x(y+z)=32 , y(x+z)=27 , z(y+x)=35 hỏi xyz=?
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x\left(y+z\right)=32;y\left(x+z\right)=27;z\left(x+y\right)=35\\ \Rightarrow\left(xy+xz\right)+\left(xy+yz\right)+\left(xz+yz\right)=32+27+35\\ \Rightarrow2\left(xy+yz+zx\right)=94\\ \Rightarrow xy+yz+xz=47\\ \Rightarrow yz=15;xz=20;xy=12\\ \Rightarrow\left(x.y.z\right)^2=3600\)
Ta có : x;y;z khác 0 nên x.y.z khác 0
=> x.y.z=60
\(xy=\dfrac{1}{2};yz=\dfrac{3}{5};xz=\dfrac{27}{10}\)
\(\Rightarrow xy.yz.xz=\dfrac{1}{2}.\dfrac{3}{5}.\dfrac{27}{10}\)
\(\Rightarrow\left(xyz\right)^2=\dfrac{81}{100}\)
\(\Rightarrow\left\{{}\begin{matrix}xyz=\dfrac{9}{10}\\xyz=\dfrac{-9}{10}\\xyz=\dfrac{9}{-10}\\xyz=\dfrac{-9}{-10}\end{matrix}\right.\)
a: Sửa đề: xy=35; yz=5
\(\left(xy\cdot yz\cdot xz\right)=\left(35\cdot7\cdot5\right)^2=35^2=35\)
=>z=1; x=5; y=7
b: \(x+y+z=\dfrac{-1+6+1}{2}=3\)
=>z=3+1=4; y=3-6=-3; x=3-1=2
Ta có:
\(x\left(y+z\right)=32\Rightarrow xy+xz=32\)
\(y\left(x+z\right)=27\Rightarrow xy+yz=27\)
\(z\left(y+x\right)=yz+xz=35\)
\(\Rightarrow xy+xz+xy+yz+yz+xz=32+27+35\)
\(\Rightarrow2\left(xy+yz+zx\right)=94\)
\(\Rightarrow xy+yz+zx=47\)
Mà \(xy+yz=27\)
\(\Rightarrow27+zx=47\)
\(\Rightarrow zx=20\)
Tương tự ta được : \(xy=12\) ; \(yz=15\)
\(\Rightarrow zx.xy=20.12\)
\(\Rightarrow x^2.yz=240\)
Mà \(yz=15\)
\(\Rightarrow x^2=240:15\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=\pm4\)
+)Nếu \(x=4\Rightarrow xyz=4.yz=4.15=60\)
+)Nếu \(x=-4\Rightarrow xyz=-4.yz=-4.15=-60\)
Vậy \(xyz=60;xyz=-60\)
xyz = 60