Question

rút gọn biểu thức :

A=\(\dfrac{\left(^{1+\sqrt{x}}\right)^2-4\sqrt{x}}{\sqrt{x}-1}\)

Answer 1

\(A=\dfrac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{\sqrt{x}-1}\left(dkxd:x\ge0,x\ne1\right)\\ =\dfrac{1+2\sqrt{x}+x-4\sqrt{x}}{\sqrt{x}-1}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\\ =\sqrt{x}-1\)

Vậy \(A=\sqrt{x}-1\) với \(x\ge0,x\ne1\)

Answer 2

A=\(\dfrac{1+2\sqrt{x}+x-4\sqrt{x}}{\sqrt{x}-1}\)

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

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

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