Question

giải dùm mình phương trình sao:(2x+1)(x² -1)=4x²-2x-2

Answer 1

\(\left(2x+1\right)\left(x+1\right)\left(x-1\right)=4x^2-4x+2x-2\)

\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)\left(x-1\right)=4x\left(x-1\right)+2\left(x-1\right)\)

\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)\left(x-1\right)=\left(x-1\right)\left(4x+2\right)\)

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

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

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

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

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

Answer 2

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

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

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

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

KL vậy....

Answer 3

Cám ơn  2 ng nhiều thank you so much

Answer 4

giải các phương trình sau 

5(x-1)+2/6 -7x-1/4=2(2x+1)/7-5 

3x-2/6-5=3-2(x+7)/4