Giải phương trình ạ a) (2x – 5)^2 – (x + 2)^2 = 0 b) (3x^2 + 10x – 8)^2 = (5x^2 – 2x + 10)^2 c) (x^2 – 2x + 1) – 4 = 0 d) 4x^2 + 4x + 1 = x^2

Đáp án:

$a) (2x-5)^2-(x+2)^2=0$

$⇔ (2x-5-x-2).(2x-5+x+2)=0$

$⇔ (x-7).(3x-3)=0$

$⇔$ \(\left[ \begin{array}{l}x-7=0\\3x-3=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=7\\3x=3\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=7\\x=1\end{array} \right.\) 

$\text{Vậy S={7;1}}$

$b) (3x^2+10x-8)^2= (5x^2-2x+10)^2$

$⇔ (3x²+10x-8)²-(5x²-2x+10)²=0$

$⇔ (3x²+10x-8-5x²+2x-10).(3x^2+10x-8+5x²-2x+10)=0$

$⇔ (-2x²+12x-18).(8x²+8x+2)=0$

$⇔ -2.(x-3)^2.2.(2x+1)^2=0$

$⇔ -4.(x-3)^2.(2x+1)^2=0$

$⇔$ \(\left[ \begin{array}{l}(x-3)^2=0\\(2x+1)^2=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x-3=0\\2x+1=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=3\\2x=-1\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=3\\x=-0,5\end{array} \right.\) 

$\text{Vậy S={3; -0,5}}$

$c) (x^2-2x+1)-4=0$

$⇔ (x-1)²-2²=0$

$⇔ (x-1-2).(x-1+2)=0$

$⇔ (x-3).(x+1)=0$

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

$\text{Vậy S={3;-1}}$

$d) 4x^2+4x+1=x^2$

$⇔ (2x+1)²-x²=0$

$⇔ (2x+1-x).(2x+1+x)=0$

$⇔ (x+1).(3x+1)=0$

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

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