Đáp án: $a.x\in\{3,4\}$
$b. x=\dfrac{-23\pm\sqrt{553}}{2}$
$c.x=-\dfrac13$
Giải thích các bước giải:
Bài 1:
a.$x(x-4)-3x+12=0$
$\to x^2-4x-3x+12=0$
$\to x^2-7x+12=0$
$\to x^2-4x-3x+12=0$
$\to x(x-4)-3(x-4)=0$
$\to (x-3)(x-4)=0$
$\to x\in\{3,4\}$
b.Ta có :
$\dfrac{2x-1}{x}+2=\dfrac{1-x}{6}$
$\to 6\left(2x-1\right)+12x=x\left(1-x\right)$
$\to 24x-6=x-x^2$
$\to x^2+23x-6=0$
$\to \left(x+\dfrac{23}{2}\right)^2-\dfrac{553}{4}=0$
$\to \left(x+\dfrac{23}{2}\right)^2=\dfrac{553}{4}$
$\to x=\dfrac{-23\pm\sqrt{553}}{2}$
c.Ta có :
$\dfrac{2x-1}{x}+\dfrac{x+3}{x-1}=3$
$\to 2-\dfrac{1}{x}+\dfrac{x-1+4}{x-1}=3$
$\to 2-\dfrac{1}{x}+1+\dfrac{4}{x-1}=3$
$\to 3-\dfrac{1}{x}+\dfrac{4}{x-1}=3$
$\to \dfrac{1}{x}=\dfrac{4}{x-1}$
$\to 4x=x-1$
$\to 3x=-1$
$\to x=-\dfrac13$
