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3 tháng 8 2018

de dung ko vay ban

22 tháng 8 2017

1)

\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)

\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)

22 tháng 8 2017

dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)

\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)

\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)

2: \(\left(\dfrac{7}{a+7}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right):\dfrac{a+1}{2}\)

\(=\dfrac{7a-49+a^2+49-7a-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}\)

\(=\dfrac{a^2-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}=\dfrac{2}{a+1}\)

3: \(=\dfrac{x^4-4x^2+4x^2}{x^2-4}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x^2-4\right)}\cdot\dfrac{x^2-4}{x-2}\right)\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x-2\right)}\right)\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x^2-4\right)+\left(2-3x\right)\left(x-4\right)}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-4x+2x-8-3x^2+12x}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-3x^2+10x-8}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-x^2-2x^2+2x+8x-8}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^3\left(x-1\right)\left(x^2-2x+8\right)}{\left(x-2\right)^2\cdot\left(x+2\right)\left(x-4\right)}\)