F(-1)+g(-1)=`10

F(1)-g(1)=23

\(f\left(x\right)=4x^2-11x+8x^3+4x^4+5+2x^2\)

\(=4x^4+8x^3+6x^2-11x+5\)

\(g\left(x\right)=-5x^3-6x^2-4x^4+9+5x-3x^3\)

\(=-4x^4-8x^3-6x^2+5x+9\)

\(f\left(-1\right)=4\cdot\left(-1\right)^4+8\cdot\left(-1\right)^3+6\cdot\left(-1\right)^2-11\cdot\left(-1\right)+5\)

=4-8+6+11+5

=-4+17+5

=1+17

=18

g(-1)=\(-4\cdot\left(-1\right)^4-8\cdot\left(-1\right)^3-6\cdot\left(-1\right)^2+5\cdot\left(-1\right)+9\)

=-4+8-6-5+9

=4-11+9

=13-11

=2

f(-1)+g(-1)

=18+2

=20

\(f\left(1\right)=4\cdot1^4+8\cdot1^3+6\cdot1^2-11\cdot1+5\)

=4+8+6-11+5

=12+6-11+5

=1+11

=12

g(1)=\(-4\cdot1^4-8\cdot1^3-6\cdot1^2+5\cdot1+9\)

=-4-8-6+5+9

=-12-6+14

=-18+14

=-4

f(1)-g(1)

=12-(-4)

=12+4

=16

Đặt A(x)= \(f\left(x\right)-g\left(x\right)\)

\(=x^{2n}-x^{2n-1}+\cdots+x^2-x+1+x^{2n+1}-x^{2n}+x^{2n-1}-\cdots-x^2+x-1\)

\(=x^{2n+1}\)

Khi x=1/10 thì \(A\left(x\right)=\left(\frac{1}{10}\right)^{2n+1}=\frac{1}{10^{2n+1}}\)

\(#HaimeeOkk\)

\(a)\)

\(f ( x ) + g ( x ) = ( x ^3 − 2 x + 1 ) + ( 2 x ^2 − x ^3 + x − 3 ) \)

\(f ( x ) + g ( x ) = x ^3 − 2 x + 1 + 2 x ^2 − x ^3 + x − 3 \)

\(f ( x ) + g ( x ) = x ^3 − x ^3 + 2 x ^2 − 2 x + x + 1 − 3 \)

\(f ( x ) + g ( x ) = 2 x ^2 − x − 2\)

\(f ( x ) − g ( x ) = ( x ^3 − 2 x + 1 ) − ( 2 x ^2 − x ^3 + x − 3 ) \)

\(f ( x ) − g ( x ) =x^3- 2 x + 1 −2x^2+x^3-x+3\)

\(f ( x ) − g ( x ) = x ^3 + x ^3 − 2 x ^2 − 2 x − x + 1 + 3 \)

\(f ( x ) − g ( x ) = 2 x ^3 − 2 x ^2 − 3 x + 4\)

\(-----------------------------\)

\(b)\)

Thay \(x=-1\) vào \(f ( x ) + g ( x )\)

\(f ( x ) + g ( x ) = 2 x ^2 − x − 2\)

\(⇒ 2 ( − 1 ) ^2 − ( − 1 ) − 2 = 1\)

Thay \(x=-2\) vào \(f ( x ) + g ( x )\)

\(f ( x ) + g ( x ) = 2 x ^2 − x − 2\)

\(⇒ 2 ( − 2 ) ^2 − ( − 2 ) − 2 = 8\)

Tham khảo:

f(x)=x^3-2x^2+3x+1

g(x)=x^3+x^2-5x+3

a: f(-1/3)=-1/27-2/9-1+1=-1/27-6/27=-7/27

g(-2)=-8+4+10+3=17-8=9

b: f(x)-g(x)=x^3-2x^2+3x+1-x^3-x^2+5x-3

=x^2+8x-2

f(x)+g(x)

=x^3-2x^2+3x+1+x^3+x^2-5x+3

=2x^3-x^2-2x+4

\(f\left(x\right)+g\left(x\right)=\left[x\left(1-2x\right)+\left(2x^2-x+4\right)\right]+\left[x\left(x-5\right)-x\left(x+2\right)+7x\right]\)

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

                          \(=4\)

\(f\left(x\right)-g\left(x\right)=\left[x\left(1-2x\right)+\left(2x^2-x+4\right)\right]-\left[x\left(x-5\right)-x\left(x+2\right)+7x\right]\)

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

                          \(=4\)

a) f(x)-g(x)+h(x)= (2x^2-3x^3)-(3x-3x^3+2x-2)+(2x^2+1)

                      =2x^2-3x^3-3x+3x^3-2x+2+2x^2+1

                      =(2x^2+2x^2)+(-3x^3-3x^3)+(2x+3x)+(-2+1)

                      =4x^2-6x^3+5x-1

b)g(x)-f(x)+h(x)=3x-3x^3+2x-2-2x^2+3x^3+2x^2+1

                      =(3x+2x)+(-3x^3+3x^3)+(-2x^2+2x^2)+(-2+1)

                      =5x-1

bạn ơi, cái chỗ mình bỏ trống là như trên nha

cám ơ bạn