thử làm:))

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

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

\(\Rightarrow2\cdot f\left(x\right)=6x^2-4x+8\)

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

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

\(\Rightarrow\hept{\begin{cases}g\left(x\right)=2x^2-1\\g\left(x\right)=2x^2-1\end{cases}}\)

Vậy ...

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 ! 

`@` `\text {Ans}`

`\downarrow`

`a,`

` F(x)=3x^2-7+5x-6x^2-4x^2+8`

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

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

Bậc của đa thức: `2`

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

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

`= 3x^4 + 3x^3 + x + 1`

Bậc của đa thức: `4`

`b,`

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

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

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

`= 3x^4 + 3x^3 - 7x^2 + 6x + 2`

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

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

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

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

a/

\(F\left(x\right)=\left(3-6-4\right)x^2+5x+\left(-7+8\right)=-7x^2+5x+1\) -> Đa thức bậc 2

\(G\left(x\right)=\left(1+2\right)x^4+3x^3+\left(2-1\right)x+\left(-1+2\right)=3x^4+3x^3+x+1\) -> Đa thức bậc 4

b/

\(F\left(x\right)+G\left(x\right)=-7x^2+5x+1+3x^4+3x^3+x+1\\ =3x^4+3x^3-7x^2+6x+2\)

\(F\left(x\right)-G\left(x\right)=-7x^2+5x+1-3x^4-3x^3-x-1\\ =-3x^4-3x^3-7x^2+4x\)

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

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 

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