1) $[(x-1)(x+2)][x(x+1)]=(x^2+x-2)(x^2+x)=24$
Đặt: $x^2+x=t$
pttt: t(t-2)=24⇔\(\left[ \begin{array}{l}t=6\\t=-4\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x^2+x=6\\x^2+x=-4\end{array} \right.\)
⇔x={2;-3}
2) $4x^3+7x^2+7x+4 $= $4(x+1)(x^2-x+1)+7x(x+1)$=$4(x+1)[(x+1)^2-3x]+7x(x+1)=0$
Đặt x+1=t
pttt: $4t(t^2-3x)+7xt=0⇔4t^3-5xt=0$
⇔\(\left[ \begin{array}{l}t=0\\4t^2=5x\end{array} \right.\)
⇔\(\left[ \begin{array}{l}4(x+1)^2=5x\\x=-1\end{array} \right.\)
⇔x=-1
3) $x^4-3x^3+4x^2-3x+1$=$x^4-3x^3+4x^2-3x+1=(x^2+1)^2-3x(x^2+1)+2x^2$
Đặt $x^2+1=t$
pttt: $t^2-3xt+2x^2=0$⇔\(\left[ \begin{array}{l}t=2x\\t=x\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x^2+1=2x\\x^2+1=x\end{array} \right.\) ⇔x=1
4) $2x^5-7x^4+5x^3+5x^2-7x+2=(2x^2-7x+5)(x^3+1)+3(x^2-1)=(x^2-1)(2x-5)(x^2-x+1)+3(x^2-1)=0$
TH1: $x^2-1=0$⇔x=±1
TH2: $(2x-5)(x^2-x+1)+3=0$
⇔$[2(x-1/2)-4][(x-1/2)^2+3/4]+3=0$
đặt x-1/2=t
pttt: $(2t-4)(t^2+3/4)+3=0$⇔\(\left[ \begin{array}{l}t=3/2\\t=1/2\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\)
vậy x={±1;2}
