Bài 1:
a: \(\left(5x-4\right)\left(4x+6\right)=0\)
=>\(\left[{}\begin{matrix}5x-4=0\\4x+6=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}5x=4\\4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b: (3,5x-7)(2,1x-6,3)=0
=>\(\left[{}\begin{matrix}3,5x-7=0\\2,1x-6,3=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3,5x=7\\2,1x=6,3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3,5}=2\\x=\dfrac{6.3}{2.1}=3\end{matrix}\right.\)
c: \(\left(4x-10\right)\left(5x+24\right)=0\)
=>\(\left[{}\begin{matrix}4x-10=0\\5x+24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{10}{4}=\dfrac{5}{2}\\x=-\dfrac{24}{5}\end{matrix}\right.\)
d: \(\left(x-3\right)\left(2x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
e: (5x-10)(8-2x)=0
=>\(\left[{}\begin{matrix}5x-10=0\\8-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=10\\2x=8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{10}{5}=2\\x=\dfrac{8}{2}=4\end{matrix}\right.\)
f: (9-3x)(15+3x)=0
=>\(\left[{}\begin{matrix}-3x+9=0\\3x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-9\\3x=-15\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Bài 2:
a: (x+5)(2x-3)=0
=>\(\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(\left(x^2+1\right)\left(6x+3\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên 6x+3=0
=>6x=-3
=>\(x=-\dfrac{3}{6}=-\dfrac{1}{2}\)
c: \(\left(\dfrac{3}{4}x-2\right)\left(\dfrac{5}{3}x+1\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{3}{4}x-2=0\\\dfrac{5}{3}x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=2\\\dfrac{5}{3}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2:\dfrac{3}{4}=2\cdot\dfrac{4}{3}=\dfrac{8}{3}\\x=-1:\dfrac{5}{3}=-\dfrac{3}{5}\end{matrix}\right.\)
d: 2(x+3)(x-4)=0
=>(x+3)(x-4)=0
=>\(\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
