Có A(x) + B(x)= 5x² +5x +1

Suy ra: B(x)= (5x² +5x +1) - A(x)

(Xong rồi thay đa thức A(x) vào rồi tính là ra đó nha!!!!)

a: \(=\dfrac{x^3-3x^2-7x+x^2-3x-7}{x^2-3x-7}=x+1\)

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

c:\(=\dfrac{x^3-3x^2-7x+2x^2-6x-14}{x^2-3x-7}=x+2\)

d: \(=\dfrac{x^2\left(x+5\right)+5x+25-25}{x+5}=x^2+5-\dfrac{25}{x+5}\)

a: 5x-20xy

\(=5x\cdot1-5x\cdot4y=5x\left(1-4y\right)\)

b: \(x^2-9=\left(x-3\right)\left(x+3\right)\)

c: \(x^2-2xy+y^2-z^2\)

\(=\left(x-y\right)^2-z^2\)

=(x-y-z)(x-y+z)

d: \(5x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(5x-2\right)\)

e; \(x^2+4x+3=x^2+x+3x+3\)

=x(x+1)+3(x+1)

=(x+1)(x+3)

f: \(x^3-x+3x^2y+3xy^2+y^3-y\)

\(=\left(x+y\right)^3-\left(x+y\right)\)

\(=\left(x+y\right)\left\lbrack\left(x+y\right)^2-1\right\rbrack\)

=(x+y)(x+y-1)(x+y+1)

g: \(x^2-x-y^2-y\)

\(=\left(x^2-y^2\right)-\left(x+y\right)\)

=(x-y)(x+y)-(x+y)

=(x+y)(x-y-1)

h: \(16x-5x^2-3\)

\(=-5x^2+15x+x-3\)

=-5x(x-3)+(x-3)

=(x-3)(-5x+1)

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

j: \(2x^2-6x=2x\cdot x-2x\cdot3=2x\left(x-3\right)\)

k: \(x^3-3x^2-4x+12\)

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

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

l: \(x^2-y^2-5x+5y\)

=(x-y)(x+y)-5(x-y)

=(x-y)(x+y-5)

\(x^3-5x^2-5x+1\\ =x^3-6x^2+x+x^2-6x+1\\ =x\left(x^2-6x+1\right)+\left(x^2-6x+1\right)\\ =\left(x+1\right)\left(x^2-6x+1\right)\)

x - \(\frac{9}{5}\)= 2.x

x + \(\frac{-9}{5}\)= 2.x

x - x + \(\frac{-9}{5}\)= 2.x - x

\(\frac{-9}{5}\)= x

Bài làm 

\(x-\frac{9}{5}=2x\Leftrightarrow-x=\frac{9}{5}\Leftrightarrow x=-\frac{9}{5}\)

@Hoc tot@

Ta có:

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

Do đó: \(P\left(a\right).P\left(b\right).P\left(c\right)=\left(a^2-4\right)\left(b^2-4\right)\left(c^2-4\right)\)

Ta có:

\(\left(x-a\right)\left(x-b\right)\left(x-c\right)=x^3-3x+1\)

\(\Rightarrow\left\{{}\begin{matrix}a+b+c=0\\ab+ac+bc=-3\\abc=-1\end{matrix}\right.\)

C1: \(\left(a^2-4\right)\left(b^2-4\right)\left(c^2-4\right)=\left(abc\right)^2-4\left(a^2b^2+b^2c^2+c^2a^2\right)+16\left(a^2+b^2+c^2\right)-4^3\)

\(=1-4.9+16.6-4^3=-3\)\(\Rightarrow P\left(a\right).P\left(b\right).P\left(c\right)=-3\)

C2: Biến đổi thêm một chút

Ta có: \(a,b,c\ne0\) nên 

 \(a^3-3a+1=0\Leftrightarrow a\left(a^2-3\right)+1=0\)\(\Rightarrow a^2-3=\dfrac{-1}{a}\)

Tương tự...

 \(\Rightarrow P\left(a\right).P\left(b\right).P\left(c\right)=\left(-\dfrac{1}{a}-1\right)\left(-\dfrac{1}{b}-1\right)\left(-\dfrac{1}{c}-1\right)\)

\(=-\left(\dfrac{1}{a}+1\right)\left(\dfrac{1}{b}+1\right)\left(\dfrac{1}{c}+1\right)\)\(=-\dfrac{a+1}{a}.\dfrac{b+1}{b}.\dfrac{c+1}{c}=abc+ac+bc+ab+a+b+c+1=-1-3+1=-3\)

\(\frac{x^2}{5x+25}-\frac{10-2x}{x}+\frac{5x+50}{5x+x^2}=\frac{x^2}{5\left(x+5\right)}-\frac{10-2x}{x}+\frac{5x+50}{x\left(x+5\right)}\)

\(=\frac{x^3}{5x\left(x+5\right)}-\frac{5\left(x+5\right)\left(10-2x\right)}{5x\left(x+5\right)}+\frac{5\left(5x+50\right)}{5x\left(x+5\right)}\)

\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)

Đề là gì vậy bn ????

Mik có ghi mà bạn câu 1,2,... đó