Violympic toán 7

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2. cho đa thức

f(x)=\(2x^5-4x^4+3x^3-x^2+5x-1\)

g(x)= \(-x^5+2x^4-3x^3-x^2-2x+7\)

h(x)=\(x^5-2x^4-2x^2-x-3\)

tính

a, f(x)+g(x)

b, f(x)+h(x)

c, g(x)+h(x)

d, f(x)-g(x)

e, f(x)-h(x)

h, g(x)-h(x)

f, f(x)+g(x)+h(x)

g, f(x)+g(x)-h(x)

n, f(x)-g(x)+h(x)

m,f(x)-g(x)-h(x)

Bro_Jeon_Downy
29 tháng 3 2019 lúc 22:21

a. f(x)+g(x)=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)

=2x5-x5-4x4+2x4+3x3-3x3-x2-x2+5x-2x-1+7

=x5-2x4-2x2+3x+6

b. f(x)+h(x)=2x5−4x4+3x3−x2+5x−1+x5−2x4−2x2−x−3

=2x5+x5-4x4-2x4+3x3-x2-2x2+5x-x-1-3

=3x5-6x4+3x3-3x2+6x-4

c. g(x)+h(x)=−x5+2x4−3x3−x2−2x+7+x5−2x4−2x2−x−3

=-x5+x5+2x4-2x4-3x3-x2-2x2-2x-x+7-3

=-3x3-3x2-3x+4

d. f(x)-g(x)=2x5−4x4+3x3−x2+5x−1-(−x5+2x4−3x3−x2−2x+7)

=2x5−4x4+3x3−x2+5x−1-x5-2x4+3x3+x2+2x-7

=2x5-x5-4x4-2x4+3x3+3x3-x2+x2+5x+2x-1-7

=x5-6x4+6x3+7x-8

e. f(x)-h(x)=2x5−4x4+3x3−x2+5x−1-(x5−2x4−2x2−x−3)

=2x5−4x4+3x3−x2+5x−1-x5+2x4+2x2+x+3

=2x5-x5-4x4+2x4+3x3-x2+2x2+5x+x-1+3

=x5-2x4+3x3+x2+6x-4

h. g(x)-h(x)=−x5+2x4−3x3−x2−2x+7-(x5−2x4−2x2−x−3)

=−x5+2x4−3x3−x2−2x+7-x5+2x4+2x2+x+3

=-x5-x5+2x4+2x4-3x3-x2+2x2-2x+x+7+3

=-2x5+4x4-3x3+x2-x+10

f. f(x)+g(x)+h(x)=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)+x5−2x4−2x2−x−3

=2x5-x5+x5-4x4+2x4-2x4+3x3-3x3-x2-x2-2x2+5x-2x-x-1+7-3

=2x5-4x4-4x2+2x+3

g. f(x)+g(x)-h(x)=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)-(x5−2x4−2x2−x−3)

=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)-x5+2x4+2x2+x+3

=2x5-x5-x5-4x4+2x4+2x4+3x3-3x3-x2-x2+2x2+5x-2x+x-1+7+3

=4x+9

n. f(x)-g(x)+h(x)=2x5−4x4+3x3−x2+5x−1-(−x5+2x4−3x3−x2−2x+7)+x5−2x4−2x2−x−3

=2x5−4x4+3x3−x2+5x−1-x5-2x4+3x3+x2+2x-7+x5−2x4−2x2−x−3

=2x5-x5+x5-4x4-2x4-2x4+3x3+3x3-x2+x2-2x2+5x+2x-x-1-7-3

=2x5-8x4+6x3-2x2+6x-11

m. f(x)-g(x)-h(x)=2x5−4x4+3x3−x2+5x−1-(−x5+2x4−3x3−x2−2x+7)-(x5−2x4−2x2−x−3)

=2x5−4x4+3x3−x2+5x−1-x5-2x4+3x3+x2+2x-7-x5+2x4+2x2+x+3

=2x5-x5-x5-4x4-2x4+2x4+3x3+3x3-x2+x2+2x2+5x+2x+x-1-7+3

=-4x4+6x3+2x2+8x-5


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