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a) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
\(=\left(\frac{x+2-\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right):\left(\frac{\left(\sqrt{x}-4\right)\left(x+1\right)-\sqrt{x}\left(1-x\right)}{1-x^2}\right)\)
\(=\left(\frac{x+2-x-\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{x\sqrt{x}+\sqrt{x}-4x-4-\sqrt{x}+x\sqrt{x}}{1-x^2}\right)\)
\(=\frac{2-\sqrt{x}}{\sqrt{x}+1}:\frac{2x\sqrt{x}-3x-4}{\left(1-x\right)\left(1+x\right)}\)
\(=\frac{2-\sqrt{x}}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(1+x\right)}{2x\sqrt{x}-3x-4}\)
\(=\frac{\left(2-\sqrt{x}\right)\left(\sqrt{x}+x\sqrt{x}-1+x\right)}{2x\sqrt{x}-3x-4}\)
\(=\frac{2\sqrt{x}+2x\sqrt{x}-2+2x-x-x^2+\sqrt{x}-x\sqrt{x}}{2x\sqrt{x}-3x-4}\)
tới đêy tự xử đi
a) A có nghĩa\(\Leftrightarrow x-y\ne0\Leftrightarrow x\ne y\)
b) \(A=\frac{x+y-2\sqrt{xy}}{x-y}=\frac{\left(\sqrt{x-\sqrt{y}}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
1) \(\left(a-b\right)\cdot\sqrt{\frac{ab}{\left(a-b\right)^2}}=\left(a-b\right)\cdot\frac{\sqrt{ab}}{a-b}=\sqrt{ab}\)
2) \(\frac{x-y}{y}\cdot\sqrt{\frac{y^4}{x^2-2xy+y^2}}=\frac{x-y}{y}\cdot\frac{\sqrt{y^4}}{\sqrt{\left(x-y\right)^2}}=\frac{x-y}{y}\cdot\frac{y^2}{x-y}=y\)
B=\(\left(\frac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)\(-\frac{\sqrt{x}}{x+\sqrt{x}+1}\))\(\left(\frac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)=\(\left(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)\(\left(x-2\sqrt{x}+1\right)\)=\(\sqrt{x}-1\)
a) Xét \(x^2-4=\left(\sqrt{\frac{a}{b}}\right)^2+\left(\sqrt{\frac{b}{a}}\right)^2+2-4\)
\(=\left(\sqrt{\frac{a}{b}}\right)^2+\left(\sqrt{\frac{b}{a}}\right)^2-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\ge0\)
b) \(\sqrt{x^2-4}=\sqrt{\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2}=\left|\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right|\)
a) Vì a<0 , b<0 => \(\frac{a}{b}>0;\frac{b}{a}>0\Rightarrow\sqrt{\frac{a}{b}}>0;\sqrt{\frac{b}{a}}>0\)
Áp dụng bất đẳng thức cô si ta có:
\(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\ge2\sqrt{\sqrt{\frac{a}{b}}\cdot\sqrt{\frac{b}{a}}}=2\)
=> \(\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)^2\ge4\)
Hay \(x^2\ge4\)