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 

\(a,f\left(1\right)=3\cdot1^2+1+1=5\\ f\left(-\dfrac{1}{3}\right)=3\cdot\left(-\dfrac{1}{3}\right)^2-\dfrac{1}{3}+1=\dfrac{1}{3}-\dfrac{1}{3}+1=1\\ f\left(\dfrac{2}{3}\right)=3\cdot\left(\dfrac{2}{3}\right)^2-\dfrac{2}{3}+1=\dfrac{4}{3}-\dfrac{2}{3}+1=\dfrac{5}{3}\\ f\left(-2\right)=3\cdot\left(-2\right)^2-2+1=11\\ f\left(-\dfrac{4}{3}\right)=3\cdot\left(-\dfrac{4}{3}\right)^2-\dfrac{4}{3}+1=\dfrac{16}{3}-\dfrac{4}{3}+1=5\)

\(b,f\left(\dfrac{2}{3}\right)=\left|2\cdot\dfrac{2}{3}-9\right|-3=\dfrac{23}{3}-3=\dfrac{14}{3}\\ f\left(-\dfrac{5}{4}\right)=\left|2\cdot\left(-\dfrac{5}{4}\right)-9\right|-3=\dfrac{23}{2}-3=\dfrac{17}{2}\\ f\left(-5\right)=\left|2\left(-5\right)-9\right|-3=19-3=16\\ f\left(4\right)=\left|2\cdot4-9\right|-3=1-3=-2\\ f\left(-\dfrac{3}{8}\right)=\left|2\cdot\left(-\dfrac{3}{8}\right)-9\right|-3=\dfrac{39}{4}-3=\dfrac{27}{4}\)

\(c,x=0\Rightarrow y=2\cdot0^2-7=-7\\ x=-3\Rightarrow y=2\cdot\left(-3\right)^2-7=11\\ x=-\dfrac{1}{2}\Rightarrow y=2\cdot\left(-\dfrac{1}{2}\right)^2-7=\dfrac{-13}{2}\\ x=\dfrac{2}{3}\Rightarrow y=2\cdot\left(\dfrac{2}{3}\right)^2-7=-\dfrac{55}{9}\)

`1,`

`f(x)+g(x)=(5+4-2+7)+(4x^4-2x^3+3x^2+4x-1)`

`= 5+4-2+7+4x^4-2x^3+3x^2+4x-1`

`=(5x^4+4x^4)-2x^3+(4x^2+4x^2)+(-2x+4x)+(7-1)`

`= 9x^4-2x^3+8x^2+2x+6`

Đề phải là `f(x)-g(x)` chứ nhỉ :v?

`f(x)-g(x)=(5+4-2+7)-(4x^4-2x^3+3x^2+4x-1)`

`= 5+4-2+7-4x^4+2x^3-3x^2-4x+1`

`= (5x^4-4x^4)+2x^3+(-2x-4x)+(4x^2-3x^2)+(7+1)`

`= x^4+2x^3-6x+x^2+8`

`2,`

`a, (x+3)(x-1)`

`= x(x-1)+3(x-1)`

`= x*x+x*(-1)+3*x+3*(-1)`

`=x^2-x+3x-3`

`= x^2+2x-3`

`b, (4x+3)(x-2)`

`= 4x(x-2)+3(x-2)`

`= 4x*x+4x*(-2)+3*x+3*(-2)`

`= 4x^2-8x+3x-6`

`c, (2x+3)(x+1)`

`= 2x(x+1)+3(x+1)`

`= 2x*x+2x*1+3*x+3*1`

`= 2x^2+2x+3x+3`

`= 2x^2+5x+3`

`d, (5x-2)(x^2-3x+1)`

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

`= 5x*x^2+5x*(-3x)+5x*1+(-2)*x^2+(-2)*(-3x)+(-2)*1`

`= 5x^3-15x^2+5x-2x^2+6x-2`

`= 5x^3-17x^2+11x-2`

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

a) Ta có: f(x)=-3

<=>x⁵-2x²+x⁴-x⁵+3x²-x⁴-3+2x=-3

<=>(x⁵-x⁵)+(-2x²+3x²)+(x⁴-x⁴)+2x-3=-3

<=>x²+2x-3=-3

<=>x²+2x=0

<=>x(x+2)=0

<=>x=0 hoặc x+2=0

<=>x=0 hoặc x=-2

Vậy..........

b)đa thức f(x) có nghiệm

<=>f(x)=0

<=>x²+2x-3=0

<=>x²+3x-x-3=0

<=>x(x+3)-(x+3)=0

<=>(x-1)(x+3)=0

<=>x-1=0 hoặc x+3=0

<=>x=1 hoặc x=-3

Vậy nghiệm của đa thức f(x) là x=-3;x=1

a, f(x)=0

b, x=1 và x=-3