Question

Bài 1:Rút gọn A=( √x/ √x -1 + 2/x- √x ).( √x -1)

Bài 2:Rút gọn P=(1/ √x +2 +1/ √x -2).x-2 √x

Giúp mk vs chiều nộp r

Answer 1

Bài 1:

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

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

\(=\frac{x+2}{\sqrt{x}}\)

Bài 2:

Sửa đề: \(P=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right)\cdot\left(x-2\sqrt{x}\right)\)

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

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

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

\(=\frac{2x}{\sqrt{x}+2}\)

Answer 2

1,

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

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

2

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

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

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