Question

Giải phương trình

(3x+2)(x-5) = (2x-5)(3x+2)

(2x-1)² + (2-x)(2x-1) = 0

Answer 1

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

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\3x+2=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-\frac{2}{3}\end{cases}}\)

Vậy...

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

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

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

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

Vậy...

Answer 2

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

(2x-1)² + (2-x)(2x-1) = 0 \(\Rightarrow\)( 2x - 1 )( 2x - 1 + 2 - x ) \(\Rightarrow\)( 2x - 1 )( x + 1 ) = 0

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