2.

a)

\(\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(2-\frac{2\sqrt{a}-a}{\sqrt{a}-2}\right)\\ =\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(2+\frac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\right)\\ =\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)\\ =2^2-\left(\sqrt{a}\right)^2\\ =4-a\)

b)

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

c)

\(\left(\frac{1-x\sqrt{x}}{1-x}+\sqrt{x}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\\ =\left(\frac{1-\sqrt{x^3}}{1-x}+\sqrt{x}\right)\cdot\frac{\left(1-\sqrt{x}\right)^2}{\left(1-x\right)^2}\\ =\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\sqrt{x}\right)\cdot\frac{\left(1-\sqrt{x}\right)^2}{\left[\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\right]^2}\\ =\left(\frac{1+\sqrt{x}+x+\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\right)\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\\ =\frac{2x+2\sqrt{x}+1}{1+\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)

\(=\frac{2x+2\sqrt{x}+1}{\left(1+\sqrt{x}\right)^3}\)

1. (Ko viết lại đề nha :v)

a)

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

\(=\frac{2}{x-1}\)

b) Để A đạt giá trị nguyên thì \(2⋮x-1\Leftrightarrow x-1\inƯ\left(2\right)\)

\(\Leftrightarrow x-1\in\left\{-1;1;-2;2\right\}\\ \Leftrightarrow x\in\left\{0;2;-1;3\right\}\)

Vậy......

5/ĐK: \(\left[{}\begin{matrix}x\le-1\\x\ge5\end{matrix}\right.\)

PT \(\Leftrightarrow2\left(x^2-4x-6\right)+\sqrt{x^2-4x-5}-1=0\)

\(\Leftrightarrow\left(x^2-4x-6\right)\left(2+\frac{1}{\sqrt{x^2-4x-5}+1}\right)=0\)

\(\Leftrightarrow x^2-4x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{10}\\x=2-\sqrt{10}\end{matrix}\right.\)

Vậy..

\(M=\frac{\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}+1\right)}{\left(2\sqrt{x}+1\right)}+\frac{x-\sqrt{x}-5}{\sqrt{x}+3}=1+\frac{x-\sqrt{x}-5}{\sqrt{x}+3}\)

\(M=\frac{\sqrt{x}+3+x-\sqrt{x}-5}{\sqrt{x}+3}=\frac{x-2}{\sqrt{x}+3}=\sqrt{x}-3+\frac{7}{\sqrt{x}+3}\)

Để M nguyên \(\Leftrightarrow x\) chính phương và \(\sqrt{x}+3=Ư\left(7\right)=7\)

\(\Rightarrow\sqrt{x}+3=7\Rightarrow\sqrt{x}=4\Rightarrow x=16\)

(\(\sqrt{x}+3\ge3\) nên chỉ cần xét các ước lớn hơn 3 của 7)

các bạn ơi mình ghi nhầm câu b

câu b của mình là:tìm x nguyên để M=\(A.\frac{\sqrt{x}+1}{2\sqrt{x}+1}+\frac{x-\sqrt{x}-5}{\sqrt{x}+3}\)có giá trị nguyên

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

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

Ko thấy biểu thức B để làm câu b

a)

\(N=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\\ =\frac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ =\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}+1\)

b)

\(N=x-\sqrt{x}+1=x-2\cdot\sqrt{x}\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy Min N = \(\frac{3}{4}\)khi x=\(\frac{1}{4}\)

Câu c) mk ko bt, sorry nha :<

\(M=\frac{2\sqrt{x}}{x-\sqrt{x}+1}=\frac{2\sqrt{x}}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}>0\) \(\forall x>0\)

\(M-2=\frac{2\sqrt{x}}{x-\sqrt{x}+1}=\frac{-2\left(x-2\sqrt{x}+1\right)}{x-\sqrt{x}+1}=\frac{-2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}< 0\) \(\forall x\ne1;x>0\)

\(\Rightarrow0< M< 2\Rightarrow\) để M nguyên thì \(M=1\)

\(\Rightarrow\frac{2\sqrt{x}}{x-\sqrt{x}+1}=1\Rightarrow x-3\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3\pm\sqrt{5}}{2}\) \(\Rightarrow x=\frac{7\pm3\sqrt{5}}{2}\)