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: ĐKXĐ: x<>2; x<>-2

\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{2}{x+2}\)

\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\right)\cdot\frac{x+2}{2}\)

\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{2}=\frac{2x-2-2x-4}{2\left(x-2\right)}=\frac{-6}{2\left(x-2\right)}\)

\(=\frac{-3}{x-2}\)

b: Để A nguyên thì -3⋮x-2

=>x-2∈{1;-1;3;-3}

=>x∈{3;1;5;-1}

\(1-\left(\frac{a+x}{ax-x^2}+\frac{2a+3x}{x^2-a^2}\right):\frac{a^4-4x^4}{a^4x-a^2x^3}\)

\(=1-\left(\frac{a+x}{x\left(a-x\right)}+\frac{2a+3x}{\left(x-a\right)\left(x+a\right)}\right):\frac{\left(a^2-2x^2\right)\left(a^2+2x^2\right)}{a^2x\left(a^2-x^2\right)}\)

\(=1-\frac{\left(a+x\right)\left(a+x\right)-\left(2a+3x\right)\cdot x}{x\left(a-x\right)\left(a+x\right)}\cdot\frac{a^2x\left(a-x\right)\left(a+x\right)}{\left(a^2-2x^2\right)\left(a^2+2x^2\right)}\)

\(=1-\frac{a^2+2\cdot ax+x^2-2ax-3x^2}{\left(a^2-2x^2\right)\left(a^2+2x^2\right)}\cdot a^2=1-a^2\cdot\frac{a^2-2x^2}{\left(a^2-2x^2\right)\left(a^2+2x^2\right)}\)

\(=1-\frac{a^2}{a^2+2x^2}=\frac{a^2+2x^2-a^2}{a^2+2x^2}=\frac{2x^2}{a^2+2x^2}\)

a/ \(16a^4+48a^3+4a^2-120a-72\)

b.\(2b^2+2a^2=4a\)

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