Solve the following equation and verify your answer:

$\left(\frac{x+1}{x+2}\right)^{2}=\frac{x+2}{x+4}$

or $\frac{x^{2}+2 x+1}{x^{2}+4 x+4}=\frac{x+2}{x+4}$

or $\mathrm{x}^{3}+2 \mathrm{x}^{2}+\mathrm{x}+4 \mathrm{x}^{2}+8 \mathrm{x}+4=\mathrm{x}^{3}+4 \mathrm{x}^{2}+4 \mathrm{x}+2 \mathrm{x}^{2}+8 \mathrm{x}+8[$ After c ross multiplication $]$

or $\mathrm{x}^{3}-\mathrm{x}^{3}+6 \mathrm{x}^{2}-6 \mathrm{x}^{2}+9 \mathrm{x}-12 \mathrm{x}=8-4$

or $-3 \mathrm{x}=4$

or $\mathrm{x}=\frac{4}{-3}=\frac{-4}{3}$

Thus, $\mathrm{x}=\frac{-4}{3}$ is the solution of the given equation.

Check:

Substituting $\mathrm{x}=\frac{-4}{3}$ in the given equation, we get:

L.H.S. $=\left(\frac{\frac{-4}{3}+1}{\frac{4}{3}+2}\right)^{2}=\left(\frac{-4+3}{-4+6}\right)^{2}=\frac{1}{4}$

R.H.S. $=\frac{\frac{-4}{3}+2}{\frac{4}{3}+4}=\frac{-4+6}{-4+12}=\frac{2}{8}=\frac{1}{4}$

$\therefore$ L.H.S. $=$ R.H.S. for $x=\frac{-4}{3}$