Question

Tìm x biết 2x^2+3x-5=0

 

 

Answer 1

\(2x^2+3x-5=0\)

\(\Leftrightarrow2x^2-2x+5x-5=0\)

\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}}\)

Answer 2

Giải:

2x² + 3x - 5 = 0 

=> (2x² - 2) + (3x - 3) = 0 

=> (2x + 5)(x-1) = 0

\(\Rightarrow\orbr{\begin{cases}x=-2,5\\x=1\end{cases}}\)

Answer 3

\(2x^2+3x-5=0\)

\(\leftrightarrow2x^2-2x+5x-5=0\)

\(\leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)

\(\leftrightarrow\left(x-1\right)\left(2x+5\right)=0\)

\(\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}\leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}}\)