a) f(x) = 3x6 + 4x4 - 2x4 + 5x3 - 4x3 - x3 + 3x2 - 2x2 + 1
= 3x6 +2x4 + x2 + 1
b) f(-1) = 3(-1)6 + 2(-1)4 + (-1)2 + 1
= 3 +2 +1 +1 = 7
f(1) = 3.16 + 2.14 + 12 +1
= 3 +2 +1 +1 = 7
c) có 3x6 \(\ge\) 0 với mọi x
2x4 \(\ge\) 0 với mọi x
x2 \(\ge\) 0 với mọi x
1 > 0
=> 3x6 + 2x4 +x2 + 1 > 0 với mọi x
vậy đa thức f(x) ko có nghiệm
TICK CHO MK NHA!!!
Cũng dễ
a) \(f\left(x\right)=3x^6+3x^2+5x^3-2x^2+4x^4-x^3+1-4x^3-2x^4\)
\(f\left(x\right)=3x^6+\left(4x^4-2x^4\right)+\left(5x^3-x^3-4x^3\right)+\left(3x^2-2x^2\right)+1\)
\(f\left(x\right)=3x^6+2x^4+x^2+1\)
b) \(f\left(1\right)=3.1^6+2.1^4+1^2+1=7\)
\(f\left(-1\right)=3.\left(-1\right)^6+2.\left(-1\right)^4+\left(-1\right)^2+1=7\)
c) \(f\left(x\right)=3x^6+2x^4+x^2+1=3x^6+x^4+\left(x^4+\dfrac{1}{2}x^2+\dfrac{1}{2}x^2+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=3x^6+x^4+\left[x^2.\left(x^2+\dfrac{1}{2}\right)+\dfrac{1}{2}.\left(x^2+\dfrac{1}{2}\right)\right]+\dfrac{3}{4}\)
\(=3x^6+x^4+\left(x^2+\dfrac{1}{2}\right).\left(x^2+\dfrac{1}{2}\right)+\dfrac{3}{4}\)
\(=3x^6+x^4+\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Có \(3x^6\ge0,x^4\ge0,\left(x^2+\dfrac{1}{2}\right)^2>0\Rightarrow f\left(x\right)>0\)
Do đó \(f\left(x\right)\) vô nghiệm