2x-3=3(x-1)+x+2
=> 2x-3=3x-3+x+2
=> 2x-3-3x+3-x-2=0
=> -2x-2=0 => -2x= 2 => x=-1
\(\left(3x-2\right)\left(4x+5\right)=0\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=2\\4x=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
\(2x\left(x-3\right)-5\left(x-3\right)=0\Leftrightarrow\left(x-3\right)\left(2x-5\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
câu 3:
Gọi quãng đg AB là\(x\) đk: x>0
thời gian đi: \(\dfrac{x}{30}\)
thời gian về là: \(\dfrac{x}{24}\)
đổi: 5h30' =11/2 h
theo bài trên ta có phương trình:
\(\dfrac{x}{30}+\dfrac{x}{24}+1=\dfrac{11}{2}\)
\(\dfrac{4x}{120}+\dfrac{5x}{120}+\dfrac{120}{120}=\dfrac{660}{120}\Leftrightarrow4x+5x+120=660\Leftrightarrow9x=660-120\Leftrightarrow9x=540\Rightarrow x=\dfrac{540}{9}=60km\)
quãng đg AB dài 60 km
