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15 tháng 1 2017
a,\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\Leftrightarrow\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)=3
TN
2
HH
1
5 tháng 1 2018
b)
\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{5x-5}{10}=\dfrac{3y+9}{12}=\dfrac{4z-20}{24}\)
\(\Rightarrow\dfrac{\left(5x-3y-4z\right)-\left(5+9-20\right)}{10-12-24}=\dfrac{46+6}{-26}=-2\)
\(\Rightarrow x-1=-4\Rightarrow x=-3\)
\(\Rightarrow y+3=-8\Rightarrow y=-11\)
\(\Rightarrow z-5=-12\Rightarrow-7\)
15 tháng 6 2016
a) Cộng cả 3 đẳng thức trên ta có:
2(x + y + z) = 1/2 +1/3 + 1/4 = 13/12 => x + y + z = 13/24 (*)
z = 13/24 - 1/2 = 1/24
x = 13/24 - 1/3 = 5/24
y = 13/24 - 1/4 = 7/24.
b) Nhân cả 3 đẳng thức ta có: x2y2z2 = 1/16 => xyz = 1/4 hoặc -1/4
- Nếu xyz = 1/4 thì: z = -1/2; x = 1/2; y = -1
- Nếu xyz = -1/4 thì: z = 1/2; x = -1/2; y = 1
MM
3
24 tháng 7 2017
\(\left\{{}\begin{matrix}xy=\dfrac{3}{5}\\yz=\dfrac{4}{5}\\zx=\dfrac{3}{4}\end{matrix}\right.\Rightarrow x^2y^2z^2=\dfrac{3}{5}.\dfrac{4}{5}.\dfrac{3}{4}=\dfrac{9}{25}\)
\(\Rightarrow xyz=\pm\dfrac{3}{5}\)
+) \(xyz=\dfrac{3}{5}\Rightarrow\left\{{}\begin{matrix}z=1\\x=\dfrac{3}{4}\\y=\dfrac{4}{5}\end{matrix}\right.\)
+) \(xyz=\dfrac{-3}{5}\Rightarrow\left\{{}\begin{matrix}z=-1\\x=\dfrac{-3}{4}\\y=\dfrac{-4}{5}\end{matrix}\right.\)
Vậy...
24 tháng 7 2017
\(\text{Ta có : }xy=\dfrac{3}{5}\\ yz=\dfrac{4}{5}\\ zx=\dfrac{4}{4}\\ \Rightarrow xy\cdot yz\cdot zx=\dfrac{3}{5}\cdot\dfrac{4}{5}\cdot\dfrac{3}{4}\\ \Rightarrow x^2\cdot y^2\cdot z^2=\dfrac{9}{25}\Rightarrow\left(xyz\right)^2=\dfrac{9}{25}\\ \Rightarrow xyz=\dfrac{-3}{5}\text{hoặc : }\\ xyz=\dfrac{3}{5}\)
\(\text{+) Xét }xyz=-\dfrac{3}{5}\Leftrightarrow\left\{{}\begin{matrix}x\cdot\left(yz\right)=-\dfrac{3}{5}\\y\cdot\left(xz\right)=-\dfrac{3}{5}\\z\cdot\left(xy\right)=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\cdot\dfrac{4}{5}=-\dfrac{3}{5}\\y\cdot\dfrac{3}{4}=-\dfrac{3}{5}\\z\cdot\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=-\dfrac{4}{5}\\z=-1\end{matrix}\right.\)
\(\text{+) Xét }xyz=\dfrac{3}{5}\Leftrightarrow\left\{{}\begin{matrix}x\cdot\left(yz\right)=\dfrac{3}{5}\\y\cdot\left(xz\right)=\dfrac{3}{5}\\z\cdot\left(xy\right)=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\cdot\dfrac{4}{5}=\dfrac{3}{5}\\y\cdot\dfrac{3}{4}=\dfrac{3}{5}\\z\cdot\dfrac{3}{5}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=\dfrac{4}{5}\\z=1\end{matrix}\right.\)
Vậy \(x;y;z=-\dfrac{3}{4};-\dfrac{4}{5};-1\) hoặc \(x;y;z=\dfrac{3}{4};\dfrac{3}{5};1\)
CD
1
24 tháng 7 2017
Theo bài ra: x.y=\(\frac{3}{5}\)(1)
y.z=\(\frac{4}{5}\)(2)
z.x=\(\frac{3}{4}\)(3)
Ta có: x.y.y.z.z.x=\(\frac{3}{5}.\frac{4}{5}.\frac{3}{4}\)\(\Leftrightarrow\)(x.y.z)\(^2\)=\(\frac{9}{25}\)\(\Rightarrow\)x.y.z=\(\frac{3}{5}\)
Từ (1), ta có:x.y=\(\frac{3}{5}\), mà x.y.z=\(\frac{3}{5}\)\(\Rightarrow\)z=1
Từ (2), ta có:y.z=\(\frac{4}{5}\), mà x.y.z=\(\frac{3}{5}\)\(\Rightarrow\)x=\(\frac{3}{4}\)
Ta có: x.y.z=\(\frac{3}{5}\), mà z=1;x=\(\frac{3}{4}\)\(\Rightarrow\)y=\(\frac{4}{5}\)
ko hiểu đề cho lắm
a)Ta có:
\(\left\{{}\begin{matrix}x+y=\frac{1}{3}\\y+z=\frac{-1}{4}\\z+x=\frac{1}{5}\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=\frac{1}{3}+\frac{-1}{4}+\frac{1}{5}\)
\(\Rightarrow2\left(x+y+z\right)=\frac{17}{60}\)
\(\Rightarrow x+y+z=\frac{17}{60}:2=\frac{17}{120}\)
\(\Rightarrow\left\{{}\begin{matrix}z=\frac{-23}{120}\\x=\frac{47}{120}\\y=\frac{-7}{120}\end{matrix}\right.\)
b)Ta có:
\(\left\{{}\begin{matrix}xy=\frac{3}{5}\\yz=\frac{4}{5}\\zx=\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow xyyzzx=\frac{3}{5}.\frac{4}{5}.\frac{3}{4}=\frac{9}{25}\)
\(\Rightarrow\left(xyz\right)^2=\frac{9}{25}\Rightarrow\left[{}\begin{matrix}xyz=\frac{3}{5}\\xyz=-\frac{3}{5}\end{matrix}\right.\)
TH1: \(xyz=\frac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}z=1\\x=\frac{3}{4}\\y=\frac{4}{5}\end{matrix}\right.\)
TH2:
\(xyz=-\frac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}z=-1\\x=-\frac{3}{4}\\y=-\frac{4}{5}\end{matrix}\right.\)