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11 tháng 1 2018

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

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

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

= 7x5 + 5x4 + x3 +x2 + 6x

f(x) - g(x) - h(x)

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

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

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

7 tháng 7 2018

a)f(x)+g(x)=\(x^5-4x^4-2x^2-7-2x^5+6x^4-2x^2+6.\)

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

f(x)-g(x)=\(x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)

=\(3x^5-10x^4-13\)

b)f(x)+g(x)=\(5x^4+7x^3-6x^2+3x-7-4x^4+2x^3-5x^2+4x+5\)

=\(x^4+9x^3-11x^2+7x-2\)

f(x)-g(x)=\(5x^4+7x^3-6x^2+3x-7+4x^4-2x^3+5x^2-4x-5\)

=\(9x^4+5x^3-x^2-x-12\)

7 tháng 7 2018

a ) 

\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)

\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)

\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)

\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)

1 tháng 5 2017

bài 3:

a) f(x)= x2+2x4-2x3+x2+5x4+4x3-x+5

= (2x4+5x4)+(4x3-2x3)+(x2+x2)-x+5

= 7x4+2x3+2x2-x+5

g(x)= -2x2+8x4+x-x4-3x3+3x2+5+4x3

=(8x4-x4)+(4x3-3x3)+(3x2-2x2)+x+5

= 7x4+x3+x2+x+5

b) h(x)=f(x)-g(x)

=(7x4+2x3+2x2-x+5)-(7x4+x3+x2+x+5)

=7x4+2x3+2x2-x+5-7x4-x3-x2-x-5

=(7x4-7x4)+(2x3-x3)+(2x2-x2)-(x+x)+(5-5)

=x3+x2-2x

Bài 4:

a) f(x)=5x4+x3-x+11+x4-5x3

=(5x4+x4)+(x3-5x3)-x+11

=6x4-4x3-x+11

g(x)=2x3+3x4+9-4x3+2x4-x

=(3x4+2x4)+(2x3-4x3)-x+9

=5x4-2x3-x+9

b) h(x)=f(x)-g(x)

=(6x4-4x3-x+11)-(5x4-2x3-x+9)

=6x4-4x3-x+11-5x4-2x3-x+9

=(6x4-5x4)-(4x3+2x3)-(x+x)+(11+9)

= x4-6x3-2x+20

c) Với x = -2

Ta có: h(-2)=(-2)4-6.(-2)3-2.(-2)+20=88\(\ne\)0

Vậy x = -2 không phải là nghiệm của đa thức h(x)

đúng thì tặng 1 tick cho mk nk các pn!!!

2 tháng 5 2017

giải câu c ở bài 3 với

a: \(f\left(x\right)+g\left(x\right)-h\left(x\right)\)

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

\(=6x^5-6x^4+4x^3-x^2-4x+11\)

f(x)-g(x)-h(x)

\(=15x^5-12x^4+9x^3-7x^2+7x+x^5-2x^4+x^3-x+7\)

\(=16x^5-14x^4+10x^3-7x^2+6x+7\)

b: f(x)+2g(x)=0

\(\Leftrightarrow10x^5-8x^4+6x^3-4x^2+2x+2-10x^5+8x^4-6x^3+6x^2-10x+4=0\)

\(\Leftrightarrow2x^2-8x+6=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

=>x=1 hoặc x=3

29 tháng 3 2019

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

11 tháng 5 2020

Trình bày đề bài cho dễ nhìn bạn eyy :v 

Khó nhìn như này thì God cũng chịu -.-

11 tháng 5 2020

mù mắt xD ghi rõ đề đi bạn ơi !

19 tháng 3 2017

a, f(x)+g(x)= (\(x^5-3\) + 7\(x^4-9x^3+x^2-\dfrac{1}{4}x\))+(\(5x^4-x^5\)+\(x^2\)\(-2x^3+3x^2-\dfrac{1}{4})\)

= \(12x^4-12x^3+5x^2-\dfrac{1}{4}x-\dfrac{13}{4}\)

b, f(x)\(-\)g(x)= (\(x^5-3\) + 7\(x^4-9x^3+x^2-\dfrac{1}{4}x\))\(-\)(\(5x^4-x^5\)+\(x^2\)\(-2x^3+3x^2-\dfrac{1}{4})\)

= f(x)+g(x)= \(x^5-3\) + 7\(x^4-9x^3+x^2-\dfrac{1}{4}x\)\(-\)\(5x^4+x^5\)\(-\)\(x^2\)\(+2x^3-3x^2+\dfrac{1}{4}\)

=2x\(^5\)+2x\(^4\)\(-7x^3\)\(-2x^2\)\(-\dfrac{1}{4}x\) \(-\) \(\dfrac{11}{4}\)

c,Ta có:h(x)+f(x)=f(x) \(\Rightarrow\)h(x)=f(x)\(-\)f(x)=0