a, Sắp xếp các hạng tử của mỗi đa thức theo lũy thừa giảm dần của biến là:
\(P\left(x\right)=8x^4+\dfrac{1}{4}x^2-x-\dfrac{7}{2}\)
\(Q\left(x\right)=-8x^4+\dfrac{1}{4}x^2+x+\dfrac{7}{2}\)
b,\(M\left(x\right)=P\left(x\right)+Q\left(x\right)=\left(8x^4+\dfrac{1}{4}x^2-x-\dfrac{7}{2}\right)+\left(-8x^4+\dfrac{1}{4}x^2+x+\dfrac{7}{2}\right)\)
\(=8x^4+\dfrac{1}{4}x^2-x-\dfrac{7}{2}+-8x^4+\dfrac{1}{4}x^2+x+\dfrac{7}{2}\)
\(=\left(8x^4-8x^4\right)+\left(\dfrac{1}{4}x^2+\dfrac{1}{4}x^2\right)-\left(x-x\right)-\left(\dfrac{7}{2}-\dfrac{7}{2}\right)\)
\(=\dfrac{1}{2}x^2\) \(\Rightarrow M\left(x\right)=\dfrac{1}{2}x^2\)
\(N\left(x\right)=P\left(x\right)-Q\left(x\right)=\left(8x^4+\dfrac{1}{4}x^2-x-\dfrac{7}{2}\right)-\left(-8x^4+\dfrac{1}{4}x^2+x+\dfrac{7}{2}\right)\)
\(=8x^4+\dfrac{1}{4}x^2-x-\dfrac{7}{2}+8x^4-\dfrac{1}{4}x^2-x-\dfrac{7}{2}\)
\(=\left(8x^4+8x^4\right)+\left(\dfrac{1}{4}x^2-\dfrac{1}{4}x^2\right)-\left(x+x\right)-\left(\dfrac{7}{2}+\dfrac{7}{2}\right)\)
\(=16x^4-2x-7\) \(\Rightarrow N\left(x\right)=16x^4-2x-7\)
c, \(\text{Thay }x=0\text{ vào }N\left(x\right)=16x^4-2x-7\)
\(\Rightarrow N\left(0\right)=16.0^4-2.0-7=16.0-2.0-7=0-0-7=-7\)
\(\Rightarrow\text{Khi }x=0\text{ thì }N\left(x\right)=-7\Rightarrow x=0\text{ không phải là nghiệm của đa thức }N\left(x\right)\)
d,\(\text{Cho }M\left(x\right)=0\Rightarrow\dfrac{1}{2}x^2=0\)
\(\Leftrightarrow x^2=-\dfrac{1}{2}\left(1\right)\)
\(\text{Ta có: }x^2\ge0,\forall x\in R\left(2\right)\)
\(\text{Từ (1) và (2)}\Rightarrow x\in\varnothing\Rightarrow M\left(x\right)\text{ vô nghiệm}\)
