Ta có:

x-y+x+z+y-x=8+10+12=30

=>2x            = 30

=>x             = 15 

=> y = 15-8=7

     z = 15-12=3

Vậy x=15;y=7;z=3

Đăt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\Rightarrow x=2k,y=3k,z=4k\)

\(\Rightarrow M=\frac{y+x-z}{x-y+z}=\frac{3k+2k-4k}{2k-3k+4k}=\frac{k}{3k}=\frac{1}{3}\)

Thank you!!!!

bạn St sai rồi phải là bằng \(\dfrac{5k}{3k}\)=\(\dfrac{5}{3}\)

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

Ta có

\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=1\)

Đạt được khi a = b = c = \(\frac{2}{3}\)