Nếu xy = 3, xz = 4 và yz = 6. Giá trị của biểu thức \(A=x^2+y^2+z^2\) là ...
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x . y . x . z . y . z = 3 . 4 . 6
(x . x) . (y . y) . (z . z) = 72
x2 . y2 . z2 = 72
=>A=72
\(xy=3;xz=4;yz=6\Rightarrow xy.xz.yz=3.4.6\Leftrightarrow\left(xyz\right)^2=72\)\(\Leftrightarrow xyz=\pm6\sqrt{2}\)
+)\(xyz=-6\sqrt{2}\) => \(x=-\sqrt{2};y=-\frac{3\sqrt{2}}{2};z=-2\sqrt{2}\)
Thay vào A
+))\(xyz=6\sqrt{2}\) => \(x=\sqrt{2};y=\frac{3\sqrt{2}}{2};z=2\sqrt{2}\)
Thay vào A
![](https://rs.olm.vn/images/avt/0.png?1311)
xy=4 , xz=6
=> \(x^2zy\)= 3x4=12
=>\(x^2\)=\(12:yz=12:6=2\)
\(xz=4,yz=6\)
=>\(z^2xy=6x4=24\)
=>\(z^2=24:xy=8\)
\(xy=3,yz=6\)
=>\(xy^2z=6x3=18\)
=>\(y^2=18:xz=18:4=4.5\)
Vậy \(x^2+y^2+z^2=2+8+4.5=14.5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(x^2+y^2-z^2\)
\(=\left(x+y\right)^2-z^2-2xy\)
\(=\left(x+y+z\right)\left(x+y-z\right)-2xy\)
\(=-2xy\)
Ta có: \(x^2+z^2-y^2\)
\(=\left(x+z\right)^2-y^2-2xz\)
\(=\left(x+y+z\right)\left(x+z-y\right)-2xz\)
\(=-2xz\)
Ta có: \(y^2+z^2-x^2\)
\(=\left(y+z\right)^2-x^2-2yz\)
\(=\left(x+y+z\right)\left(y+z-x\right)-2yz\)
\(=-2yz\)
Ta có: \(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{xz}{x^2+z^2-y^2}+\dfrac{yz}{y^2+z^2-x^2}\)
\(=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}\)
\(=\dfrac{1}{-2}+\dfrac{1}{-2}+\dfrac{1}{-2}\)
\(=\dfrac{-3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2-z^2=x^2+\left(y-z\right)\left(y+z\right)=x^2-x\left(y-z\right)=x\left(x-y+z\right)=x\left(-y-y\right)=-2xy\)
Tương tự \(x^2+z^2-y^2=-2xz;y^2+z^2-x^2=-2yz\)
Cộng VTV:
\(\Leftrightarrow\text{Biểu thức }=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}=-\dfrac{1}{8}\)
\(\left\{\begin{matrix}xy=3\left(1\right)\\xz=4\left(2\right)\\yz=6\left(3\right)\end{matrix}\right.\).Từ \(yz=6\Rightarrow z=\frac{6}{y}\) thay vào (2) ta có:
\(xz=4\Rightarrow x\cdot\frac{6}{y}=4\)\(\Leftrightarrow\frac{6x}{y}=4\Leftrightarrow6x=4y\Leftrightarrow y=\frac{6x}{4}=\frac{3x}{2}\) thay vào (1) ta có:
\(x\cdot\frac{3x}{2}=3\Leftrightarrow\frac{3x^2}{2}=3\Leftrightarrow3x^2=6\Leftrightarrow x^2=2\)
Từ \(\left(1\right)\Rightarrow x^2y^2=9\Rightarrow y^2=\frac{9}{x^2}=\frac{9}{2}\)
Từ \(\left(2\right)\Rightarrow x^2z^2=16\Rightarrow z^2=\frac{16}{x^2}=\frac{16}{2}=8\)
Khi đó \(A=x^2+y^2+z^2=2+\frac{9}{2}+8=\frac{29}{2}\)