a, (3x-2)(4x+5) =0
⇔ \(\left[ \begin{array}{l}3x-2=0\\4x+5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{2}{3} \\x=\frac{-5}{4} \end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=\frac{2}{3} \\x=\frac{-5}{4} \end{array} \right.\)
b, (2x+7)(x-5)(5x+1)
⇔ 2x+7 =0
hoặc x-5=0
hoặc 5x+1 = 0
⇔ x=$\frac{-7}{2}$
hoặc x =5
hoặc x= $\frac{-1}{5}$
Vậy x =$\frac{-7}{2}$ hoặc x =5 hoặc x= $\frac{-1}{5}$
c, (4x-10)(24+5x)=0
⇔\(\left[ \begin{array}{l}4x-10=0\\24-5x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{10}{4} \\x=\frac{24}{5} \end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=\frac{10}{4} \\x=\frac{24}{5} \end{array} \right.\)
d, (5x+2)(x-7)=0
⇔\(\left[ \begin{array}{l}5x+2=0\\x-7=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{-2}{5} \\x=7\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=\frac{-2}{5} \\x=7\end{array} \right.\)
e, (4x+2)(x² +1)=0
⇔ \(\left[ \begin{array}{l}4x+2=0\\x²+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-\frac{1}{2} \\x²=-1 (vô lý)\end{array} \right.\)
⇔ x= -$\frac{1}{2}$
Vậy x= -$\frac{1}{2}$
Chúc bạn học tốt nha :>
p/s : bạn chỉ cần cho mỗi tích =0 là xong hết
