Bạn thử tham khảo bài của tôi nhé :
b) M=A-B hay
M=\(\frac{1}{\sqrt{x}+1}-\frac{1-\sqrt{x}}{1+\sqrt{x}}\)
đkxd\(x\ge0,x\pm25\)
=\(\frac{1-1+\sqrt{x}}{\sqrt{x}+1}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(\frac{\sqrt{x}}{\sqrt{x}+1}=1+\frac{-1}{\sqrt{x}+1}\)
Để \(\frac{\sqrt{x}}{\sqrt{x}+1}\) nguyên thì \(\frac{-1}{\sqrt{x+1}}\) nguyên \(hay\sqrt{x}+1\inƯ\left(-1\right)\)
\(\rarr\sqrt{x}+1\in\left.\begin{cases}\\ \end{cases}-1;1\right)\)
+) \(Với\) \(\sqrt{x}+1=-1,\lrArr\sqrt{x}=-2\) (vô lí)
+)\(Với\) \(\sqrt{x}+1=1\lrArr\sqrt{x}=0\rarr x=0\)
Vậy để M nguyên thì \(x=0\)

a) A=
\(\left(\frac{15-\sqrt{x}}{x-25}+\frac{2}{\sqrt{x}+5}\right)\colon\frac{\sqrt{x}+1}{\sqrt{x}-5},x\pm25,x\ge0\), MTC=x-25
\(=\left(\frac{15-\sqrt{x}}{x-25}+\frac{2\left(\sqrt{x}-5\right)}{x-25}\right)\colon\frac{\sqrt{x}+1}{\sqrt{x}-5}\)
=\(\left(\frac{15-\sqrt{x}}{x-25}+\frac{2\sqrt{x}-10}{x-25}\right)\colon\frac{\sqrt{x}+1}{\sqrt{x}-5}\)
=\(\left(\frac{15-\sqrt{x}+2\sqrt{x}-10}{x-25}\right)\colon\frac{\sqrt{x}+1}{\sqrt{x}-5}\)
=\(\frac{5+\sqrt{x}}{x-25}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+1}\)
=\(\frac{5+\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+1}\)
=\(\frac{1}{\sqrt{x}+1}\)