Ta có: \(n^2\left(n+1\right)+2n\left(n+1\right)\)

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

\(=n\left(n+1\right)\left(n+2\right)\)

Vì n,(n+1),(n+2) là 3 số lên tiếp nên chúng luôn chia hết cho 6

a)  \(\left(a+b\right)^2-m^2+a+b-m\)

\(=\left(a+b-m\right)\left(a+b+m\right)+\left(a+b-m\right)\)

\(=\left(a+b-m\right)\left(a+b+m+1\right)\)

b)  \(x^3-6x^2+12x-8=\left(x-2\right)^3\)

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

\(x^3+6x^2+11x+6=x^3+x^2+5x^2+5x+6x+6\)

\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)=\left(x+1\right)\left(x^2+5x+6\right)\)

\(=\left(x+1\right)\left(x^2+2x+3x+6\right)=\left(x+1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)

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

=(x+1)(x+2)(x+3)

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

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

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

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

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

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

to cho nick

ai choi bb2 thi kb va k dung cho mik

Bài 2:

a)  \(A=ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

\(=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)

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

\(=\left(b-a\right)\left(c-a\right)\left(c-b\right)\)

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

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

p/s: từ sau bn đăng 1-2 bài thôi nhé, nhiều thế này người lm bài cx hơi bất tiện để đọc đề

      còn mấy câu nữa bn đăng lại nhé

Bài 1: 

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

b)   \(x^4+4x^2-5=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

c)  \(x^3-19x-30=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)

a) Ta có: \(x^2-x-6\)

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

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

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

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

b) Sử dụng phương pháp Hệ số bất định

\(A=4x^2+4x+11\)

\(=\left(4x^2+4x+1\right)+10\)

\(=\left(2x+1\right)^2+10\ge10\)

Min A = 10 khi:  2x + 1 = 0

                      <=> x = -1/2

jbdgvsvvsgvhvhb

Bài 1:

a)  \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=100+99+98+97+...+2+1=5050\)

b)  \(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

\(=2^{128}-1+1=2^{128}\)

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

\(=2c^2\)

ket ban bang bang 2 ko ban

\(A=2\left(x^2+\frac{2.3x}{2}+\frac{9}{4}\right)-5-\frac{9}{4}.\)

\(A=2\left(x+\frac{3}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)

dấu = xảy ra khi   , x= -3/2