Đáp án:
Giải thích các bước giải:
$a,(3x-1)(2x+7)-(x+1)(6x-5)=16$
$⇔6x^2+21x-2x-7-6x^2+5x-6x+5=16$
$⇔18x=16+7-5$
$⇔18x=18$
$⇔x=1$
$b,(2x+3)^2-2(2x+3)(2x-5)+(2x-5)^2=x^2+6x+64$
$⇔(2x+3-2x+5)^2=x^2+6x+64$
$⇔8^2=x^2+6x+64$
$⇔x^2+6x+64-64=0$
$⇔x^2+6x=0$
$⇔x(x+6)=0$
⇔\(\left[ \begin{array}{l}x=0\\x=-6\end{array} \right.\)
$c,(x^4+2x^3+10x-25):(x^2+5)=3$
$⇔x^2+2x-5=3$
$⇔x^2+2x-5-3=0$
$⇔x^2+2x-8=0$
$⇔x^2+4x-2x-8=0$
$⇔(x-2)(x+4)=0$
⇔\(\left[ \begin{array}{l}x=2\\x=-4\end{array} \right.\)
