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

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

a ) 

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

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

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

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

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

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

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

b ) Làm tương tự 

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!!!

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

a. f(x)+g(x)=2x⁵−4x⁴+3x³−x²+5x−1+

2x⁵−4x⁴+3x³−x²+5x−1+

−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+x⁵−2x⁴−2x²−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+x⁵−2x⁴−2x²−x−3)

=2x⁵−4x⁴+3x³−x²+5x−1+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-x⁵−2x⁴−2x²−x−3

=2x⁵−4x⁴+3x³−x²+5x−1-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-x⁵−2x⁴−2x²−x−3)

=2x⁵−4x⁴+3x³−x²+5x−1-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

Trình bày đề bài cho dễ nhìn bạn eyy :v 

Khó nhìn như này thì God cũng chịu -.-

mù mắt xD ghi rõ đề đi bạn ơi !

Dịch:

Cho \(\hept{\begin{cases}f\left(x\right)=4x^3-2x^2+x-5\\g\left(x\right)=x^3+4x^2-3x+2\\h\left(x\right)=-3x^2+x^2+x-2\end{cases}}\)

Tính a) \(f\left(x\right)+g\left(x\right)\)

b) \(g\left(x\right)-h\left(x\right)\)

2. Tìm nghiệm của đa thức

a) \(7-2x\)

b) (x+1)(x-2)(2x-1)

c) 2x+5

d) 3x²+x

3. CMR các đa thức sau không có nghiệm

\(a,f\left(x\right)=x^2+1\)

\(b,\left(2x+1\right)^2+3\)

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