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Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} \frac{1}{z}=-\left(\frac{1}{x}+\frac{1}{y}\right)\\ \frac{2}{xy}-\frac{1}{z^2}=4\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} \frac{1}{z^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}\\ \frac{2}{xy}-\frac{1}{z^2}=4\end{matrix}\right.\)
\(\Rightarrow \frac{2}{xy}-\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}\right)=4\)
\(\Leftrightarrow -\left(\frac{1}{x^2}+\frac{1}{y^2}\right)=4>0\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}< 0\) (vô lý)
Do đó không tồn tại $x,y,z$ kéo theo không tồn tại giá trị của P
\(\left\{{}\begin{matrix}x-2\sqrt{y}+1=0\\y-2\sqrt{z}+1=0\\z-2\sqrt{x}+1=0\end{matrix}\right.\)
Cộng theo vế 3 pt trên ta có:
\(\left(x-2\sqrt{x}+1\right)+\left(y-2\sqrt{y}+1\right)+\left(z-2\sqrt{z}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2+\left(\sqrt{z}-1\right)^2=0\)
Dễ thấy: \(VT=\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2+\left(\sqrt{z}-1\right)^2\ge0=VP\)
Xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{y}-1=0\\\sqrt{z}-1=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{y}=1\\\sqrt{z}=1\end{matrix}\right.\)\(\Rightarrow x=y=z=1\)
1. Giải phương trình, hệ phương trình:
a) 2x2 - 5x + 3 = 0
\(\Leftrightarrow2x^2-2x-3x+3=0\)
\(\Leftrightarrow2x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
b) x2 - 3x = 0
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}2\left(x+1\right)-5\left(y+1\right)=5\\3\left(x+1\right)-2\left(y+1\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6\left(x+1\right)-15\left(y+1\right)=15\\6\left(x+1\right)-4\left(y+1\right)=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-11\left(y+1\right)=13\\3\left(x+1\right)-2\left(y+1\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=\dfrac{-13}{11}\\3\left(x+1\right)-2.\left(-\dfrac{13}{11}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{24}{11}\\3\left(x+1\right)=-\dfrac{15}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{24}{11}\\x=-\dfrac{16}{11}\end{matrix}\right.\)
Hix ,mệt quá.
\(d,\left\{{}\begin{matrix}\dfrac{15}{x}-\dfrac{7}{y}=9\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{60}{x}-\dfrac{28}{y}=36\\\dfrac{60}{x}+\dfrac{135}{y}=525\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{163}{y}=-489\\\dfrac{60}{x}+\dfrac{135}{y}=525\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\\dfrac{60}{x}+405=525\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
+ \(\left(1\right)\Leftrightarrow x^3+1+2\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow x^3+1+2\left(y-1\right)^2=0\)
Với \(\forall y\in R\Rightarrow\left(y-1\right)^2\ge0\Rightarrow x^3+1\le0\)
\(\Rightarrow x^3\le-1\Leftrightarrow x\le-1\)(*)
+ \(\left(2\right)\Leftrightarrow x^2y^2-2y+x^2=0\)
Có \(\Delta'_y=1-x^4\) \(\ge0\) thì \(\left(2\right)\) có nghiệm
\(\Leftrightarrow x^4\le1\Leftrightarrow-1\le x\le1\)(**)
Từ (*) và (**) => \(x=-1\Rightarrow\) Thay vào (1) \(\Rightarrow y=1\)
Vậy \(B=x^2+y^2=\left(-1\right)^2+1^2=2\)
Có: \(\left\{{}\begin{matrix}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{matrix}\right.\)
\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)
\(\Rightarrow\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)
\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0^{\left(1\right)}\)
Lại có:
\(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\\\left(z+1\right)^2\ge0\forall z\end{matrix}\right.\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2\ge0\forall x;y;z^{\left(2\right)}\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x+1=0\\y+1=0\\z+1=0\end{matrix}\right.\Rightarrow x=y=z=-1\)
Thay \(x=y=z=-1\) vào \(A\), ta được:
\(A=\left(-1\right)^{2000}+\left(-1\right)^{2000}+\left(-1\right)^{2000}\)
\(=1+1+1=3\)
\(\text{#}\mathit{Toru}\)