Question

B= (\(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\))^2.(\(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\))

a) Rút gọn B

b)Tìm giá trị của x để B=-2

Giúp em với bí quá.

Answer 1

ĐKXĐ ...

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

\(=\left(\frac{x-1}{2\sqrt{x}}\right)^2\left(\frac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\)

\(=\frac{\left(x-1\right)^2}{4x}.\frac{-4\sqrt{x}}{\left(x-1\right)}=\frac{1-x}{\sqrt{x}}\)

\(B=-2\Leftrightarrow\frac{1-x}{\sqrt{x}}=-2\Leftrightarrow1-x=-2\sqrt{x}\)

\(\Leftrightarrow x-2\sqrt{x}-1=0\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1+\sqrt{2}\\\sqrt{x}=1-\sqrt{2}< 0\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}\)