\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{2.4}=\frac{y}{3.4}\Rightarrow\frac{x}{8}=\frac{y}{12}\left(1\right)\)

\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{4.3}=\frac{z}{5.3}\Rightarrow\frac{y}{12}=\frac{z}{15}\left(2\right)\)

Từ 1 và 2 

=> \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau 

\(\Rightarrow\frac{x}{8}=\frac{z}{15}\Rightarrow\frac{x}{8}=\frac{-z}{-15}=\frac{x+\left(-z\right)}{8+\left(-15\right)}=\frac{10}{-7}\)

Do đó

\(\frac{x}{8}=\frac{10}{-7}\Rightarrow x=\frac{80}{-7}\)

\(\frac{y}{12}=\frac{10}{-7}\Rightarrow y=\frac{120}{-7}\)

\(\frac{z}{15}=\frac{10}{-7}\Rightarrow z=\frac{150}{-7}\)

\(Ta có: x-y=0 => x=y \)

\(Khi đó, ta có: \)

\(a) M=7x-7y+4ax-4ay-5 =(7x-7x) +(4ax-4ax)-5 \)

\(M=0+0-5=-5 \)

\(Vậy M=-5\)

\(b) N=x(x^2+y^2)-y(x^2+y^2)+3\)

\(=> N=[x(x^2+x^2)-x(x^2+x^2)]+3\)

\(=> N=0+3=3\)

\(Vậy N=3\)

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{2.4}=\frac{y}{3.4}\Rightarrow\frac{x}{8}=\frac{y}{12}\left(1\right)\)

\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{4.3}=\frac{z}{5.3}\Rightarrow\frac{y}{12}=\frac{z}{15}\left(2\right)\)

Từ (1) và (2) 

\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

Do đó

\(\frac{x}{8}=2\Rightarrow x=16\)

\(\frac{y}{12}=2\Rightarrow y=24\)

\(\frac{z}{15}=2\Rightarrow z=30\)