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1, Ta có: \(x+y=9\Rightarrow\left(x+y\right)^2=81\)
\(\Rightarrow x^2+2xy+y^2=81\)
\(\Rightarrow x^2+y^2=45\)
\(\Rightarrow x^2+y^2-2xy=9\)
\(\Rightarrow\left(x-y\right)^2=9\Rightarrow\left[{}\begin{matrix}x-y=3\\x-y=-3\end{matrix}\right.\)
\(A=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(\Rightarrow\left[{}\begin{matrix}A=3.63=189\\A=-3.63=-189\end{matrix}\right.\)
Vậy...
1)
\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)
dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
1) Đặt \(B=x^2+y^2+z^2\)
\(C=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)
Ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)
\(\Leftrightarrow-2\left(xy+yz+xz\right)=x^2+y^2+z^2\)
Suy ra: \(C=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow A=\dfrac{B}{C}=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
2) \(x^2-2y^2=xy\Leftrightarrow x^2-xy-2y^2=0\)
\(\Leftrightarrow x^2+xy-2xy-2y^2=0\)
\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)
\(\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)
Do \(x+y\ne0\) nên \(x-2y=0\Leftrightarrow x=2y\)
Do đó: \(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
Lời giải:
Ta có:
\(x^3+y^3+z^3=3xyz\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0\)
Vì \(x+y+z\neq 0\Rightarrow x^2+y^2+z^2-xy-yz-xz=0\)
\(\Leftrightarrow 2(x^2+y^2+z^2-xy-yz-xz)=0\)
\(\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0\)
Ta thấy \((x-y)^2; (y-z)^2;(z-x)^2\geq 0\)
\(\Rightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0\). Dấu bằng xảy ra khi
\((x-y)^2=(y-z)^2=(z-x)^2=0\Leftrightarrow x=y=z\)
Khi đó:
\(P=(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=(1+1)(1+1)(1+1)=8\)
\(\)\(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\rightarrow\left(a;b;c\right)\)
Viết lại đề: \(\left\{{}\begin{matrix}a+b+c=2\\2ab-c^2=4\end{matrix}\right.\) . Tính \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^{2018}\)
\(\Leftrightarrow\left(a+b+c\right)^2-2ab+c^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-2ab+c^2=0\)
\(\Leftrightarrow a^2+b^2+2c^2+2bc+2ac=0\)
\(\Leftrightarrow\left(a^2+c^2+2ac\right)+\left(b^2+c^2+2bc\right)=0\)
\(\Leftrightarrow\left(a+c\right)^2+\left(b+c\right)^2=0\)
\(\Leftrightarrow....\)