(×+1)(×+2)(×+4)(×+5)=40

Đáp án:

Giải thích các bước giải:

    (x+1)(x+2)(x+4)(x+5)=40

⇔(x² + x + 5x + 5)(x² + 2x + 4x + 8)=40

⇔(x² + 6x + 5)(x² + 6x + 8)=40 (1)

Đặt x²+6x+5=a

⇒(1)⇔a(a+3)=40

⇔a²+3a-40=0

⇔a²+2×a×$\frac{3}{2}$ +$\frac{9}{4}$ -$\frac{169}{4}$ =0

⇔(a+$\frac{3}{2}$)$^{2}$ =$\frac{169}{4}$ 

⇔\(\left[ \begin{array}{l}a+\frac{3}{2}=\frac{13}{2}\\a+\frac{3}{2}=\frac{-13}{2}\end{array} \right.\)

⇔\(\left[ \begin{array}{l}a=5\\a=-8\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x²+6x+5=5\\x²+6x+5=-8\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x²+6x=0\\x²+6x+9=-4\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x(x+6)=0\\(x+3)²=-4(loại vì(x+3)²\geq0) \end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=0\\x=-6\end{array} \right.\) 

⇒S={0;-6}