Question

Answer 1

\(a,\left(2x+3\right)\left(x-9\right)-5\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x-9-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x-14=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=14\end{matrix}\right.\)

\(b,\dfrac{3x+5}{x+2}-\dfrac{10}{x}=\dfrac{3x^2-15}{x\left(x+2\right)}\left(dkxd:x\ne0,x\ne-2\right)\)

\(\Leftrightarrow\dfrac{3x+5}{x+2}-\dfrac{10}{x}-\dfrac{3x^2-15}{x\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x\left(3x+5\right)-10\left(x+2\right)-3x^2+15}{x\left(x+2\right)}=0\)

\(\Leftrightarrow3x^2+5x-10x-20-3x^2+15=0\)

\(\Leftrightarrow-5x-5=0\)

\(\Leftrightarrow-5x=5\)

\(\Leftrightarrow x=-1\left(tmdk\right)\)

Vậy \(S=\left\{-1\right\}\)

Answer 2

a: =>(2x+3)(x-14)=0

=>x=14 hoặc x=-3/2

b: =>3x^2+5x-10x-20=3x^2-15

=>-5x-20=-15

=>5x+20=15

=>5x=-5

=>x=-1