Đáp án + Giải thích các bước giải:
Bài 1:
$a)(y+1/2)^2=y^2+2.y.1/2+(1/2)^2=y^2+y+1/4$
$b)(4x-1)^2=(4x)^2-2.4x.1+1^2=16x^2-8x+1$
$c)(x/3-y)(y+2x/6)=(x-3y/3)(6y+2x/6)=(x-3y/3)[2(x+3y)]/6$
$=(x-3y/3)(x+3y)/3$=x^2-9y^2/9$
$d)(3a+2)^3=(3a)^3+3.(3a^2).2+3.3a.4+2^3=27a^3+54a^2+36a+8$
$e)(2-a)^3=2^3-3.2^2.a+3.2.a^2-a^3=8-12a+6a^2-a^3$
$f)(x-2)(x^2+2x+4)=x^3-8$
$g)(3+2x)(4x^2-6x+9)=(2x+3)[(2x)^2-2x.3+3^2]=8x^3+27$
Bài 2:
$a)x^2(x-3)=1-3x$
$⇔x^3-3x^2=1-3x$
$⇔x^3-3x^2+3x-1=0$
$⇔(x-1)^3=0$
$⇔ x-1=0$
$⇔x=1$
VẬy S={1}
$b)(x-1)(x^2+x+1)=x^3-x^2$
$⇔x^3-1-x^3+x^2=0$
$⇔x^2-1=0$
$⇔(x-1)(x+1)=0$
$⇔\left[ \begin{array}{l}x+1=0\\x-1=0\end{array} \right.$
$⇔\left[ \begin{array}{l}x=-1\\x=1\end{array} \right.$
Vậy S={1;-1}
Mk xin hay nhất ạ, mk rất cần ạ
