

`A=[3-3\sqrt{x}-\sqrt{x}-x+6\sqrt{x}-4]/[5(1-\sqrt{x})^2]`

`A=[-x+2\sqrt{x}-1]/[5(\sqrt{x}-1)^2]`

`A=[-(\sqrt{x}-1)^2]/[5(\sqrt{x}-1)^2]`

`A=[-1]/5`