Ta có : \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)

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

\(x^2+x+1=x^2+x+1\)

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

Quy đồng :  

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

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

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

\(\frac{x}{x^3-1};\frac{x+1}{x^2+x};\frac{x-1}{x^2+x+1}\)

Ta có:\(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)

         \(x^2+x=x\left(x+1\right)\)

        \(x^2+x+1=x^2+x+1\)

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

Quy đồng:

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

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

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

\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^2-3x}{2x^2-x^3}\)

\(=\left(-\frac{x+2}{x-2}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{2-x}{x+2}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\left(-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(2-x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\left(\frac{x^2-4x-4-4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\left(\frac{-3x^2-4x-4+x^2-4x-4}{\left(x-2\right)\left(x+2\right)}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\left(\frac{-2x^2-8}{\left(x-2\right)\left(x+2\right)}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(=\frac{-2x^2-8}{\left(x-2\right)\left(x+2\right)}.\frac{-x^2\left(x-2\right)}{x\left(x-3\right)}=\frac{2x^4+8x^2}{x\left(x+2\right)\left(x-3\right)}\)

Ta có a + b + 8 = 0

=> x³ + 3x² + 6x + y³ + 3y² + 6y + 8 = 0

=> (x³ + 3x² + 3x + 1) + (y³ + 3y² + 3y + 1) + (3x + 3y + 6) = 0

=> (x + 1)³ + (y + 1)³ + 3(x + y + 2) = 0

=> (x + y + 2)[(x + 1)² + (x + 1)(y + 1) + (y + 1)² + 3] = 0

Vì (x + 1)² + (x + 1)(y + 1) + (y + 1)² + 3 \(>0\forall x;y\)

=> x + y + 2 = 0

=> x + y = -2

Vậy A = -2

xyz bạn ơi! tại sao từ dòng 3 lại thành dòng 4 vậy

thank you bạn!!! <3

Ý đề bài là \(\left(2^n+1\right)\left(2^n+2\right)\) hay \(\left(2^{n+1}+2^{n+2}\right)\) vậy?

Để phân thức A được xác định thì x khác -2 x khác 3 

Mk có tâm rút gọn hộ bạn luôn rồi nè =)) 

a, ĐK : \(x\ne-2;3\)

b, \(A=\frac{8-x}{\left(x+2\right)\left(x-3\right)}+\frac{2}{x+2}\)

\(=\frac{8-x}{\left(x+2\right)\left(x-3\right)}+\frac{2\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\frac{8-x+2x-6}{\left(x+2\right)\left(x-3\right)}\)

\(=\frac{x-2}{\left(x-2\right)\left(x-3\right)}=\frac{1}{x-3}\)

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

    \(\frac{x^2+y^2}{\left(x-y\right)^3}-\frac{2xy}{\left(x-y\right)^3}\)

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

\(=\frac{x^2-2xy+y^2}{\left(x-y\right)\left(x^2-2xy+y^2\right)}\) 

\(=\frac{1}{x-y}\)