Ta có:\(\frac{2x+2}{3x-6}=\frac{2x-6}{3x-15}\)

\(\Rightarrow\frac{2\left(x+1\right)}{3\left(x-2\right)}=\frac{2\left(x-3\right)}{3\left(x-5\right)}\)

\(\Rightarrow\frac{2}{3}\cdot\frac{x+1}{x-2}=\frac{2}{3}\cdot\frac{x-3}{x-5}\)

\(\Rightarrow\frac{x+1}{x-2}=\frac{x-3}{x-5}\)

\(\Rightarrow\frac{x+1}{x-2}-1=\frac{x-3}{x-5}-1\)

\(\Rightarrow\frac{x+1-x+2}{x-2}=\frac{x-3-x+5}{x-5}\)

\(\Rightarrow\frac{3}{x-2}=\frac{2}{x-5}\)

\(\Rightarrow3\left(x-5\right)=2\left(x-2\right)\)

\(\Rightarrow3x-15=2x-4\)

\(\Rightarrow3x-2x=-4+15\)

\(\Rightarrow x=11\)

x=11

Ta có: \(\frac{3x}{4}\)= \(\frac{y}{2}\)= \(\frac{3z}{5}\)

=> \(\frac{1}{3}.\frac{3x}{4}=\frac{1}{3}.\frac{y}{2}=\frac{1}{3}.\frac{3z}{5}\)

\(\Rightarrow\frac{3x}{12}=\frac{y}{6}=\frac{3z}{15}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)

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

\(\frac{x}{4}=\frac{y}{6}=\frac{z}{5}=\frac{y-z}{6-5}=15\)

Suy ra:  

\(\Rightarrow\)x + y + z = 225

a) \(2x^2-2x-x^2+6=0\) 

\(\Leftrightarrow x^2-2x+1+5=0\)

\(\Leftrightarrow\left(x-1\right)^2=-5\) ( vô lý)

Vậy không có x thoả mãn \(2x.\left(x-1\right)-x^2+6=0\)

b) \(x^4-2x^2.\left(3+2x^2\right)+3x^2.\left(x^2+1\right)=-3\) 

\(\Leftrightarrow x^4-6x^2-4x^4+3x^4+3x^2+3=0\)

\(\Leftrightarrow3-3x^2=0\)

\(\Leftrightarrow3x^2=3\Leftrightarrow x^2=1\) \(\Leftrightarrow x\in\left\{-1;1\right\}\)

Vậy \(x\in\left\{-1;1\right\}\)

c) \(\left(x+1\right).\left(x^2-x+1\right)-2x=x.\left(x-2\right).\left(x+2\right)\)

\(\Leftrightarrow x^3+1-2x-x.\left(x^2-4\right)=0\)

\(\Leftrightarrow x^3+1-2x-x^3+4x=0\)

\(\Leftrightarrow1+2x=0\Leftrightarrow x=\dfrac{-1}{2}\)

Vậy x=\(\dfrac{-1}{2}\)

d) \(\left(x+3\right).\left(x^2-3x+9\right)-x.\left(x-2\right).\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x.\left(x^2-4\right)-15=0\)

\(\Leftrightarrow x^3-27-x^3+4x-15=0\)

\(\Leftrightarrow4x-42=0\)

\(\Leftrightarrow x=10,5\)

Vậy x=10,5