Question

\(\left(3x-5\right).\left(-2x-7\right)=0\)

\(9x^2-1=\left(1+3x\right)\left(2x-3\right)\)

\(x^2-9=\left(1-4x\right)\left(x+3\right)\)

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

\(9x^2-16-x\left(3x+16\right)=0\)

\(\frac{2+4+...+2016+2018}{1019090}=-3x^2-4x\)

mình cần gấp mong các bạn giúp đỡ

Answer 1

\(\left(3x-5\right)\left(-2x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\-2x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=5\\-2x=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{-7}{2}\end{cases}}}\)

Answer 2

\(9x^2-1=\left(1+3x\right)\left(2x-3\right)\)

\(\Leftrightarrow9x^2-1=2x-3+6x^2-9x\)

\(\Leftrightarrow9x^2-1=-7x-3+6x^2\)

\(\Leftrightarrow9x^2-1+7x+3-6x^2=0\)

\(\Leftrightarrow3x^2+2+7x=0\)

\(\Leftrightarrow3x^2+6x+x+2=0\)

\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}\)