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\(=>P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\sqrt{3}\)
CHÚC BẠN HỌC TỐT..........
\(\)\(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\rightarrow\left(a;b;c\right)\)
Viết lại đề: \(\left\{{}\begin{matrix}a+b+c=2\\2ab-c^2=4\end{matrix}\right.\) . Tính \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^{2018}\)
\(\Leftrightarrow\left(a+b+c\right)^2-2ab+c^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-2ab+c^2=0\)
\(\Leftrightarrow a^2+b^2+2c^2+2bc+2ac=0\)
\(\Leftrightarrow\left(a^2+c^2+2ac\right)+\left(b^2+c^2+2bc\right)=0\)
\(\Leftrightarrow\left(a+c\right)^2+\left(b+c\right)^2=0\)
\(\Leftrightarrow....\)
\(A=\left(xy+yz+xz\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-xyz\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\\ =y+x+\dfrac{xy}{z}+y+z+\dfrac{yz}{x}+x+z+\dfrac{xz}{y}-\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\\ =2\left(x+y+z\right)=2.2018=4036\)
Do \(x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}\)
=> \(\left\{{}\begin{matrix}x+\dfrac{1}{y}=y+\dfrac{1}{z}\Leftrightarrow x-y=\dfrac{1}{z}-\dfrac{1}{y}\Leftrightarrow x-y=\dfrac{y-z}{yz}\\y+\dfrac{1}{z}=z+\dfrac{1}{x}\Leftrightarrow y-z=\dfrac{1}{x}-\dfrac{1}{z}\Leftrightarrow y-z=\dfrac{z-x}{xz}\\z+\dfrac{1}{x}=x+\dfrac{1}{y}\Leftrightarrow z-x=\dfrac{1}{y}-\dfrac{1}{x}\Leftrightarrow z-x=\dfrac{x-y}{xy}\end{matrix}\right.\)
=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)=\dfrac{\left(y-z\right)\left(z-x\right)\left(x-y\right)}{x^2y^2z^2}\)
<=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)x^2y^2z^2=\left(y-z\right)\left(z-x\right)\left(x-y\right)\)
<=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x^2y^2z^2-1\right)=0\)
=> \(\left(x-y\right)\left(y-z\right)\left(z-x\right)=0\) hoặc \(x^2y^2z^2-1=0\)
=> x=y=z hoặc xyz=1 hoặc xyz=-1
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\sqrt{3}\)
\(\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2=3\)
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{xz}=3\)
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=3\)
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.\left(\dfrac{x+y+z}{xyz}\right)=3\)
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.1=3\) ( Do x+y+z=xyz )
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=3-2=1\)
Vậy P = 1
Lời giải:
Từ \(x+y+z=xyz\Rightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)
Do đó:
\(M=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)
\(\Leftrightarrow M=3^2-2=7\)
Vậy \(M=7\)
Lời giải:
Ta có:
\(\left\{\begin{matrix} x=by+cz\\ y=ax+cz\\ z=ax+by\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x-y=by-ax\\ z=ax+by\end{matrix}\right.\)
\(\Rightarrow x-y+z=2by\Rightarrow b=\frac{x+z-y}{2y}\)
Hoàn toàn tương tự ta nhận được:
\(a=\frac{y+z-x}{2x};c=\frac{x+y-z}{2z}\)
Suy ra:
\(\left\{\begin{matrix} a+1=\frac{x+y+z}{2x}\\ b+1=\frac{x+y+z}{2y}\\ c+1=\frac{x+y+z}{2z}\end{matrix}\right.\)
\(\Rightarrow \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2\) (ĐPCM)