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Ta có : x - y = 0 => x = y
Vì x = y => xy = x2 = y2 ≥ 0
=> xy ≥ 0 ( đpcm )
Lời giải:
Khi $x-y+z=0\Rightarrow y=x+z$. Thay vào biểu thức $xy+yz-xz$ thì:
$xy+yz-xz=x(x+z)+(x+z)z-xz=x^2+xz+z^2=x^2+\frac{xz}{2}+\frac{xz}{2}+\frac{z^2}{4}+\frac{3}{4}z^2$
$=(x+\frac{z}{2})^2+\frac{3}{4}z^2$
Dễ thấy $(x+\frac{z}{2})^2\geq 0; \frac{3}{4}z^2\geq 0$ với mọi $x,y,z$ nên $xy+yz-xz\geq 0$
Ta có đpcm.
a, Sửa đề \(xy=\dfrac{2}{7}\)
Ta có: \(xy=\dfrac{2}{7};yz=\dfrac{3}{2};zx=\dfrac{3}{7}\Rightarrow xy.yz.zx=\dfrac{2}{7}.\dfrac{3}{2}.\dfrac{3}{7}\)
\(\Rightarrow\left(xyz\right)^2=\dfrac{9}{49}\Leftrightarrow\left(xyz\right)^2=\left(\pm\dfrac{3}{7}\right)^2\Rightarrow\left[{}\begin{matrix}xyz=\dfrac{3}{7}\\xyz=-\dfrac{3}{7}\end{matrix}\right.\)
+) Xét trường hợp \(xyz=\dfrac{3}{7}\)\(\Rightarrow\dfrac{2}{7}.z=\dfrac{3}{7}\Rightarrow z=\dfrac{3}{7}:\dfrac{2}{7}=\dfrac{3}{2}\)
\(\Rightarrow y.\dfrac{3}{2}=\dfrac{3}{2}\Rightarrow y=1\Rightarrow x.1=\dfrac{2}{7}\Rightarrow x=\dfrac{2}{7}\)
+) Xét trường hợp \(xyz=-\dfrac{3}{7}\Rightarrow\dfrac{2}{7}.z=-\dfrac{3}{7}\Rightarrow z=-\dfrac{3}{2}\)
\(\Rightarrow y.\dfrac{-3}{2}=\dfrac{3}{2}\Rightarrow y=-1\Rightarrow x.\left(-1\right)=\dfrac{2}{7}\Rightarrow x=-\dfrac{2}{7}\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=1\\z=\dfrac{2}{7}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-1\\z=-\dfrac{2}{7}\end{matrix}\right.\)
b, Ta có: \(xy=9z;yz=4x;zx=16y\Rightarrow\dfrac{xy}{z}=9;\dfrac{yz}{x}=4;\dfrac{zx}{y}=16\)
\(\Rightarrow\dfrac{xy}{z}.\dfrac{yz}{x}.\dfrac{zx}{y}=9.4.16\Rightarrow xyz=576\)
\(\Rightarrow xy=\dfrac{576}{z};yz=\dfrac{576}{x};zx=\dfrac{576}{y}\)
\(\Rightarrow\dfrac{576}{z}=9z\Rightarrow9z^2=576\Rightarrow z^2=64\Rightarrow z=\pm8\)
\(\dfrac{576}{x}=4x\Rightarrow4x^2=576\Rightarrow x^2=144\Rightarrow x=\pm12\)
\(\dfrac{576}{y}=16y\Rightarrow16y^2=576\Rightarrow y^2=36\Rightarrow y=\pm6\)
Vì xyz=156 => x;y;z dương hoặc trong x;y;z có 2 số âm
\(\Rightarrow\left\{{}\begin{matrix}x=12\\y=6\\z=8\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=12\\y=-6\\z=-8\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-12\\y=6\\z=-8\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-12\\y=-6\\z=8\end{matrix}\right.\)
Vậy...
a) \(xy=\dfrac{3}{7};yz=\dfrac{3}{2};zx=\dfrac{3}{7}\)
từ \(xy=\dfrac{3}{7}vàzx=\dfrac{3}{7}\) \(\Rightarrow\) \(z=y\)
\(yz=\dfrac{3}{2}\) \(\Leftrightarrow\) \(y^2=\dfrac{3}{2}\) \(\Leftrightarrow\) \(y=\sqrt{\dfrac{3}{4}}\) \(\Leftrightarrow\) \(y=z=\dfrac{\sqrt{3}}{2}\)
\(\Rightarrow\) \(xy=\dfrac{3}{7}\) \(\Leftrightarrow\) \(x.\dfrac{\sqrt{3}}{2}=\dfrac{3}{7}\) \(\Leftrightarrow\) \(x=\dfrac{3}{7}:\dfrac{\sqrt{3}}{2}\) = \(\dfrac{3}{7}.\dfrac{2}{\sqrt{3}}=\dfrac{6}{7\sqrt{3}}\) = \(\dfrac{2\sqrt{3}}{7}\)
vậy \(x=\dfrac{2\sqrt{3}}{7}\) ; \(y=\dfrac{\sqrt{3}}{2}\) ; \(z=\dfrac{\sqrt{3}}{2}\)
Ta có :
\(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{zx}{z+x}=\frac{xyz}{z\left(x+y\right)}=\frac{xyz}{x\left(y+z\right)}=\frac{xyz}{y\left(x+z\right)}\)
\(\Rightarrow z\left(x+y\right)=x\left(y+z\right)=y\left(z+x\right)\)
Từ \(z\left(x+y\right)=x\left(y+z\right)\Leftrightarrow xz+yz=xy+xz\Leftrightarrow yz=xy\Rightarrow x=z\) (1)
Từ \(x\left(y+z\right)=y\left(x+z\right)\Leftrightarrow xy+xz=xy+yz\Leftrightarrow xz=yz\Rightarrow x=y\) (2)
Từ \(z\left(x+y\right)=y\left(z+x\right)\Leftrightarrow xz+yz=yz+xy\Leftrightarrow xz=xy\Rightarrow z=y\) (3)
Từ (1) ; (2) ; (3) \(\Rightarrow x=y=z\) (đpcm)
\(\left\{{}\begin{matrix}\left|x\right|\ge3\\\left|y\right|\ge3\\\left|z\right|\ge3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|\dfrac{1}{x}\right|\le\dfrac{1}{3}\\\left|\dfrac{1}{y}\right|\le\dfrac{1}{3}\\\left|\dfrac{1}{z}\right|\le\dfrac{1}{3}\end{matrix}\right.\)
\(\left|A\right|=\left|\dfrac{xy+yz+xz}{xyz}\right|=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\le\left|\dfrac{1}{x}\right|+\left|\dfrac{1}{y}\right|+\left|\dfrac{1}{z}\right|\le\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)
\(\Rightarrow A\le\left|A\right|\le1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=3\)