Question

tìm x,y,z

a,\(x\left(x+y+z\right)=7,y\left(x+y+z\right)=3,z\left(x+y+z\right)\)

b,\(xy=\dfrac{6}{7},yz=\dfrac{7}{12},xz=2\)

giúp mình nhé mình

Answer 1

\(\left\{{}\begin{matrix}x\left(x+y+z\right)=7\\y\left(x+y+z\right)=3\\z\left(x+y+z\right)=15\end{matrix}\right.\) (đoán là đề vậy thôi vì bạn viết thiếu)

\(\Rightarrow x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=7+3+15\)\(\Rightarrow\left(x+y+z\right)\left(x+y+z\right)=25\)

\(\Rightarrow\left(x+y+z\right)^2=25\)

\(\Rightarrow x+y+z=\pm5\)

....

\(\left\{{}\begin{matrix}xy=\dfrac{6}{7}\\yz=\dfrac{7}{12}\\xz=2\end{matrix}\right.\)

\(\Rightarrow xy.yz.xz=\dfrac{6}{7}.\dfrac{7}{12}.2\)

\(\Rightarrow xyz^2=1\)

\(\Rightarrow xyz=\pm1\)

...

Answer 2

tìm x,y,z

a,x(x+y+z)=7,y(x+y+z)=3,z(x+y+z)x(x+y+z)=7,y(x+y+z)=3,z(x+y+z)=6

b,xy=67,yz=712,xz=2