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 ! 

a: f(a)=g(a)

=>5a-3=-1/2a+1

=>5,5a=4

=>\(a=\dfrac{4}{5.5}=\dfrac{8}{11}\)

b: f(b-2)=g(2b+4)

=>\(5\left(b-2\right)-3=-\dfrac{1}{2}\left(2b+4\right)+1\)

=>\(5b-13=-b-2+1=-b-1\)

=>6b=12

=>b=2

f(a) = g(a)

⇔ 5a - 3 = -a/2 + 1

⇔ 5a + a/2 = 1 + 3

⇔ 11a/2 = 4

⇔ 11a = 8

⇔ a = 8/11

Vậy a = 8/11 thì f(a) = g(a)

b) f(b - 2) = g(2b + 4)

⇔ 5.(b - 2) - 3 = -(2b + 4)/2 + 1

⇔ 5b - 10 - 3 = -b - 2 + 1

⇔ 5b + b = 1 + 13

⇔ 6b = 14

⇔ b = 7/3

Vậy b = 7/3 thì f(b - 2) = g(2b + 4)

a)\(f\left(x\right)=5x^3-9x^2+2x+m=5x^2\left(x+2\right)-19x\left(x+2\right)+40\left(x+2\right)-80+m=\left(x+2\right)\left(5x^2-19x+40\right)+m-80\)

Để \(f\left(x\right)⋮g\left(x\right)\) thì \(m-80=0\Leftrightarrow m=80\)

b) \(f\left(x\right)=\left(x+2\right)\left(5x^2-19x+40\right)+m-80\)

Để f(x) chia g(x) có số dư bằng 3 thì \(m-80=3\Leftrightarrow m=83\)

Bài 9:

a: f(-4)=0

=>-4(m-1)+3m-1=0

=>-4m+4+3m-1=0

=>-m+3=0

=>m=3

b: f(-5)=-1

=>-5(m-1)+3m-1=-1

=>-5m+5+3m-1=-1

=>-2m+4=-1

=>-2m=-5

=>m=5/2

d: Ta có: f(x):g(x)

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

\(=\dfrac{x^3+x^2-3x^2-3x+6x+6-1}{x+1}\)

\(=x^2-3x+6+\dfrac{-1}{x+1}\)

Để f(x) chia hết cho g(x) thì \(x+1\in\left\{1;-1\right\}\)

hay \(x\in\left\{0;-2\right\}\)

a: \(h\left(x\right)=7x^5+x^4-2x^3+4+x^4+6x^3-9x^2-2x-1=7x^5+2x^4+4x^3-9x^2-2x+3\)

b: \(h\left(x\right)=7x^5+x^4-2x^3+4-x^4-6x^3+9x^2+2x+1=7x^5-8x^3+9x^2+2x+5\)

Chọn C.

Đặt  u   =   G ( x ) d v   =   f ( x ) d x ⇒ d u   =   G ( x ) ' d x   =   g ( x )   d x v   =   ∫ f ( x ) d x   =   F ( x )

Suy ra: I =  G ( x ) F ( x ) 2 0   - ∫ 0 2 F ( x ) g ( x ) d x  

= G(2)F(2) – G(0)F(0) – 3 = 1 – 0 – 3 = -2.