\(A=x^3-3xy-y^3\)

\(A=x^3-y^3-3xy\)

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

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

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

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

\(A=1^2\)

\(A=1\)

gọi số hạng thứ 2007 là x ta có:

\(\left(x-11\right)\div3+1=2007\)

\(\Leftrightarrow\left(x-11\right)\div3=2006\)

\(\Leftrightarrow x-11=6018\)

\(\Leftrightarrow x=6029\)

Vậy số hạng thứ 2007 là 6029

Câu 1:

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

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

\(=\left(5x\right)^2\)

\(=25x^2\)

Câu 2:

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

\(=9x^2-6x+1+5\)

\(=\left[\left(3x\right)^2-2.3x.1+1^2\right]+5\)

\(=\left(3x+1\right)^2+5\)

Vì \(\left(3x+1\right)^2\ge0\) với mọi x

Nên \(\left(3x+1\right)^2+5>0\) với mọi x

Vậy \(9x^2-6x+6>0\) với mọi x ( điều phải chứng minh)

a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc-a^3-b^3-c^3\)

\(=3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\)

\(=3\left(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\right)\)

\(=3\left(ab\left(a+b\right)+b^2c+abc+bc^2+c^2a+ca^2+abc\right)\)

\(=3\left(ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\right)\)

\(=3\left(a+b\right)\left(ab+bc+c^2+ac\right)\)

\(=3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)

\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)

4)

b) \(A=4+4^3+4^5+.....+4^{2015}\)

\(A=\left(4+4^3\right)+\left(4^5+4^7\right)+.....+\left(4^{2013}+4^{2015}\right)\)

\(A=4\left(1+4^2\right)+4^5\left(1+4^2\right)+....+4^{2013}\left(1+4^2\right)\)

\(A=4.17+4^5.17+....+4^{2013}.17\)

\(A=\left(4+4^5+......+4^{2013}\right)17\) chia hết cho 5

Vậy A chia hết cho 5

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

\(=\left(4x^2+4x+1\right)+\left(4x^2-4x+1\right)-\left(2-8x^2\right)\)

\(=4x^2+4x+1+4x^2-4x+1-2+8x^2\)

\(=16x^2\)

2. Chứng minh rằng:

a) \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)

biến đổi vế phải

\(\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)

\(=a^3+b^3\) bằng vế trái ( điều phải chứng minh)

b) \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

biến đổi vế phải

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

\(=a^3+ab^2+ac^2-a^2b-abc-ca^2+ba^2+b^3+bc^2-ab^2-b^2c-abc+ca^2+cb^2+c^3-abc-bc^2-c^2a\)

\(=a^3+b^3+c^3-3abc\) bằng vế trái ( điều phải chứng minh )

3. Phân tích thành nhân tử:

c) \(81x^4+4\)

\(=\left(9x^2\right)^2+2.9x^2.2+2^2-36x^2\)

\(=\left(9x^2+2\right)^2-\left(6x\right)^2\)

\(=\left(9x^2+2-6x\right)\left(9x^2+2+6x\right)\)

d) \(x^5+x+1\)

\(=x^5-x^2+x^2+x+1\)

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

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

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