Bài 1:
\(A=4x^2-28x+49\)
\(=4x^2-16x-12x+48+1\)
\(=4x\left(x-4\right)-12\left(x-4\right)+1\)
\(=\left(4x-12\right)\left(x-4\right)+1\)
Thay x = 4
\(\Leftrightarrow A=1\)
Vậy A= 1 tại x = 4
câu b
\(x^2-x=24\Leftrightarrow x^2-2\dfrac{1}{2}x+\dfrac{1}{4}=24+\dfrac{1}{4}=\dfrac{97}{4}\)
\(\left(x-\dfrac{1}{2}\right)^2=\left(\dfrac{\sqrt{97}}{2}\right)^2\)
\(\left[{}\begin{matrix}x_1=\dfrac{1+\sqrt{97}}{2}\\x_2=\dfrac{1-\sqrt{97}}{2}\end{matrix}\right.\)
a)
2x= t
t^2 -2.7t+49 =(t-7)^2
x=4 => t= 8 => t-7 =1 => A=1
64-28.4
