\(y\left(y+3\right)^2-\left(y+2\right)\left(y^2-2y+4\right)-6\left(y+5\right)\left(y-5\right)=97\)

\(y\left(y^2+6y+9\right)-\left(y^3+8\right)-6\left(y^2-25\right)=97\)

\(y^3+6y^2+9y-y^3-8-6y^2+150=97\)

\(9y=97+8-150\)

\(9y=-45\Rightarrow y=-5\)

a: \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3-1-3\left(4x^2-4x+1\right)=15\)

=>\(12x^2+6x-12x^2+12x-3=15\)

=>18x=18

=>x=1

b: \(\Leftrightarrow y^3+6y^2+9y-y^3-8-6y^2+150=97\)

=>9y+142=97

=>9y=-45

=>y=-5

1.
\(\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}\right)-x:\frac{3}{2}=\frac{7}{3}\\ \left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right):\frac{3}{2}-x:\frac{3}{2}=\frac{7}{3}\\\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x\right]:\frac{3}{2}=\frac{7}{3}\\ \left(1-\frac{1}{99}\right)-x=\frac{7}{3}\times\frac{3}{2}\\ \frac{98}{99}-x=\frac{7}{2}\\ x=\frac{98}{99}-\frac{7}{2}=\frac{-497}{198}\)

2.\(\frac{x}{y}=\frac{4}{3}\Rightarrow\hept{\begin{cases}x=4a\\y=3a\\x-y=4a-3a=a\end{cases}}\\ \left(x-y\right)^{2015}=5^{2015}\Rightarrow x-y=5\\ \Rightarrow a=5\Rightarrow\hept{\begin{cases}x=4\times5=20\\y=3\times5=15\end{cases}}\)

1.