\(2x^3+x^2-8x-4=0\)
\(\Leftrightarrow\)\(x^2\left(2x+1\right)-4\left(2x+1\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(x-2\right)\left(x+2\right)=0\)
đến đây bạn làm tiếp nha
\(2x^3+x^2-8x-4=0\)
\(x^2\left(2x+1\right)-4\left(2x+1\right)=0\)
\(\left(x^2-4\right)\left(2x+1\right)=0\)
\(1.x^2-4=0\)
\(\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow x=\pm2\)
\(2.2x+1=0\)
\(x=-\frac{1}{2}\)
\(2x^3+x^2-8x-4=0\)
\(\Leftrightarrow x^2.\left(2x+1\right)-4\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\)2x + 1 = 0 => x = -1/2
Hoặc x - 2 = 0 => x = 2
Hoặc x + 2 = 0 => x = -2
Vậy x = -1/2 hoặc x = 2 hoặc x = -2
2x3 + x2 - 8x - 4 = 0
<=> (2x3 - 8x) + (x2 - 4) = 0
<=> 2x(x2 - 4) + (x2 - 4) = 0
<=> (2x + 1) (x2 - 4) = 0
\(< =>\orbr{\orbr{\begin{cases}2x+1=0\\x^2-4=0\end{cases}< =>\orbr{\begin{cases}x=\frac{-1}{2}\\\orbr{\begin{cases}x=2\\x=-2\end{cases}}\end{cases}}}}\)\(< =>\orbr{\begin{cases}2x+1=0\\x^2-4\end{cases}}\)\(< =>\orbr{\begin{cases}x=\frac{-1}{2}\\\orbr{\begin{cases}x=2\\x=-2\end{cases}}\end{cases}}\)<=> x = \(\frac{-1}{2}\)hoặc -2 hoặc 2
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