Đáp án+Giải thích các bước giải:
$\begin{array}{l}1,(x-1).x.(x+1)(x+2)=24\\↔[x(x+1)][(x-1)(x+2)]=24\\↔(x^2+x)(x^2+x-2)=24\\↔(x^2+x-1)^2=25\\↔(x^2+x-1-5)(x^2+x-1+5)=0\\↔(x^2+x-6)(x^2+x+4)=0\\↔x^2+x-6=0(do \,\, x^2+x+4>0)\\↔x^2-2x+3x-6=0\\↔x(x-2)+3(x-2)=0\\↔(x-2)(x+3)=0\\↔\left[ \begin{array}{l}x-2=0\\x+3=0\end{array} \right.\\↔\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\\vậy \,\, x=2 \,\, hoặc \,\, x=-3\\2,4x^3+7x^2+7x+4=0\\↔(4x^3+4)+(7x^2+7x)=0\\↔4(x^3+1)+7x(x+1)=0\\↔4(x+1)(x^2-x+1)+7x(x+1)=0\\↔(x+1)(4x^2-4x+4+7x)=0\\↔(x+1)(4x^2+3x+4)=0\\↔4(x+1)(x^2+\dfrac{3}{4}+1)=0\\↔x+1=0(do \,\, x^2+\dfrac{3}{4}x+1>0)\\↔x=-1\\vậy \,\, x=-1\\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$
