Mình làm gộm 2 ý luôn nhé

Ta có : \(Q\left(x\right)=5x+3x^2+5+x^2+2x^4=5x+4x^2+5+2x^4\)

Ta có : \(M\left(x\right)=P\left(x\right)+Q\left(x\right)=\left(x^4-5x+2x^2+1\right)+\left(5x+4x^2+5+2x^4\right)\)

\(=x^4-5x+2x^2+1+5x+4x^2+5+2x^4\)

\(=5x^4+6x^2+6\)

Mà : \(5x^4+6x^2\ge0\forall x\)

Nên : \(5x^4+6x^2+6\ge6\forall x\)

Suy ra : M(x) > 0 với mọi x

Vậy M(x) vô nghiệm

a) P(x) = x⁴ - 5x + 2x² + 1 = x⁴ + 2x² - 5x + 1 

Q(x) = 5x + 3x² + 5 + 1x² + x⁴.2 = 2x⁴ + 4x² + 5x + 5

        P(x) = x⁴ + 2x² - 5x + 1
+
        Q(x) = 2x⁴ + 4x² + 5x + 5
_________________________
P(x)+Q(x) = 3x⁴ + 6x² + 6

b) Ta có: \(\hept{\begin{cases}3x^4\ge0\\6x^2\ge0\end{cases}}\forall x\)

\(\Rightarrow3x^4+6x^2\ge0\forall x\)

\(\Rightarrow M\left(x\right)=3x^4+6x^2+6\ge6>0\forall x\)

Vậy M(x) không có nghiệm

a)

P(x) +Q(x)=(X⁴-5X+2x²+1)+(5x+3x²+5+1x²+x⁴2)

=2x⁴.2 +6x² +6

a) \(M\left(x\right)=P\left(x\right)+Q\left(x\right)=\left(x^4-5x+2x^2+1\right)+\left(5x+3x^2+5+\frac{1}{2}x^2+x\right)\)

\(M\left(x\right)=x^4-5x+2x^2+1+5x+3x^2+5+\frac{1}{2}x^2+x\)

\(M\left(x\right)=x^4+\left(2x^2+3x^2+\frac{1}{2}x^2\right)+\left(5x-5x\right)+\left(1+5\right)\)\(=x^4+5\frac{1}{2}x^2+6\)

b) Đặt  \(M\left(x\right)=x^4+5\frac{1}{2}x^2+6=0\Leftrightarrow x^4+5\frac{1}{2}x^2=0-6=-6\)

Mà \(x^4\ge0;5\frac{1}{2}x^2\ge0\forall x\Rightarrow x^4+5\frac{1}{2}x^2\ne-6\Rightarrow M\left(x\right)\) vô nghiệm

a, M(x)=P(x)+Q(x)

=(x\(^4\)-5x+2x\(^2\)+1)+(5x+3x\(^2\)+5+\(\dfrac{1}{2}x^2\) +x)

= x\(^4\)-5x+2x\(^2\)+1 + 5x+3x\(^2\)+5+\(\dfrac{1}{2}x^2\) +x

= x\(^4\) +(2x\(^2+3x^2+\dfrac{1}{2}x^2\))+(-5x+5x+x)+(1+5)

=x\(^4\) + \(\dfrac{11}{2}x^2\) + x + 6