Đáp án + Giải thích các bước giải:
$1.(2x+1)^2-4(x+1)=3$
$⇔(2x)^2+2.2x.1+1^2-(4x+4)=3$
$⇔4x^2+4x+1-4x-4=3$
$⇔4x^2=6$
$⇔x^2=\dfrac{3}{2}$
$⇔x=±\dfrac{\sqrt6}{2}$
Vậy `x∈{\frac{\sqrt6}{2};-\frac{\sqrt6}{2}}`
$2. 3(x-1)(x+1)-2(x+1)^2-(x-1)^2=0$
$⇔3(x^2-1)-2(x^2+2x+1)-(x^2-2x+1)=0$
$⇔3x^2-3-2x^2-4x-2-x^2+2x-1=0$
$⇔-2x=6$
$⇔x=-3$
Vậy `x∈{-3}`
$3. (3x-1)^2+(2+x)^2-10(x-2)(x+2)=0$
$⇔(9x^2-6x+1)+(x^2+4x+4)-10(x^2-4)=0$
$⇔9x^2-6x+1+x^2+4x+4-10x^2+40=0$
$⇔-2x=-45$
$⇔x=\dfrac{45}{2}$
Vậy `x∈{\frac{45}{2}}`