\(a,5x^3+x=0\)\(\Rightarrow x\left(5x^2+1\right)=0\)
Vì \(5x^2+1>0\Rightarrow x=0\)
\(b,x^3+3x^2+3x+2=0\)
\(\Rightarrow x^3+2x^2+x^2+2x+x+2=0\)
\(\Rightarrow x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x^2+x+1\right)=0\)
Mà \(x^2+x+1=x^2+2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow x+2=0\Leftrightarrow x=-2\)
a) x + 5x3 = 0
=> x.( 1 + 5x2 ) = 0
=> x = 0 hoặc 1 + 5x2 = 0 ( Vô lí )
Vậy : x = 0
a) \(x+5x^3=0\)
\(\Leftrightarrow x\left(1+5x^2\right)=0\)
nhưng vì \(1+5x^2\ne0\)nên:
\(\Leftrightarrow x=0\)
\(\Rightarrow x=0\)
b) \(x^3+3x^2+3x+2=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x+2\right)=0\)
nhưng vì \(x^2+x+1\ne0\)nên:
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
\(\Rightarrow x=-2\)
