a: Ta có: \(\left(x+2\right)^2-x\left(x+1\right)=7\)
\(\Leftrightarrow x^2+4x+4-x^2-x=7\)
hay x=1
b: Ta có: \(\left(5x+3\right)^2-\left(x-8\right)^2=0\)
\(\Leftrightarrow\left(5x+3-x+8\right)\left(5x+3+x-8\right)=0\)
\(\Leftrightarrow\left(4x+11\right)\left(6x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{4}\\x=\dfrac{5}{6}\end{matrix}\right.\)


