`Answer:`

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

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

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

\(g\left(x\right)=-2x^3+10x-1-7x^2+x^4-15x+10x^2\)

\(=x^4-2x^3+\left(-7x^2+10x^2\right)+\left(10x-15x\right)-1\)

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

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

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

\(=2x^4-2x^3-3x-6\)

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

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

Ta có: \(g\left(x\right)=2x^3+10x-1-7x^2-15x+10x^2\)

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

b) Ta có: f(x)+g(x)

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

\(=x^4-2x^3-3x-6\)

Ta có: f(x)-g(x)

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

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

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?

a)F(x)=5x²-7+6x-8x³-x⁴=\(x^4-8x^3+5x^2+6x-7\)

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

b)\(F\left(x\right)+G\left(x\right)=x^4-8x^3+5x^2+6x-7+x^4+8x^3-5x^2+5\)

\(=x^4+x^4-8x^3+8x^3+5x^2-5x^2+6x-7+5\)

=\(2x^4+6x-2\)

\(F\left(x\right)-G\left(x\right)=x^4-8x^3+5x^2+6x-7-x^4-8x^3+5x^2-5\)

\(=x^4-x^4-8x^3-8x^3+5x^2+5x^2+6x-7-5\)

=-16x³+10x²+6x-12

\(f\left(x\right)=8x^4-7x^3+7x^2+\frac{29}{5}x-\frac{1}{3}\)

\(g\left(x\right)=-8x^4-7x^3-3x^2+\frac{82}{3}\)

\(f\left(x\right)+g\left(x\right)=-14x^3+4x^2+\frac{29}{5}x+27\)

a, \(f\left(x\right)=2x^2+6x^4-3x^3+2011\)

\(=6x^4-3x^3+2x^2+2011\)

\(g\left(x\right)=2x^3-5x^2-3x^4-2012\)

\(=-3x^4+2x^3-5x^2-2012\)

b, \(f\left(x\right)+g\left(x\right)=6x^4-3x^3+2x^2+2011-3x^4+2x^3-5x^2-2012\)

\(=\left(6x^4-3x^4\right)+\left(2x^3-3x^3\right)+\left(2x^2-5x^2\right)+\left(2011-2012\right)\)

\(=3x^4-x^3-3x^2-1\)

\(f\left(x\right)-g\left(x\right)=6x^4-3x^3+2x^2+2011-\left(-3x^4+2x^3-5x^2-2012\right)\)

\(=6x^4-3x^3+2x^2+2011+3x^4-2x^3+5x^2+2012\)

\(=\left(6x^4+3x^4\right)-\left(3x^3+2x^3\right)+\left(2x^2+5x^2\right)+\left(2011+2012\right)\)

\(=9x^4-5x^3+7x^2+4023\)

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!!!