\(8x^3+12x^2+6x+1=0.\)
\(\Leftrightarrow8x^2\left(x+\frac{1}{2}\right)+8x\left(x+\frac{1}{2}\right)+2\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)\left(8x^2+8x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\2\left(4x^2+4x+1\right)=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2\left(2x+1\right)^2=0\Leftrightarrow x=-\frac{1}{2}\end{cases}}\)
Vậy pt có 1 No là...
\(2\left(x+5\right)-x^2-5x=0.\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)
nhầm
\(\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)