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29 tháng 6 2016

Ta có: f(x) = 2x3 - 3x + x2 - 5x - 2x2 + 1

            g(x) = 3x2 - 2x -x3 + 2

=> f(x) + g(x) = ( 2x3 - 3x + x2 - 5x - 2x2 + 1 ) + ( 3x2 - 2x - x3 + 2 )

= 2x3 - 3x + x2 - 5x - 2x2 + 1 + 3x2 - 2x - x+ 2

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

= x3 + 2x2 - 10x +3

=> f(x) - g(x) = ( 2x3 - 3x + x2 - 5x - 2x2 + 1 ) - ( 3x2 - 2x - x3 + 2 )

= 2x3 - 3x + x2 - 5x - 2x2 + 1 - 3x2 + 2x + x- 2

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

= 3x3 - 4x2 - 6x - 1

Chuk bn hk tốt! vui

5 tháng 11 2017

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

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

25 tháng 1 2017

f(x) + g(x)

= (x5 - 3x2 + 7x4 - 9x3 + x2 - 1/4x) + (5x4 - x5 +x2 - 2x3 + 3x2 - 1/4)

= x5​ - 3x2 + 7x4 - 9x3 + x2 - 1/4x + 5x4 - x5 +x2 - 2x3 + 3x2 - 1/4

=12x4 - 11x3 + 2x2 - 1/4x - 1/4

f(x) - g(x)

= (x5 - 3x2 + 7x4 - 9x3 + x2 - 1/4x) - (5x4 - x5 +x2 - 2x3 + 3x2 - 1/4)

=​ = x5​ - 3x2 + 7x4 - 9x3 + x2 - 1/4x - 5x4 + x5 - x2 + 2x3 - 3x2 + 1/4

= 2x5 + 2x4 - 7x3 - 6x2 - 1/4x + 1/4

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

30 tháng 3 2017

a) \(P\left(x\right)=f\left(x\right)-g\left(x\right)\)

\(P\left(x\right)=\left(2x^3+x^2-3x-4\right)-\left(-x^3+3x^2+5x-1\right)\)

\(P\left(x\right)=2x^3+x^2-3x-4+x^3-3x^2-5x+1\)

\(P\left(x\right)=2x^3+x^3+x^2-3x^2-3x-5x-4+1\)

\(P\left(x\right)=3x^3-2x^2-8x-3\)

b) \(R\left(x\right)=f\left(x\right)-h\left(x\right)\)

\(R\left(x\right)=\left(2x^3+x^2-3x-4\right)-\left(-3x^3+2x^2-x-3\right)\)

\(R\left(x\right)=2x^3+x^2-3x-4+3x^3-2x^2+x+3\)

\(R\left(x\right)=2x^3+3x^3+x^2-2x^2-3x+x-4+3\)

\(R\left(x\right)=5x^3-x^2-2x-1\)

c) Mình chưa học ạ nên không biết làm.

30 tháng 3 2017

a)P(x)=(2x3+x2-3x-4) - (-x3+3x2+5x-1)
= 2x3+x2-3x-4 - x3-3x2-5x+1
= (2x3-x3)+(x2-3x2) +(-3x-5x)+(-4+1)
= x3-2x2-8x-3

b) R(x)=(2x3+x2-3x-4) - (-3x3+2x2-x-3)
= 2x3+x2-3x-4 - 3x3-2x2+x+3
=(2x3-3x3)+(x2-2x2)+(-3x+x)+(-4+3)
= -x3-x2-2x-1

a: \(f\left(x\right)=x^5-3x^4+2x^3-x^2-4x+1\)

\(g\left(x\right)=x^4-5x^3+2x-1\)

\(f\left(x\right)+g\left(x\right)=x^5-2x^4-3x^3-x^2-2x\)

b: \(A\left(x\right)=f\left(x\right)+g\left(x\right)=x^5-2x^4-2x^3-x^2-2x\)

\(A\left(0\right)=0^5-2\cdot0^4-2\cdot0^3-0^2-2\cdot0=0\)

=>x=0 là nghiệm của A(x)