4.
\((2x+7)(x+3)^2(2x+5)=18\)
\(\Leftrightarrow [(2x+7)(2x+5)](x+3)^2=18\)
\(\Leftrightarrow (4x^2+24x+35)(x^2+6x+9)=18\)
\(\Leftrightarrow [4(x^2+6x+9)-1](x^2+6x+9)=18\)
\(\Leftrightarrow (4a-1)a=18\) (đặt \(x^2+6x+9=a\) )
\(\Leftrightarrow 4a^2-a-18=0\)
\(\Leftrightarrow (4a-9)(a+2)=0\Rightarrow \left[\begin{matrix} a=\frac{9}{4}\\ a=-2\end{matrix}\right.\)
Nếu \(a=x^2+6x+9=\frac{9}{4}\Leftrightarrow (x+3)^2=\frac{9}{4}\)
\(\Rightarrow \left[\begin{matrix} x+3=\frac{3}{2}\\ x+3=\frac{-3}{2}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{-9}{2}\end{matrix}\right.\)
Nếu \(a=x^2+6x+9=-2\Leftrightarrow (x+3)^2=-2< 0\) (vô lý)
Vậy ............
5.
PT \(\Leftrightarrow (x-1)(x-2)(2x-3)(2x-5)=30\)
\(\Leftrightarrow [(x-1)(2x-5)][(x-2)(2x-3)]=30\)
\(\Leftrightarrow (2x^2-7x+5)(2x^2-7x+6)=30\)
Đặt \(2x^2-7x+5=a\) thì:
PT \(\Leftrightarrow a(a+1)=30\)
\(\Leftrightarrow a^2+a-30=0\)
\(\Leftrightarrow (a-5)(a+6)=0\Rightarrow \left[\begin{matrix} a-5=0\\ a+6=0\end{matrix}\right.\)
Nếu \(a-5=0\Leftrightarrow 2x^2-7x=0\Leftrightarrow x(2x-7)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ x=\frac{7}{2}\end{matrix}\right.\)
Nếu \(a+6=0\Leftrightarrow 2x^2-7x+11=0\)
\(\Leftrightarrow 2(x-\frac{7}{4})^2+\frac{39}{8}=0\Leftrightarrow 2(x-\frac{7}{4})^2=-\frac{39}{8}<0\) (vô lý)
Vậy...........
6.
PT\(\Leftrightarrow (x+1)(x+2)(x+3)(x+4)=24\)
\(\Leftrightarrow [(x+1)(x+4)][(x+2)(x+3)]=24\)
\(\Leftrightarrow (x^2+5x+4)(x^2+5x+6)=24\)
\(\Leftrightarrow (a+4)(a+6)=24\) (đặt \(x^2+5x=a\) )
\(\Leftrightarrow a^2+10a=0\Leftrightarrow a(a+10)=0\Rightarrow \left[\begin{matrix} a=0\\ a+10=0\end{matrix}\right.\)
Nếu \(a=0\Leftrightarrow x^2+5x=0\Leftrightarrow x(x+5)=0\Leftrightarrow \left[\begin{matrix} x=0\\ x=-5\end{matrix}\right.\)
Nếu \(a+10=0\Leftrightarrow x^2+5x+10=0\Leftrightarrow (x+\frac{5}{2})^2+\frac{15}{4}=0\)
\(\Leftrightarrow (x+\frac{5}{2})^2=-\frac{15}{4}\) (vô lý)
Vậy...........
