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Khách
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TN
2
V
1
LD
29 tháng 12 2017
\(\left\{{}\begin{matrix}xy=\dfrac{1}{2}\\yz=\dfrac{3}{5}\\zx=\dfrac{27}{10}\end{matrix}\right.\Rightarrow xyyzzx=\dfrac{1}{2}\cdot\dfrac{3}{5}\cdot\dfrac{27}{10}\Leftrightarrow\left(xyz\right)^2=\dfrac{81}{100}\)
\(\Rightarrow\left[{}\begin{matrix}xyz=-\dfrac{9}{10}\\xyz=\dfrac{9}{10}\end{matrix}\right.\)
+ Khi \(xyz=-\dfrac{9}{10}\)
\(\Rightarrow\left\{{}\begin{matrix}z=-\dfrac{9}{10}:\dfrac{1}{2}=-\dfrac{9}{5}\\x=-\dfrac{9}{10}:\dfrac{3}{5}=-\dfrac{3}{2}\\y=-\dfrac{9}{10}:\dfrac{27}{10}=-\dfrac{1}{3}\end{matrix}\right.\)
+ Khi \(xyz=\dfrac{9}{10}\)
\(\Rightarrow\left\{{}\begin{matrix}z=\dfrac{9}{10}:\dfrac{1}{2}=\dfrac{9}{5}\\x=\dfrac{9}{10}:\dfrac{3}{5}=\dfrac{3}{2}\\y=\dfrac{9}{10}:\dfrac{27}{10}=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(\dfrac{3}{2};\dfrac{1}{3};\dfrac{9}{5}\right);\left(-\dfrac{3}{2};-\dfrac{1}{3};-\dfrac{9}{5}\right)\)
NT
29 tháng 12 2017
\(\left(x.y\right).\left(y.z\right)\left(z.x\right)=\dfrac{1}{2}.\dfrac{3}{5}.\dfrac{27}{10}\\ \Rightarrow\left(x.y.z\right)^2=\dfrac{81}{100}\\ \Rightarrow\left[{}\begin{matrix}x.y.z=\dfrac{9}{10}\\x.y.z=-\dfrac{9}{10}\end{matrix}\right.\)
Nếu x.y.z=9/10
\(\Rightarrow z=\dfrac{9}{10}:\dfrac{1}{2}=\dfrac{9}{5};x=\dfrac{9}{10}:\dfrac{3}{5}=\dfrac{3}{2};y=\dfrac{9}{10}:\dfrac{27}{10}=\dfrac{1}{3}\)
Nếu x.y.z=-9/10
\(\Rightarrow z=-\dfrac{9}{5};x=-\dfrac{3}{2};y=-\dfrac{1}{3}\)
5 tháng 1 2018
b)
\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{5x-5}{10}=\dfrac{3y+9}{12}=\dfrac{4z-20}{24}\)
\(\Rightarrow\dfrac{\left(5x-3y-4z\right)-\left(5+9-20\right)}{10-12-24}=\dfrac{46+6}{-26}=-2\)
\(\Rightarrow x-1=-4\Rightarrow x=-3\)
\(\Rightarrow y+3=-8\Rightarrow y=-11\)
\(\Rightarrow z-5=-12\Rightarrow-7\)
26 tháng 8 2018
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=x+y+z\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{x}{x+y+z}\\b=\dfrac{y}{x+y+z}\\c=\dfrac{z}{x+y+z}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x^2}{\left(x+y+z\right)^2}+\dfrac{y^2}{\left(x+y+z\right)^2}+\dfrac{z^2}{\left(x+y+z\right)^2}=1\)
\(\Rightarrow x^2+y^2+z^2=\left(x+y+z\right)^2\)
\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2-2\left(xy+yz+zx\right)\)
\(\Rightarrow2\left(xy+yz+zx\right)=0\)
\(\Rightarrow xy+yz+zx=0\) (đpcm)
TM
1
2 tháng 4 2018
\(\dfrac{5z-6y}{4}=\dfrac{6x-4z}{5}=\dfrac{4y-5x}{6}\)
\(\Rightarrow\dfrac{5z-6y}{4}=\dfrac{6x-4z}{5}=\dfrac{4y-5x}{6}=\dfrac{5z-6y+6x-4z+4y-5x}{4+5+6}=\dfrac{x-2y+z}{4+5+6}\)
\(\Rightarrow\dfrac{x}{4}=\dfrac{-2y}{5}=\dfrac{z}{6}\)
\(\Rightarrow\dfrac{3x}{12}=\dfrac{-2y}{5}=\dfrac{5z}{30}\)
\(\Rightarrow\dfrac{3x}{12}=\dfrac{-2y}{5}=\dfrac{5z}{30}=\dfrac{3x-2y+5z}{12-5+30}=\dfrac{96}{37}\)
NH
4
14 tháng 1 2015
\(\frac{x}{y.z}:\frac{y}{x.z}=\frac{x}{y.z}.\frac{x.z}{y}=\frac{x.x.z}{y.z.y}=\frac{x^2}{y^2}\)mà ta có 3x=2y=.\(\frac{x^2}{y^2}=\frac{2^2}{3^2}=\frac{4}{9}\)
vậy kết quả bài này là \(\frac{4}{9}\)
14 tháng 1 2015
x/y.z:y/z.x=x/y.z . z.x/y=x.z.x/y.z.x =x.z.x/y.s.y
=x.x/y.y=x2/y2 mà 3x=2y => x2/y2=22/32=4/9
vậy kết quả bì đó bằng 4/9
Ta có: \(\dfrac{x}{y.z}.\dfrac{y}{z.x}=\dfrac{xy}{xyz^2}=\dfrac{1}{z^2}\)
cho bạn ấy GP đi