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a) \(\sqrt{12}-3\sqrt{75}+0,5\sqrt{\left(-6\right)^2\cdot3}\)
\(=2\sqrt{3}-15\sqrt{3}+0,5\sqrt{108}\)
\(=-13\sqrt{3}+3\sqrt{3}\)
\(=-10\sqrt{3}\)
b) \(3\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{4+2\sqrt{3}}\)
\(=3\left|\sqrt{2}-\sqrt{3}\right|-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=3\left(\sqrt{3}-\sqrt{2}\right)-\left|\sqrt{3}+1\right|\)
\(=3\sqrt{3}-3\sqrt{2}-\sqrt{3}-1\)
\(=2\sqrt{3}-3\sqrt{2}-1\)
c) \(\left(\frac{2x+1}{x\sqrt{x}-1}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right)\div\frac{1}{x-2\sqrt{x}+1}\)
\(=\frac{2x+1-\left(\sqrt{x}-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\div\frac{1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\)
\(=\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\)
\(=\sqrt{x}-1\)
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
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ĐKXĐ: \(x\ge-15\)
Ta có: \(\sqrt{16\left(x-1\right)^2}=2x+\sqrt{12}\cdot\sqrt{75}\)
\(\Leftrightarrow\sqrt{\left[4\left(x-1\right)\right]^2}=2x+30\)
\(\Leftrightarrow\left|4x-4\right|=2x+30\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-4=2x+30\\4x-4=-2x-30\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x-4-2x-30=0\\4x-4+2x+30=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-34=0\\6x+26=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=34\\6x=-26\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\left(nhận\right)\\x=-\frac{26}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{17;\frac{-26}{6}\right\}\)
\(\sqrt{16\left(x-1\right)^2}=2x+\sqrt{12}.\sqrt{75}\)
\(\Leftrightarrow\)\(4\sqrt{\left(x-1\right)^2}=2x+\sqrt{12.75}\)
\(\Leftrightarrow4x-4=2x+\sqrt{900}\)
\(\Leftrightarrow4x-2x=4+30\) (chuyển vế đổi dấu)
\(\Leftrightarrow2x=34\)
\(\Leftrightarrow x=17\)