\(=A^{2k}+A^{2k-1}B+...+A^2B^{2k-1}+AB^{2k-1}-A^{2k-1}\cdot B-A^{2k-2}\cdot B^2-...-B^{2k}\)

\(=A^{2k}-B^{2k}\)

câu này là hằng đẳng thức thôi . nhưng nếu muốn làm chi tiết thì đây nha :))

ta có : \(\left(A+B\right)\left(A^{2K}-A^{2k-1}B+...+A^2.B^{2k-2}-AB^{2k-1}+B^{2k}\right)\)

\(=\left(A+B\right)\left(A^{2K}+B^{2k}-A^{2k-1}B+...+A^2.B^{2k-2}-AB^{2k-1}\right)\)

\(=A\left(A^{2k}+B^{2k}\right)+B\left(A^{2k}+B^{2k}\right)+A\left(-A^{2k-1}B+...+A^2B^{2k-2}-AB^{2k-1}\right)+B\left(A^{2k-1}B+...+A^2B^{2k-2}-AB^{2k-1}\right)\)

\(=A\left(A^{2k}+B^{2k}\right)+B\left(A^{2k}+B^{2k}\right)-A^{2k}B-B^{2k}A\)

\(=A^{2k+1}+AB^{2K}+BA^{2k}+B^{2k+1}-A^{2k}B-B^{2k}A\)

\(=A^{2k+1}+B^{2k+1}\)

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

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

c)\(\frac{\left(n+1\right)!-\left(n+2\right)!}{\left(n+1\right)!+\left(n+2\right)!}=\frac{n!\left(n+1\right)-n!\left(n+1\right)\left(n+2\right)}{n!\left(n+1\right)+n!\left(n+1\right)\left(n+2\right)}=\frac{n!\left(n+1\right)\left(1-n-2\right)}{n!\left(n+1\right)\left(1+n+2\right)}=\frac{-n-1}{n+3}\)

( Kí hiệu n!=1.2.3.4...n)

cảm ơn bạn nhiều nhiều nhiều lắm

a) \(\left(x^2-2x+2\right)\left(x-2\right)\left(x^2-2x+2\right)\left(x+2\right)\)

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

\(=\left(x^3-4x^2+6x-4\right)\left(x^3+8\right)\)

\(=x^6+8x^3-4x^5-32x^2+6x^4+48x-4x^3-32\)

\(=x^6-4x^5+4x^3-32x^2+48x-32\)

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

\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]+x^3-3x\left(x^2-1\right)\)

\(=2x\left[\left(x^2+2x+1\right)-\left(x^2-1\right)+\left(x^2-2x+1\right)\right]+x^3-\left(3x^3-3x\right)\)

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

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

\(=2x^3+6x-2x^3+3x\)

\(=9x\)

2 câu kia đợi tí đã nhé!

c) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+\left(2a-b\right)^2\)

\(=\left(a^2+b^2+c^2+2ab+2bc+2ca\right)+\left(a^2+b^2+c^2+2ab-2bc-2ca\right)+\left(4a^2-4ab+b^2\right)\)

\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ca+4a^2-4ab+b^2\)

\(=6a^2+3b^2+2c^2\)

d) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+2\left(a+b\right)^2\)

\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ca+2a^2+2ab+b^2\)

\(=4a^2+4b^2+2c^2+6ab.\)

a, \(\left(x^2-2x+2\right)\left(x-2\right)\left(x^2-2x+2\right)\left(x+2\right)\)

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

\(=\left(x^2+4x^2+4-4x^3+4x^2-8x\right)\left(x^2-4\right)\)

\(=\left(-4x^3+9x^2-8x+4\right)\left(x^2-4\right)\)

\(=-4x^5+16x^3+9x^4-36x^2-8x^3+32x+4x^2-16\)

\(=-4x^5+9x^4+8x^3-32x^2+32x-16\)

b, \(\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)

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

\(=2x^3+6x-3x^3+3x=-x^3+9x\)

Bạn thử giải câu này xem

NHỚ ĐỌC KỸ ĐỀ ĐẤY

https://olm.vn/hoi-dap/detail/211451950700.html?pos=476647086293

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

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

Đặt: \(x^2+2x=t\)

khi đó: \(\left(x^2+2x\right)\left(x^2+2x+2\right)+1=t\left(t+2\right)+1=\left(t+1\right)^2\)

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

b) Xét: \(\left(n+1\right)^2-n^2=\left(n+1+n\right)\left(n+1-n\right)=2n+1\)

Khi đó:

\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2.\left(n+1\right)^2}\)

\(A=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)

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