Tham khảo thanh này để soạn đề chính xác hơn nha :vvv

a) Ta có: \(M=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\right)\cdot\dfrac{x+3\sqrt{x}}{7-\sqrt{x}}\)

\(=\left(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)

\(=\dfrac{x-9-\left(x-2\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)

\(=\dfrac{x-9-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{-\left(\sqrt{x}-7\right)}\)

\(=\dfrac{\sqrt{x}-7}{\sqrt{x}-2}\cdot\dfrac{-1}{\sqrt{x}-7}\)

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

b) Ta có: \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(loại\right)\end{matrix}\right.\)

Thay x=0 vào biểu thức (1), ta được:

\(M=\dfrac{-1}{\sqrt{0}-2}=\dfrac{-1}{-2}=\dfrac{1}{2}\)

Vậy: Khi \(x^2-4x=0\) thì \(M=\dfrac{1}{2}\)

1.

a: \(A=\left(\dfrac{\sqrt{x}+4}{x-4}-\dfrac{1}{\sqrt{x}-2}\right):\left(1-\dfrac{2\sqrt{x}+5}{\sqrt{x}+2}\right)\)

\(=\dfrac{\sqrt{x}+4-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}+2-2\sqrt{x}-5}{\sqrt{x}+2}\)

\(=\dfrac{2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{-\sqrt{x}-3}\)

\(=-\dfrac{2}{\sqrt{x}-2}\)

b: Để A là số nguyên thì \(\sqrt{x}-2\inƯ\left(-2\right)\)

\(\Leftrightarrow\sqrt{x}-2\in\left\{1;-1;2;-2\right\}\)

hay \(x\in\left\{1;16;0\right\}\)