Đáp án:
a, 4x² - 28x + 49 = 25
⇔ (2x)² - 2.2.7x + 7² = 5²
⇔ (2x - 7)² = 5²
⇔ (2x - 7)² - 5² = 0
⇔(2x - 7 - 5).(2x - 7 + 5)=0
⇔ (2x - 12).(2x - 2) = 0
⇔ \(\left[ \begin{array}{l}2x-12=0\\2x-2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=6\\x=1\end{array} \right.\)
b, (3x + 1)² - (x - 2)² = 0
⇔ [3x + 1 - (x - 2)].(3x + 1 + x - 2) = 0
⇔ (3x + 1 - x + 2).(4x - 1) =0
⇔ (2x + 3).(4x - 1) = 0
⇔ \(\left[ \begin{array}{l}2x+3=0\\4x -1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{-3}{2} \\x=\frac{1}{4} \end{array} \right.\)
c, x² - 2x = 24
⇔ x² - 2x + 1 = 25
⇔ (x - 1)² - 5² = 0
⇔ (x - 1 - 5).(x - 1 + 5) = 0
⇔ (x - 6).(x + 4) = 0
⇔ \(\left[ \begin{array}{l}x-6=0\\x+4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=6\\x=-4\end{array} \right.\)
d, x² - 4 = 8(x - 2)
⇔ x² - 4 = 8x - 16
⇔ x² - 8x + 16 = 4
⇔ (x - 4)² = 2²
⇔ (x - 4)² - 2² = 0
⇔ (x - 4 - 2).(x - 4 + 2 ) = 0
⇔ (x - 6).(x - 2) = 0
⇔ \(\left[ \begin{array}{l}x=6\\x=2\end{array} \right.\)
e, 4x² - 12x + 9 = (5 - x )²
⇔ (2x)² - 2.2x.3 + 3² = ( 5 - x )²
⇔ (2x - 3)² - ( 5 - x)² = 0
⇔ (2x - 3 - 5 + x).(2x - 3 + 5 - x) =0
⇔ (3x - 8).(x +2 ) =0
⇔\(\left[ \begin{array}{l}x=\frac{8}{3} \\x=-2\end{array} \right.\)
