Đáp án:
Giải thích các bước giải:
$\left(2x-3\right)\left(x+2\right)-\left(4x-2\right)\left(x-5\right)=-16$
$⇔2x^2+x-6-\left(4x-2\right)\left(x-5\right)=-16$
$⇔2x^2+x-6-4x^2+22x-10=-16$
$⇔-2x^2+23x-16+16=0$
$⇔-2x^2+23x=0$
$⇔x(-2x+23)=0$
$⇔x=0,\:x=\dfrac{23}{2}$
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$\left(3x-1\right)\left(x+2\right)-\left(2-3x\right)\left(x+3\right)=12$
$⇔3x^2+5x-2-\left(2-3x\right)\left(x+3\right)=12$
$⇔3x^2+5x-2+3x^2+7x-6=12$
$6x^2+12x-8-12=0$
$⇔6x^2+12x-20=0$
$⇔3x^2+6x-10=0$
Kiểm tra câu nì xem -.-
