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\(=A^{2k}+A^{2k-1}B+...+A^2B^{2k-1}+AB^{2k-1}-A^{2k-1}\cdot B-A^{2k-2}\cdot B^2-...-B^{2k}\)
\(=A^{2k}-B^{2k}\)
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)
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) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ca+2a^2+2ab+b^2\)
\(=4a^2+4b^2+2c^2+6ab.\)
Bạn thử giải câu này xem
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\(x\left(x+2\right)\left(x^2+2x+2\right)+1\)
\(=\left(x^2+2x\right)\left(x^2+2x+2\right)+1\)
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:
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2.\left(n+1\right)^2}\)
\(A=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(A=1-\frac{1}{\left(n+1\right)^2}\)
Bài 1:
a, Ta có:
\(\left(a+b+c\right)^2-\left(ab+bc+ca\right)=0\Leftrightarrow a^2+b^2+c^2+ab+bc+ca=0\)\(\Leftrightarrow2a^2+2b^2+2c^2+2ab+2bc+2ca=0\)
\(\Leftrightarrow\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2=0\Leftrightarrow a+b=b+c=c+a=0\)
\(\Leftrightarrow a=b=c=0\)
Vậy điều kiện để phân thức M được xác định là a, b, c không đồng thời = 0
b, Ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
Đặt: \(a^2+b^2+c^2=x,ab+bc+ca=y\)
=> \(\left(a+b+c\right)^2=x+2y\)
Ta cũng có:
\(M=\dfrac{x\left(x+2y\right)+y^2}{x+2y-y}=\dfrac{x^2+2xy+y^2}{x+y}=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)
\(=a^2+b^2+c^2+ab+bc+ca\)
Cái này là hằng đẳng thức luôn á bạn
\(\left(A-B\right)\left(A^{n-1}+A^{n-2}\cdot B+...+A\cdot B^{n-2}+B^{n-1}\right)\)
\(=A^n-B^n\)