1.

\(a.-5x+20=5.3^2\)

\(-5x+20=5.9\)

\(-5x+20=45\)

\(-5x=45-20\)

\(-5x=25\)

\(x=25:\left(-5\right)\)

\(x=-5\)

b,\(36:\left(2x-15\right)=-12\)

\(2x-15=36:\left(-12\right)\)

\(2x-15=-3\)

\(2x=-3+15\)

\(2x=12\)

\(x=12:2\)

\(x=6\)

\(x\in N;150⋮x;120⋮x;x\) lớn nhất

\(=>x=ƯCLN\left(150,120\right)\)

Ta có: \(150=2.3.5^2\)

\(120=2^3.3.5\)

\(=>ƯCLN\left(150,120\right)=2.3.5=30\)

\(=>x=30\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x_1+x_2=-2\\x_1+x_2=2m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x_1=-2m\\x_1+x_2=2m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=-m\\x_2=2m-2+m=3m-2\end{matrix}\right.\)

\(x_1\cdot x_2=m^2-3m\)

\(\Leftrightarrow-3m^2+2m-m^2+3m=0\)

\(\Leftrightarrow-4m^2+5m=0\)

\(\Leftrightarrow m\left(4m-5\right)=0\)

=>m=0 hoặc m=5/4

Gọi vận tốc của ô tô là x

=>Vận tốc xe máy là x-10

Theo đề, ta có: 120/(x-10)-120/x=1

=>(120x-120x+1200)/x(x-10)=1

=>x^2-10x=1200

=>x^2-10x-1200=0

=>x=40