a,hđt số 3 = \(\left(a^2+2a\right)^2-9\) 

b,hđt số 3=\(\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)(đổi dấu làm ngoặc khi trước nó là dấu trừ)=\(x^2-\left(y-6\right)^2\)

a) \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)

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

\(=\left(a^2+2a\right)^2-9\)

b) \(\left(x-y+6\right)\left(x+y-6\right)\)

\(=\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)

\(=x^2-\left(y-6\right)^2\)

a) Ta có: \(N=a^2+b^2+2a-b-\dfrac{1}{4}\)

\(=a^2+2a+1+b^2-b+\dfrac{1}{4}-\dfrac{3}{2}\)

\(=\left(a+1\right)^2+\left(b-\dfrac{1}{2}\right)^2-\dfrac{3}{2}\ge-\dfrac{3}{2}\forall a,b\)

Dấu '=' xảy ra khi a=-1 và \(b=\dfrac{1}{2}\)

c: Ta có: \(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)

\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)

\(=a^4-2a^3b+2ab^3-b^4\)

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

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

a) A = 1                          b) B = -2.

b) Ta có: \(a^2+b^2\)

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

\(=3^2+2\cdot\left(-2\right)=9-4=5\)

c) Ta có: \(a^3-b^3\)

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

\(=3^3-3\cdot\left(-2\right)\cdot3\)

\(=27+18=45\)

cho mình hỏi yêu cầu đề bài là gì vậy?

Tìm được: Q = 2 a a − 3