Toán a, x^2-3x+2=0 b, -x^2+5x-6=0 c, 4x^2-12x+5=0 d, 2x^2+5x+3=0 e, x-12+4x=25+2x-1 f, x+2x+3x-19= 3x+5 g, 7-(2x+4)= -(x+4) Giúo mk vs 22/07/2021 By Allison a, x^2-3x+2=0 b, -x^2+5x-6=0 c, 4x^2-12x+5=0 d, 2x^2+5x+3=0 e, x-12+4x=25+2x-1 f, x+2x+3x-19= 3x+5 g, 7-(2x+4)= -(x+4) Giúo mk vs
Đáp án: Bên dưới Giải thích các bước giải: a) $x^2-3x+2=0$ ⇔ $x^2-x-2x+2=0$ ⇔ $x(x-1)-2(x-1)=0$ ⇔ $(x-1)(x-2)=0$ ⇔ \(\left[ \begin{array}{l}x-1=0\\x-2=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=1\\x=2\end{array} \right.\) Vậy $x=1$ hoặc $x=2$ b) $-x^2+5x-6=0$ ⇔ $-x^2+6x-x-6=0$ ⇔ $x(-x+6)-(x+6)=0$ ⇔ $(-x+6)(x-1)=0$ ⇔ \(\left[ \begin{array}{l}-x+6=0\\x-1=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}-x=-6\\x=1\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=6\\x=1\end{array} \right.\) Vậy $x=6$ hoặc $x=1$ c) $4x^2-12x+5=0$ ⇔ $4x^2-2x-10x+5=0$ ⇔ $2x(2x-1)-5(2x-1)=0$ ⇔ $(2x-1)(2x-5)=0$ ⇔ \(\left[ \begin{array}{l}2x-1=0\\2x-5=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=\frac{1}{2}\\x=\frac{5}{2}\end{array} \right.\) Vậy $x=\frac{1}{2}$ hoặc $x=\frac{5}{2}$ d) $2x^2+5x+3=0$ ⇔ $2x^2+2x+3x+3=0$ ⇔ $2x(x+1)+3(x+1)=0$ ⇔ $(x+1)(2x+3)=0$ ⇔ \(\left[ \begin{array}{l}x+1=0\\2x+3=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=-1\\x=\frac{-3}{2}\end{array} \right.\) Vậy $x=-1$ hoặc $x=\frac{-3}{2}$ e) $x-12+4x=25+2x-1$ ⇔ $x+4x-2x=25-1+12$ ⇔ $3x=36$ ⇔ $x=12$ Vậy $x=12$ f) $x+2x+3x-19=3x+5$ ⇔ $x+2x+3x-3x=5+19$ ⇔ $3x=24$ ⇔ $x=8$ Vậy $x=8$ g) $7-(2x+4)=-(x+4)$ ⇔ $7-2x-4=-x-4$ ⇔ $3-2x=-x-4$ ⇔ $-2x+x=-4-3$ ⇔ $-x=-7$ ⇔ $x=7$ Vậy $x=7$ Trả lời