Bài 1:

a: ĐKXĐ: x<>0; x<>3; x<>1

\(A=\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\right):\frac{2x-2}{x}\)

\(=\frac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}\)

\(=\frac{x^2-6x+9-x^2+9}{x\cdot\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6\left(x-3\right)}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6}{x}\cdot\frac{x}{2\left(x-1\right)}=\frac{-3}{x-1}\)

b: Để A là số nguyên thì -3⋮x-1

=>x-1∈{1;-1;3;-3}

=>x∈{2;0;4;-2}

Kết hợp ĐKXĐ, ta được: x∈{2;4;-2}

Bài 2:

a: \(x^3-2x^2=x^2\cdot x-x^2\cdot2=x^2\left(x-2\right)\)

b: \(y^2-2y-x^2+1\)

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

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

c: \(\left(x+1\right)^2-25\)

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

=(x-4)(x+6)

\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)

\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)

\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)

\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)

mấy bạn chuyên toán giải giùm mk bài b) giùm ạ, mk đaq rất cần

c)\(x^3+3xy+y^3\)

\(=x^3+y^3+3xy=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\)

\(=\left(x^2-xy+y^2\right)+3xy\)

\(=x^2-xy+y^2+3xy\)

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

\(=1^2=1\)

d) \(x^3-3xy-y^3\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

\(=\left(x^2+xy+y^2\right)-3xy\)

\(=x^2-2xy+y^2\)

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

\(=1^2=1\)

@Đoàn Đức Hiếu lm a,b đi nhé

3x²y²z² = x³y³ y³z³ z³x³ 
(3x²y²z²) / (x³y³ y³z³ z³x³) = 1
3.[(x²y²z²) / (x³y³ y³z³ z³x³)] = 1
(x²y²z²) / (x³y³ y³z³ z³x³) = 1/3
(x²y²z²) / (x³y³) (x²y²z²) / (y³z³) (x²y²z²) / (z³x³) = 1/3
z²/(xy) x/(yz) y²/(zx) = 1/3
Vậy x²/(yz) y²/(xz) z²/(xy) = 1/3