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\(A=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-x}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(A=\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)\(\div\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(A=\left(\frac{x+2\sqrt{x}+1+x-\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{2x+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}}\)
\(A=\frac{2x+1}{4\sqrt{x}}\)
c, \(A=\frac{2x+1}{4\sqrt{x}}=\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\)
ap dụng cô si ta có \(\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\ge2\sqrt{\frac{\sqrt{x}}{2}\cdot\frac{1}{4\sqrt{x}}}=\frac{\sqrt{2}}{2}\)
dấu = xảy ra khi \(\frac{\sqrt{x}}{2}=\frac{1}{4\sqrt{x}}\Leftrightarrow x=\frac{1}{2}\) (tm)
M=(\(\frac{\sqrt{x}}{\sqrt{x}+1}\)-1): \(\frac{-1}{x+\sqrt{x}+1}\)
M=\(\frac{-1}{\sqrt{x}+1}\). -(x+\(\sqrt{x}\)+1)
M=\(\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
b, x=1
M = \(\frac{3}{2}\)
c, M= 0
=> x +\(\sqrt{x}\)+1= 0
mặt khác x+\(\sqrt{x}\)+1 = (\(\sqrt{x}\)+0,5)2+0,75 >0
=> x vô nghiệm........
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......