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1)\(M=\frac{x-7}{x-4\sqrt{x}+3}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}-3}\)(ĐKXĐ : \(x\ge0;x\ne1;x\ne9\))
\(=\frac{x-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-1}\)
2) \(M>\frac{3}{4}\Leftrightarrow\frac{\sqrt{x}+3}{\sqrt{x}-1}>\frac{3}{4}\Leftrightarrow1+\frac{4}{\sqrt{x}-1}-\frac{3}{4}>0\Leftrightarrow\frac{4}{\sqrt{x}-1}+\frac{1}{4}>0\Rightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)Vậy \(M>\frac{3}{4}\Leftrightarrow\hept{\begin{cases}x>1\\x\ne9\end{cases}}\)
\(A=\frac{4xy}{y^2-x^2}:\left(\frac{1}{y^2+2xy+x^2}-\frac{x^3+y^3}{x^4-y^4}\right)\left(x\ne\pm y;y\ne0\right)\)
\(\Leftrightarrow A=\frac{4xy}{\left(y^2-x^2\right)\left(y^2+x^2\right)}:\left(\frac{1}{\left(y+x\right)^2}-\frac{x^3+y^3}{\left(x^2-y^2\right)\left(x^2+y^2\right)}\right)\)
B=\(\frac{10\sqrt{x}-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{10\sqrt{x}-2x+2\sqrt{x}+3\sqrt{x}-3-x-4\sqrt{x}-\sqrt{x}-4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)=\(\frac{-3x+3\sqrt{x}+7\sqrt{x}-7}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{\left(\sqrt{x}-1\right)\left(7-3\sqrt{x}\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)=\(\frac{7-3\sqrt{x}}{\sqrt{x}+4}\)
Vậy...
\(B=\frac{10\sqrt{x}}{x+3\sqrt{x}-4}-\frac{2\sqrt{x-3}}{\sqrt{x}+4}+\frac{\sqrt{x}+1}{1-\sqrt{x}}\)( \(x\ge0;x\ne1\)
=>\(B=\frac{10\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
=> \(B=\frac{10\sqrt{x}-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=> \(B=\frac{10\sqrt{x}-\left(2x-5\sqrt{x}+3\right)-\left(x+5\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}=\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=> \(B=\frac{\left(\sqrt{x}-1\right)\left(7-3\sqrt{x}\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}=\frac{7-3\sqrt{x}}{\sqrt{x}+4}\)( zì \(x\ge0,x\ne1\)