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a. \(x^4-5x^3+4x-5-x^4+3x^2+2x+1\)
\(=-5x^3+3x^2+6x-4\)
b. \(R\left(x\right)=x^4-5x^3+4x-5-\left(-x^4+3x^2+2x+1\right)\)
\(=x^4-5x^3+4x-5+x^4-3x^2-2x-1\)
\(=2x^4-5x^3-3x^2+2x-6\)
a) �(�)+�(�)P(x)+Q(x)
=(�4−5�3+4�−5)+(−�4+3�2+2�+1)=(x4−5x3+4x−5)+(−x4+3x2+2x+1)
=�4−5�3+4�−5−�4+3�2+2�+1=x4−5x3+4x−5−x4+3x2+2x+1
=(�4−�4)−5�3+3�2+(4�+2�)+(1−5)=(x4−x4)−5x3+3x2+(4x+2x)+(1−5)
=−5�3+3�2+6�−4=−5x3+3x2+6x−4
b) �(�)=�(�)−�(�)R(x)=P(x)−Q(x)
=(�4−5�3+4�−5)−(−�4+3�2+2�+1)=(x4−5x3+4x−5)−(−x4+3x2+2x+1)
=�4−5�3+4�−5+�4−3�2−2�−1=x4−5x3+4x−5+x4−3x2−2x−1
=(�4+�4)−5�3−3�2+(4�−2�)+(−1−5)=(x4+x4)−5x3−3x2+(4x−2x)+(−1−5)
=2�4−5�3−3�2+2�−6=2x4−5x3−3x2+2x−6
Bài 2 :
Ta có : \(S=4+4^2+4^3+...+4^{2004}\)
=> \(4S=4^2+4^3+...+4^{2005}\)
=> \(4S-S=\left(4^2+4^3+...+4^{2005}\right)-\left(4+4^2+...+4^{2004}\right)\)
=> \(3S=-4+4^{2005}\)
=> \(3S+4=-4+4^{2005}+4=4^{2005}\)
Mà \(4^{2005}:4^{2004}=4\)
=> \(4^{2005}⋮4^{2004}\)
=> \(3S+4⋮4^{2004}\) ( đpcm )
ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\Rightarrow\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}\)
\(\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}=\dfrac{2a^4}{2b^4}=\dfrac{3b^4}{3c^4}=\dfrac{4c^4}{4d^4}=\dfrac{4d^4}{4e^4}\\ =\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}\\ \dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)