Xét biểu thức tổng quát:

Khi đó ta có: 

Vậy

\(=\left(18x^{2n-3}+3x^n\right)-\left(18x^{2n-3}-2x^n\right)\)
\(=18x^{2n-3}+3x^n-18x^{2n-3}+2x^n\)
\(=\left(18x^{2n-3}-18x^{2n-3}\right)+\left(3x^n+2x^n\right)\)
\(=5x^n\)

\(=18x^{2n-3}+3x^n-18x^{2n-3}+2x^n=5x^n\)

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

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

cha này điên à hỏi bài gì vậy 

hc chưa xong kiến thức của mk mà hỏi xa vời  ahihihihi

\(P=1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

     \(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)

      \(=2-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)

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

\(P=1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{n\left(n+1\right)}=1+1-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{\left(n+1\right)}\)

\(\Rightarrow P=2-\frac{1}{\left(n+1\right)}=\frac{2n+1}{n+1}\)

5 (3xⁿ⁺¹ - y^n-1) + 3 (xⁿ⁺¹ + 5y^n-1) - 4 (- xⁿ⁺¹ - 2y^n-1)

=> 15xⁿ⁺¹ - 5y^n-1 + 3xⁿ⁺¹ + 15y^n-1 + 4xⁿ⁺¹ + 8y^n-1

=> 22xⁿ⁺¹ + 18y^n-1.