x^2-12x+11=0 x^2+8x+7=0 4x^2-4x-3=0

Đáp án:

$\text{a) x = 11 hoặc x = 1.}$

$\text{b) x = -7 hoặc x = -1.}$

$\text{c) x = $\dfrac{-1}{2}$ hoặc x = $\dfrac{3}{2}$.}$

Giải thích các bước giải:

$\text{a) x² – 12x + 11 = 0.}$

$\text{⇒ x² – x – 11x + 11 = 0.}$

$\text{⇒ x.(x – 1) – 11.(x – 1) = 0.}$

$\text{⇒ (x – 11).(x – 1) = 0.}$

$\text{⇒ \(\left[ \begin{array}{l}x-11=0.\\x-1=0.\end{array} \right.\)}$

$\text{⇒ \(\left[ \begin{array}{l}x=11.\\x=1.\end{array} \right.\)}$

$\text{Vậy x = 11 hoặc x = 1.}$

$\text{b) x² + 8x + 7 = 0.}$

$\text{⇒ x² + x + 7x + 7 = 0.}$

$\text{⇒ x.(x + 1) + 7.(x + 1) = 0.}$

$\text{⇒ (x + 7).(x + 1) = 0.}$

$\text{⇒ \(\left[ \begin{array}{l}x+7=0.\\x+1=0.\end{array} \right.\)}$

$\text{⇒ \(\left[ \begin{array}{l}x=-7.\\x=-1.\end{array} \right.\)}$

$\text{Vậy x = -7 hoặc x = -1.}$

$\text{c) 4x² – 4x – 3 = 0.}$

$\text{⇒ 4x² – 6x + 2x – 3 = 0.}$

$\text{⇒ 2x(2x – 3) + (2x – 3) = 0.}$

$\text{⇒ (2x + 1).(2x – 3) = 0.}$

$\text{⇒ \(\left[ \begin{array}{l}2x+1=0.\\2x-3=0.\end{array} \right.\)}$

$\text{⇒ \(\left[ \begin{array}{l}2x=-1.\\2x=3.\end{array} \right.\)}$

$\text{⇒ \(\left[ \begin{array}{l}x=\dfrac{-1}{2}.\\x=\dfrac{3}{2}.\end{array} \right.\)}$

$\text{Vậy x = $\dfrac{-1}{2}$ hoặc x = $\dfrac{3}{2}$.}$