Vũ Thị Ngọc Mai
Giới thiệu về bản thân
Chào mừng bạn đến với trang cá nhân của Vũ Thị Ngọc Mai
0
0
0
0
0
0
0
2026-02-22 21:32:20
\(0\)
\(\left(\right. 20 x^{3} - 20 x^{3} \left.\right) + \left(\right. - 10 x^{2} + 10 x^{2} \left.\right) + \left(\right. 5 x - 4 x \left.\right) = 0\)
\(x = 0\)
2026-02-22 21:01:27
a,Ta có:
P(x)+Q(x)=(x4−5x3+4x−5)+(−x4+3x2+2x+1)𝑃(𝑥)+𝑄(𝑥)=(𝑥4−5𝑥3+4𝑥−5)+(−𝑥4+3𝑥2+2𝑥+1)
P(x)+Q(x)=x4−5x3+4x−5−x4+3x2+2x+1𝑃(𝑥)+𝑄(𝑥)=𝑥4−5𝑥3+4𝑥−5−𝑥4+3𝑥2+2𝑥+1
P(x)+Q(x)=(x4−x4)−5x3+3x2+(4x+2x)+(-5+1)𝑃(𝑥)+𝑄(𝑥)=(𝑥4−𝑥4)−5𝑥3+3𝑥2+(4𝑥+2𝑥)+(−5+1)
P(x)+Q(x)=-5x3+3x2+6x−4𝑃(𝑥)+𝑄(𝑥)=−5𝑥3+3𝑥2 b,R(𝑥)=(𝑥4−5𝑥3+4𝑥−5)−(−𝑥4+3𝑥2+2𝑥+1)
R(x)=x4−5x3+4x−5+x4−3x2−2x−1𝑅(𝑥)=𝑥4−5𝑥3+4𝑥−5+𝑥4−3𝑥2−2𝑥−1
R(x)=(x4+x4)−5x3−3x2+(4x−2x)+(-5−1)𝑅(𝑥)=(𝑥4+𝑥4)−5𝑥3−3𝑥2+(4𝑥−2𝑥)+(−5−1)
R(x)=2x4−5x3−3x2+2x−6𝑅(𝑥)=2𝑥4−5𝑥3−3𝑥2+2𝑥−6 Vậy
P(x)+Q(x)=(x4−5x3+4x−5)+(−x4+3x2+2x+1)𝑃(𝑥)+𝑄(𝑥)=(𝑥4−5𝑥3+4𝑥−5)+(−𝑥4+3𝑥2+2𝑥+1)
P(x)+Q(x)=x4−5x3+4x−5−x4+3x2+2x+1𝑃(𝑥)+𝑄(𝑥)=𝑥4−5𝑥3+4𝑥−5−𝑥4+3𝑥2+2𝑥+1
P(x)+Q(x)=(x4−x4)−5x3+3x2+(4x+2x)+(-5+1)𝑃(𝑥)+𝑄(𝑥)=(𝑥4−𝑥4)−5𝑥3+3𝑥2+(4𝑥+2𝑥)+(−5+1)
P(x)+Q(x)=-5x3+3x2+6x−4𝑃(𝑥)+𝑄(𝑥)=−5𝑥3+3𝑥2 b,R(𝑥)=(𝑥4−5𝑥3+4𝑥−5)−(−𝑥4+3𝑥2+2𝑥+1)
R(x)=x4−5x3+4x−5+x4−3x2−2x−1𝑅(𝑥)=𝑥4−5𝑥3+4𝑥−5+𝑥4−3𝑥2−2𝑥−1
R(x)=(x4+x4)−5x3−3x2+(4x−2x)+(-5−1)𝑅(𝑥)=(𝑥4+𝑥4)−5𝑥3−3𝑥2+(4𝑥−2𝑥)+(−5−1)
R(x)=2x4−5x3−3x2+2x−6𝑅(𝑥)=2𝑥4−5𝑥3−3𝑥2+2𝑥−6 Vậy