Đáp án:
Giải thích các bước giải:
1) x^2-2x-8=0
⇒(x^2-2x+1)-9=0
⇔(x-1)^2-3^2=0
⇔(x-1-3)(x-1+3)=0
⇔(x-4)(x+2)=0
⇔\(\left[ \begin{array}{l}x-4=0\\x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=4\\x=-2\end{array} \right.\)
2) x^2+6x+5=0
⇒(x^2+x)+(5x+5)=0
⇔x(x+1)+5(x+1)=0
⇔(x+1)(x+5)=0
⇔\(\left[ \begin{array}{l}x+1=0\\x+5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x=-5\end{array} \right.\)
3) x^2+4x-21=0
⇒(x^2-3x)+(7x-21)=0
⇔x(x-3)+7(x-3)=0
⇔(x-3)(x+7)
⇔\(\left[ \begin{array}{l}x-3=0\\x+7=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=-7\end{array} \right.\)
4) x^2-14x+13=0
⇒(x^2-x)-(13x-13)=0
⇔x(x-1)-13(x-1)=0
⇔(x-1)(x-13)=0
⇔\(\left[ \begin{array}{l}x-1=0\\x-13=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=13\end{array} \right.\)
5) x^2-2x-80=0
⇒(x^2-10x)+(8x-80)=0
⇔x(x-10)+8(x-10)=0
⇔(x-10)(x+8)=0
⇔\(\left[ \begin{array}{l}x-10=0\\x+8=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=10\\x=-8\end{array} \right.\)
6) x^2+10x+16=0
⇒(x^2+8x)+(2x+16)=0
⇔x(x+8)+2(x+8)=0
⇔(x+8)(x+2)=0
⇔\(\left[ \begin{array}{l}x+8=0\\x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-8\\x=-2\end{array} \right.\)
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