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a) \(P\left(x\right)=2x^3-2x+x^2-x^3+3x+2\)\(=\left(2x^3-x^3\right)+x^2+\left(3x-2x\right)+2=x^3+x^2+x+2\)
\(Q\left(x\right)=4x^3-5x^2+3x-4x-3x^3+4x^2+1\)
Q(x) \(=\left(4x^3-3x^3\right)+\left(4x^2-5x^2\right)+\left(3x-4x\right)+1\)\(=x^3-x^2-x+1\)
b) \(P\left(x\right)+Q\left(x\right)=2x^3+3\); \(P\left(x\right)-Q\left(x\right)=2x^2+2x+1\)
a) Sắp xếp theo lũy thừa giảm dần
P(x)=x^5−3x^2+7x^4−9x^3+x^2−1/4x
=x^5+7x^4−9x^3−3x^2+x^2−1/4x
=x^5+7x^4−9x^3−2x^2−1/4x
Q(x)=5x^4−x^5+x^2−2x^3+3x^2−1/4
=−x^5+5x^4−2x^3+x^2+3x^2−1/4
=−x^5+5x^4−2x^3+4x^2−1/4
b)
P(x)+Q(x)
=(x^5+7x^4−9x^3−2x^2−1/4^x)+(−x^5+5x^4−2x^3+4x^2−1/4)
=x^5+7x^4−9x^3−2x^2−1/4x−x^5+5x^4−2x^3+4x^2−1/4
=(x^5−x^5)+(7x^4+5x^4)+(−9x^3−2x^3)+(−2x^2+4x^2)−1/4x−1/4
=12x^4−11x^3+2x^2−1/4x−1/4
P(x)−Q(x)
=(x^5+7x^4−9x^3−2x^2−1/4x)−(−x^5+5x^4−2x^3+4x^2−1/4)
=x^5+7x^4−9x^3−2x^2−1/4x+x^5−5x^4+2x^3−4x^2+1/4
=(x^5+x^5)+(7x^4−5x^4)+(−9x^3+2x^3)+(−2x^2−4x^2)−1/4x+1/4
=2x5+2x4−7x3−6x2−1/4x−1/4
c) Ta có
P(0)=0^5+7.0^4−9.0^3−2.0^2−1/4.0
⇒x=0là nghiệm của P(x).
Q(0)=−0^5+5.0^4−2.0^3+4.0^2−1/4=−1/4≠0
⇒x=0không phải là nghiệm của Q(x).
a: \(P\left(x\right)=x-2x^2+3x^5+x^4+x-1\)
\(=3x^5+x^4-2x^2+2x-1\)
\(Q\left(x\right)=3-2x-2x^2+x^4-3x^5-x^4+4x^2\)
\(=-3x^5+2x^2-2x+3\)
b: P(x)+Q(x)
\(=3x^5+x^4-2x^2+2x-1-3x^5+2x^2-2x+3\)
\(=x^4+2\)
P(x)-Q(x)
\(=3x^5+x^4-2x^2+2x-1+3x^5-2x^2+2x-3\)
\(=6x^5+x^4-4x^2+4x-4\)
\(a,Q\left(x\right)=-3x^4+4x^3+2x^2+\dfrac{2}{3}-3x-2x^4-4x^3+8x^4+1+3x\\ =\left(-3x^4-2x^4+8x^4\right)+\left(4x^3-4x^3\right)+2x^2+\left(3x-3x\right)+1\\ =3x^4+2x^2+1\\ b,Q\left(x\right)=0\\ \Leftrightarrow3x^4+2x^2+1=0\\ \Delta=b^2-4ac=2^2-4.3.1=-8< 0\)
Vậy Q(x) không có nghiệm
a)
\(P\left(x\right)=x-2x^2+3x^5+x^4+x\)
\(\Leftrightarrow P\left(x\right)=\left(x+x\right)-2x^2+x^4+3x^5\)
\(\Leftrightarrow P\left(x\right)=2x-2x^2+x^4+3x^5\)
\(Q\left(x\right)=3-2x-2x^2+x^4-3x^5-x^4+4x^2\)
\(\Leftrightarrow Q\left(x\right)=3-2x+\left(-2x^2+4x^2\right)+\left(x^4-x^4\right)-3x^5\)
\(\Leftrightarrow Q\left(x\right)=3-2x+2x^2-3x^5\)
b)
\(P\left(x\right)+Q\left(x\right)=\left(2x-2x^2+3x^5+x^4\right)+\left(3-2x+2x^2-3x^5\right)\)
\(=2x-2x^2+3x^5+x^4+3-2x+2x^2-3x^5\)
\(=\left(2x-2x\right)+\left(3x^5-3x^5\right)+\left(-2x^2+2x^2\right)+x^4+3\)
\(=x^4+3\)
\(P\left(x\right)-Q\left(x\right)=\left(2x-2x^2+3x^5+x^4\right)-\left(3-2x+2x^2-3x^5\right)\)
\(=2x-2x^2+3x^5+x^4-3+2x-2x^2+3x^5\)
\(=\left(2x+2x\right)+\left(-2x^2-2x^2\right)+\left(3x^5+3x^5\right)+x^4-3\)
\(=4x-4x^2+6x^5+x^4-3\)
\(=6x^5+x^4-4x^2+4x-3\)
a) Ta có:
\(P\left(x\right)=x-2x^2+3x^5+x^4+x=3x^5+x^4-2x^2\)
\(Q\left(x\right)=3-2x-2x^2+x^4-3x^5-x^4+4x^2\)
\(=-3x^5+2x^2-2x+3\)
b) Ta có:
\(P\left(x\right)+Q\left(x\right)=3x^5+x^4-2x^2-3x^5+2x^2-2x+3\)
\(=x^4-2x+3\)
\(P\left(x\right)-Q\left(x\right)=3x^5+x^4-2x^2+3x^5-2x^2+2x-3\)
\(=6x^5+x^4-4x^2+2x-3\)
c) Ta có: \(P\left(0\right)=3.0^5+0^4-2.0^2=0\)
=> x = 0 là nghiệm của P(x)
Mà \(Q\left(0\right)=-3.0^5+2.0^2-2.0+3=3\)
=> x = 0 không là nghiệm của đa thức Q(x)
\(P\left(x\right)=x-2x^2+3x^5+x^4+x=2x-2x^2+3x^5+x^4\)
\(Q\left(x\right)=3-2x-2x^2+x^4-3x^5-x^4-4x^2=3-2x-6x^2-3x^5\)
Tự sắp xếp nhé
\(2x-2x^2+3x^5+x^4+3-2x-6x^2-3x^5=3-8x^2+x^4\)
Tương tự vs trừ : Lưu ý nhớ đổi dấu
c, Tự làm nhé