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a/ \(\frac{2x^3}{4x^7}=\frac{1}{2x^4}\) với ĐKXĐ : \(x\ne0\)
b/ \(\frac{x-1}{\left(x+1\right)^2}.\frac{x^2+2x+1}{x^2-1}=\frac{x-1}{\left(x+1\right)^2}.\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{1}{x+1}\) với ĐKXĐ : \(x\ne\pm1\)
c/ \(\frac{x^2-7x+12}{x^2-16}=\frac{\left(x-4\right)\left(x-3\right)}{\left(x-4\right)\left(x+4\right)}=\frac{x-3}{x+4}\) với ĐKXĐ : \(x\ne\pm4\)
d/ \(\frac{x-1}{\sqrt{x}+1}:\left(\sqrt{x}-1\right)=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}.\frac{1}{\sqrt{x}-1}=1\) với ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
a/ \(2x^2-3x+1>0\Rightarrow\left[{}\begin{matrix}x>1\\x< \frac{1}{2}\end{matrix}\right.\)
b/ \(-3x^2+2x+1< 0\Rightarrow-\frac{1}{3}< x< 1\)
c/ \(\frac{x+3}{x-2}\ge0\Rightarrow\left[{}\begin{matrix}x>2\\x\le-3\end{matrix}\right.\)
d/ \(\frac{2x+1}{x+2}\ge1\Leftrightarrow\frac{2x+1}{x+2}-1\ge0\Leftrightarrow\frac{x-1}{x+2}\ge0\Rightarrow\left[{}\begin{matrix}x\ge1\\x< -2\end{matrix}\right.\)
e/ \(\frac{\sqrt{x}+3}{2-\sqrt{x}}\le0\Rightarrow\left\{{}\begin{matrix}x\ge0\\2-\sqrt{x}< 0\end{matrix}\right.\) \(\Rightarrow x>4\)
g/\(\frac{\sqrt{x}-3}{\sqrt{x}-2}\ge0\Rightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x\ge9\\x< 4\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.\)
h/ \(\frac{\sqrt{x}-3}{\sqrt{x}-1}-\frac{1}{3}< 0\Rightarrow\frac{2\left(\sqrt{x}-4\right)}{3\left(\sqrt{x}-1\right)}< 0\Rightarrow1< x< 16\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có :
\(A=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{x-1}:\frac{1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+1}\)
\(=1\)
Vậy...
b/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có :
\(B=\left(\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}+6\right)\left(\frac{x\sqrt{x}-1}{x+\sqrt{x}+1}-3\right)\)
\(=\left(\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-2}+6\right)\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-3\right)\)
\(=\left(\sqrt{x}-2+6\right)\left(\sqrt{x}-1-3\right)\)
\(=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)\)
\(=x-16\)
Vậy..
c/ ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có :
\(C=\frac{2\sqrt{x}}{x-1}+\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}-x}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{2x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x+\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-2}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{2}{\sqrt{x}}\)
Vậy..