Question

Cho biểu thức : A= \(\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\) (với \(x\ne0;x\ne1\)).

a) rút gọn A.

b) Tìm x để A=-1.

Answer 1

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

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

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

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

\(A=-1\Leftrightarrow1-x=-1\Rightarrow x=2\)

Answer 2

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

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

=1-x

b) Để A=-1 thì 1-x=-1

hay x=2