a ) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy ......
b ) \(2x^2+3x-5=0\)
\(\Leftrightarrow2x^2+5x-2x-5=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy .......
a, 2 (x+5) - x2 - 5x = 0
=> 2(x+5)-(x2+5x)=0
=> 2(x+5)-x(x+5)=0
=> (x+5)(2-x)=0
=> \(\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
vậy x=-5 và x=2
b, 2x2 + 3x - 5 = 0
=> 2x2+5x-2x-5=0
=> x(2x+5)-1(2x+5)=0
=> (x-1)(2x+5)=0
=> \(\left[{}\begin{matrix}x-1=0\\2x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\2x=-5\Rightarrow x=\dfrac{-5}{2}\end{matrix}\right.\)
2xa, 2(x+5)-x(x+5)=0
=>(x+5)(2-x)=0
=>\(\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b,2x\(^{^{ }2}\)+3x-5=0
=>2x\(^2\)+5x-2x-5=0
=>2x(x-1)+5(x-1)=0
=>(x-1)(2x+5)=0
=>\(\left[{}\begin{matrix}x-1=0\\2x+5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=1\\x=\dfrac{-5}{2}\end{matrix}\right.\)