Đáp án:
Giải thích các bước giải:
$1, (x+2)^2-(x-4)(x+5)=13$
$⇔(x^2+4x+4)-(x^2+5x-4x-20)=13$
$⇔x^2+4x+4-x^2-x+20=13$
$⇔3x=13-20-4$
$⇔3x=-11$
$⇔x=\dfrac{-11}{3}$
$2, (2x-1)^2-(1-3x)^2=0$
$⇔(2x-1-1+3x)(2x-1+1-3x)=0$
$⇔(5x-2).x=0$
$⇔\left[ \begin{array}{1}x=\dfrac{2}{5}\\x=0\end{array} \right.$
$3, 2020x^2-2019x-1=0$
$⇔2020x^2-2020x+x-1=0$
$⇔2020x(x-1)+(x-1)=0$
$⇔(2020x+1)(x-1)=0$
$⇔\left[ \begin{array}{1}x=-\dfrac{1}{2020}\\x=1\end{array} \right.$
$4,4x^2+12x-7=0$
$⇔4x^2-2x+14x-7=0$
$⇔2x(2x-1)+7(2x-1)=0$
$⇔(2x+7)(2x-1)=0$
$⇔\left[ \begin{array}{1}x=\dfrac{-7}{2}\\x=dfrac{1}{2}\end{array} \right.$
$5, 5x^2+17x=22$
$⇔5x^2-5x+22x-22=0$
$⇔5x(x-1)+22(x-1)=0$
$⇔(5x+22)(x-1)=0$
$⇔\left[ \begin{array}{1}x=\dfrac{-22}{5}\\x=1\end{array} \right.$
