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
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:
Giải thích các bước giải:
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 $x=7$