Tìm \(x\in Z\):
\(x^2=16\)
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a) \(đk:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
b) \(x=3+2\sqrt{2}\Rightarrow\sqrt{x}=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
\(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{2}+1\right)-1}{\sqrt{2}+1-2}=\dfrac{2\sqrt{2}+1}{\sqrt{2}-1}\)
c) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{1}{2}\)
\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}-2\Leftrightarrow3\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)
d) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}>2\)
\(\Leftrightarrow2\sqrt{x}-1>2\sqrt{x}-4\Leftrightarrow-1>-4\left(đúng\forall x\right)\)
e) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}-2}=2+\dfrac{3}{\sqrt{x}-2}\in Z\)
\(\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
Do \(x\ge0\)
\(\Rightarrow x\in\left\{1;9;25\right\}\)
\(\sqrt{\left(2x-\sqrt{16}\right)^2}+\left(y^2.64\right)^2+lx+y+zl=0\)
\(\Rightarrow\sqrt{2x-4}+8y^4+lx+y+zl=0\)
\(\sqrt{2x-4};8y^4;lx+y+zl\ge0\)mà \(\sqrt{2x-4}+8y^4+lx+y+zl=0\)
\(\Rightarrow\sqrt{2x-4}=8y^4=lx+y+zl=0\)
=>2x-4=y4=lx+y+zl=0
=>x=2;y=0;z=-2
Vậy x=2;y=0;z=-2
a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
b: \(M=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\dfrac{x^2-4+10-x^2}{x+2}\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\)
\(=\dfrac{-1}{x-2}\)
d: Để M nguyên thì \(x-2\in\left\{1;-1\right\}\)
hay \(x\in\left\{3;1\right\}\)
Ta có : \(4x=2y=3z\)
\(\Rightarrow\frac{4x}{12}=\frac{2y}{12}=\frac{3z}{12}\) \(\Leftrightarrow\frac{x}{3}=\frac{y}{6}=\frac{z}{4}\)
Đặt \(\frac{x}{3}=\frac{y}{6}=\frac{z}{4}=k\left(k\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}x=3k\\y=6k\\z=4k\end{cases}}\)
Mà \(2x-3y+z=16\)
\(\Rightarrow2.3k-3.6k+4k=16\)
\(\Leftrightarrow6k-18k+4k=16\)
\(\Leftrightarrow k.\left(6-18+4\right)=16\)
\(\Leftrightarrow-8k=16\)
\(\Leftrightarrow k=-2\)
\(\Rightarrow\hept{\begin{cases}x=3k=-6\\y=6k=-12\\z=4k=-8\end{cases}}\)
Vậy ...