a) 4x².(3x+5)-36.(3x+5) = 0
⇔ ( 3x + 5)(4x² - 36) = 0
⇔ \(\left[ \begin{array}{l}3x=-5\\4x^{2} = 36\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-5/3\\x^{2} = 9\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-5/3\\x = 3 ; x = - 3 \end{array} \right.\)
b,
3.(x+1)² -20.(x+1) = 0
⇔ ( x + 1 ) ( 3x + 3 - 20) = 0
⇔ ( x + 1)(3x - 17) = 0
⇔ x ∈ { - 1 ; 17/3}
c) x² -x =2-2x
⇔ x² + x - 2 = 0
⇔ $x^{2}$ + 2.$\frac{1}{2}$x + $\frac{1}{4}$ - $\frac{9}{4}$ = 0
⇔ ( x + 1/2)² = $\frac{9}{4}$
⇔ \(\left[ \begin{array}{l}x+1/2=3/2\\x+1/2=-3/2\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-2\\x=1\end{array} \right.\)
d) x² -3x+(2x-6) = 0
⇔ x² - x - 6 = 0
⇔ ( x + 1/2)² = $\frac{25}{4}$
⇔ \(\left[ \begin{array}{l}x+1/2=5/2\\x+1/2=-5/2\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
e) (x² -x)-4x+4 = 0
⇔ x² - 5x + 4 = 0
⇔ x² - x - 4x + 4 = 0
⇔ x(x - 1) - 4 ( x - 1) = 0
⇔ ( x - 4)( x - 1) = 0
⇔ \(\left[ \begin{array}{l}x=4\\x=1\end{array} \right.\)
