Question

Giúp mình phần 2) [bài rút gọn] với ạ mình tìm mãi k ra kết quả 😢

Answer 1

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

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

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

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

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

Answer 2

Bài 2: 

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

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

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

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