\(\left(2x+3\right)\left(x+2\right)^2\left(2x+5\right)=3\)
\(=4x^4+32x^3+95x^2+124x+57=0\)
\(=\left(4x^4+4x^3\right)+\left(28x^3+28x^2\right)+\left(67x^2+67x\right)+\left(57x+57\right)=0\)
\(=\left(x+1\right)\left(4x^3+28x^2+67x+57\right)=0\)
\(=\left(x+1\right)\left(\left(4x^3+12x^2\right)+\left(16x^2+48x\right)+\left(19x+57\right)\right)=0\)
\(=\left(x+1\right)\left(x+3\right)\left(4x^2+16x+19\right)=0\)
Vì \(4x^2+16x+19=\left(2x+4\right)^2+3>0\) nên ta có
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Có:(2x+3)(x+2)2(2x+5)=3
<=>4x4+32x3+95x2+124x+60=3
<=>4x4+4x3+28x3+28x2+67x2+67x+57x+57=0
<=>(x+1)(4x3+28x2+67x+57)=0
<=>x=-1 hoac x=-3
\(\left(2x+3\right)\left(x+2\right)^2\left(2x+5\right)=3\)
\(=4x64+32x^3+95x^2+124x+57=0\)
\(=\left(4x^4+4x^3\right)+\left(28x^3+28x^2\right)+\left(67x^2+67x\right)+\left(57x+57\right)=0\)
\(=\left(x+1\right)\left(4x^3-28x^2+67x+57\right)=0\)
\(=\left(x+1\right)\left[\left(4x^3+12x^2\right)+\left(16x^2+48x\right)+\left(19x+57\right)\right]=0\)
\(=\left(x+1\right)\left(x+3\right)\left(4x^2+16x+19\right)=0\)
Vì \(4x^2+16x+19=\left(2x+4\right)^2>0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy x=-1; x=-3