Đáp án:
Giải thích các bước giải:
$(x+5)^2-(x-2)(x+3)=1$
$⇔x^2+10x+25-(x^2+3x-2x-6)=1$
$⇔x^2+10x+25-x^2-3x+2x+6=1$
$⇔(x^2-x^2)+(10x-3x+2x)=1-6-25$
$⇔9x=-30$
$⇔x=\dfrac{-10}{3}$
$3(2x-1)(3x-1)-(2x-3)(9x-1)-3=-3$
$⇔(6x-3)(3x-1)-(18x^2-2x-27x+3)=0$
$⇔18x^2-6x-9x+3-18x^2+2x+27x-3=0$
$⇔14x=0$
$⇔x=0$
$5(2x-1)-4(8-3x)=-5$
$⇔10x-5-32+12x=-5$
$⇔22x=-5+5+32$
$⇔22x=32$
$⇔x=\dfrac{16}{11}$