Đáp án:
Giải thích các bước giải:
B1
Cho x=1, ta có: 3(2.1+m)(1+2)-2(2.1+1)=18
⇔ 3(2+4+m+2m)-4-2=18
⇔ 3(6+3m)-6=18
⇔ 18+9m-6=18
⇔ 9m=18-18+6
⇔ 9m=6
⇔ m=$\frac{2}{3}$
Vậy m=$\frac{2}{3}$ thì nghiệm x=1
B2
a) (x-2)²-4(x+3)=x(x-4)
⇔(x-2)²-4(x+3)-x(x-4)=0
⇔x²-4x+4-4x-12-x²+4x=0
⇔-4x-8=0
⇔-4x=8
⇔x=-2 Vậy S={-2}
b) $\frac{3}{x+1}$ +$\frac{x-1}{x-2}$ = $\frac{x}{x-2}$
⇔$\frac{3(x-2)}{MTC}$ + $\frac{(x-1)(x+1)}{MTC}$ = $\frac{x(x+1)}{MTC}$
⇒3x-6+x²-1=x²+x
⇔3x-6+x²-1-x²-x=0
⇔3x-7=0
⇔x=$\frac{7}{3}$ Vậy S={$\frac{7}{3}$ }
c) $\frac{x}{2x-6}$ + $\frac{x}{2x+2}$ = $\frac{2x^2}{x^2-2x-3}$
⇔$\frac{x}{2(x-3)}$ + $\frac{x}{2(x+1)}$ = $\frac{2x^2}{(x-3)(x+1)}$
⇔$\frac{2x(x+1)}{MTC}$ + $\frac{2x(x-3)}{MTC}$ = $\frac{4x^2}{MTC}$
⇒ 2x²+2x+2x²-6x=4x²
⇔2x²+2x+2x²-6x-4x²=0
⇔-4x=0
⇔x=4 Vậy S={4}
