Giúp mk vsbbnnkn

Giải ptr

C) 2x(x-5)=(x-5)^2

Đ)(2x-1)^2=(4-3x)^2

Đáp án:

$c)$ $\text{S = {±5}}$

$d)$ $\text{S = {3 ; 1}}$

Giải thích các bước giải:

$c)$ $2x\left(x-5\right)=\left(x-5\right)^2$

$⇔2x^2-10x=x^2-10x+25$

$⇔2x^2=x^2+25$

$⇔2x^2-x^2=25$

$⇔x^2=25$

\(⇔\left[ \begin{array}{l}x=\sqrt{25}\\x=-\sqrt{25}\end{array} \right.\)  \(⇔\left[ \begin{array}{l}x=5\\x=-5\end{array} \right.\) 

$\text{Vậy S = {±5}}$

$đ)$ $\left(2x-1\right)^2=\left(4-3x\right)^2$

$⇔4x^2-4x+1=16-24x+9x^2$

$⇔4x^2-4x+1-16+24x-9x^2=0$

$⇔-5x^2+20x-15=0$

$⇔-5\left(x^2-4x+3\right)=0$

$⇔-5\left(x-1\right)\left(x-3\right)=0$

$⇔\left(x-3\right)\left(x-1\right)=0$

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

$\text{Vậy S = {3 ; 1}}$

Đáp án:

Giải thích các bước giải:

`c) 2x(x - 5) = (x - 5)^2`

`⇔ 2x(x - 5) - (x - 5)^2 = 0`

`⇔ (2x - x + 5)(x - 5) = 0`

`⇔ (x + 5)(x - 5) = 0 `

⇔ \(\left[ \begin{array}{l}x+5=0\\x-5=0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=-5\\x=5\end{array} \right.\) 

Vậy `S = {5 ; -5}`  

`d) (2x - 1)^2 = (4 - 3x)^2`

`⇔ (2x - 1)^2 - (4 - 3x)^2 = 0`

`⇔ (2x - 1 - 4 + 3x)(2x - 1 + 4 - 3x) = 0`

`⇔ (5x - 5)(-x + 3) = 0`

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

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