A)x²-3x+2=0 B)-x²+5x-6=0 C)4x²-12x+5=0 D)2x²+5x+3=0

Đáp án:

$a,$ $\text{ S = {2 ; 1}}$

$b,$ $\text{S = {3 ; 2}}$

$c,$ $\text{S = }$ $\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}$

$d,$ $\text{S = }$ $\left\{-\dfrac{3}{2};-1\right\}$

Giải thích các bước giải:

$a,$ $x^2-3x+2=0$

$⇔\left(x−2\right)\left(x−1\right)=0$

$⇔\left[ \begin{array}{l}x-2=0\\x-1=0\end{array} \right.$ 

$⇔\left[ \begin{array}{l}x=2\\x=1\end{array} \right.$ 

$\text{Vậy S = {2 ; 1}}$

$b,$ $-x^2+5x-6=0$

$⇔x^2-5x+6=0$

$⇔\left(x−3\right)\left(x−2\right)=0$

$⇔\left[ \begin{array}{l}x-3=0\\x-2=0\end{array} \right.$ 

$⇔\left[ \begin{array}{l}x=3\\x=2\end{array} \right.$ 

$\text{Vậy S = {3 ; 2}}$

$c,$ $4x^2-12x+5=0$

$⇔4x^2-2x-10x+5=0$

$⇔2x\left(2x-1\right)-5\left(2x-1\right)=0$

$⇔\left(2x-1\right)\left(2x-5\right)=0$

$⇔\left[ \begin{array}{l}2x-1=0\\2x-5=0\end{array} \right.$ 

$⇔\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{array} \right.$ 

$\text{Vậy S = }$ $\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}$

$d,$ $2x^2+5x+3=0$

$⇔2x^2+3x+2x+3=0$

$⇔x\left(2x+3\right)+\left(2x+3\right)=0$

$⇔\left(2x+3\right)\left(x+1\right)=0$

$⇔\left[ \begin{array}{l}2x+3=0\\x+1=0\end{array} \right.$ 

$⇔\left[ \begin{array}{l}x=-\dfrac{3}{2}\\x=-1\end{array} \right.$ 

$\text{Vậy S = }$ $\left\{-\dfrac{3}{2};-1\right\}$

Đáp án:

A) $S=\text{{1;2}}$

B) $S=\text{{2;3}}$

C) $S=$ {$\dfrac12 ; \dfrac52$}

D) $S=$ {$-\dfrac32;-1$}

Giải thích các bước giải:

A) $x^2-3x+2=0$

$\Leftrightarrow (x^2-2x)-(x-2)=0$

$\Leftrightarrow x(x-2)-(x-2)=0$

$\Leftrightarrow (x-1)(x-2)=0$

$\Leftrightarrow x-1=0$ hay $x-2=0$

$\Leftrightarrow x=1$ hay $x=2$

Vậy $S=\text{{1;2}}$

B) $-x^2+5x-6=0$

$\Leftrightarrow (-x^2+3x)+(2x-6)=0$

$\Leftrightarrow -x(x-3)+2(x-3)=0$

$\Leftrightarrow (-x+2)(x-3)=0$

$\Leftrightarrow -x+2=0$ hay $x-3=0$

$\Leftrightarrow -x=-2$ hay $x=3$

$\Leftrightarrow x=2$ hay $x=3$

Vậy $S=\text{{2;3}}$

C) $4x^2-12x+5=0$

$\Leftrightarrow (4x^2-10x)-(2x-5)=0$

$\Leftrightarrow 2x(2x-5)-(2x-5)=0$

$\Leftrightarrow (2x-1)(2x-5)=0$

$\Leftrightarrow 2x-1=0$ hay $2x-5=0$

$\Leftrightarrow 2x=1$ hay $2x=5$

$\Leftrightarrow x=\dfrac12$ hay $x=\dfrac52$

Vậy $S=$ {$\dfrac12 ; \dfrac52$}

D) $2x^2+5x+3=0$

$\Leftrightarrow (2x^2+2x)+(3x+3)=0$

$\Leftrightarrow 2x(x+1)+3(x+1)=0$

$\Leftrightarrow (2x+3)(x+1)=0$

$\Leftrightarrow 2x+3=0$ hay $x+1=0$

$\Leftrightarrow 2x=-3$ hay $x=-1$

$\Leftrightarrow x=-\dfrac32$hay$x=-1$

Vậy $S=$ {$-\dfrac32;-1$}