Tìm x, biết: a) (3x + 4)^2 – (3x – 1)(3x + 1) = 49 b) x^2 -4x+4 =9(x-2)

* Hướng dẫn giải

a) 

$\begin{array}{l} {\left(3x+4\right)}^2–\left(3x–1\right)\left(3x+1\right)=49\\ \left(9x^2+24x+16\right)-\left(9x^2-1\right)=49\\ 9x^2+24x+16-9x^2+1=49\\ 24x+17=49\\ 24x=32\\ x=\frac{4}{3}\end{array}$

Vậy $x=\frac{4}{3}$

b) 

$\begin{array}{l}{x}^2–4x+4=9\left(x–2\right)\\ \left(x^2–4x+4\right)-9\left(x–2\right)=0\\ \left(x-2\right){}^2-9\left(x–2\right)=0\\ \left(x-2\right)\left[\left(x-2\right)-9\right]=0\\ \left(x-2\right)\left(x-11\right)=0\\ \Rightarrow \left[\begin{array}{l}x-2=0\\ x-11=0\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}x=2\\ x=11\end{array}\right.\end{array}$

Vậy $x\in \left\{2;11\right\}$