a: \(\left(2x+1\right)^2-\left(x+2\right)^2=0\)
=>(2x+1-x-2)(2x+1+x+2)=0
=>(x-1)(3x+3)=0
=>3(x+1)(x-1)=0
=>(x+1)(x-1)=0
=>\(\left[\begin{array}{l}x+1=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\\ x=1\end{array}\right.\)
b: \(x\left(x+5\right)\left(x+5\right)-\left(x+2\right)\left(x^2+2x+4\right)=58\)
=>\(x\left(x^2+10x+25\right)-\left(x^3+2x^2+4x+2x^2+4x+8\right)=58\)
=>\(x^3+10x^2+25x-x^3-4x^2-8x-8=58\)
=>\(6x^2+17x-66=0\)
=>\(x=\frac{-17\pm\sqrt{1873}}{12}\)
`(2x+1)^2-(x+2)^2=0`
`=> (2x+1-x+2)(2x+1+x+2)=0`
`=> [(2x -x) + (1+2)][(2x+x)(1+2)]=0`
`=> (x +3)(3x +3)=0`
\(\left[\begin{array}{l}x+3=0\\ 3x+3=0\end{array}\rArr\right.\left[\begin{array}{l}x=-3\\ 3x=-3\end{array}\rArr\left[\begin{array}{l}x=-3\\ x=-1\end{array}\right.\right.\)
Vậy `x \in{-3;-1}`
