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ĐKXĐ : \(\hept{\begin{cases}ab-2\ne0\\ab+2\ne0\\a^4b^4\ne0\end{cases}}\Rightarrow ab\ne\pm2;a\ne0;b\ne0\)
\(P=\left(\frac{1}{ab-2}+\frac{1}{ab+2}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\left(\frac{2ab}{a^2b^2-4}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\left(\frac{4a^3b^3}{a^4b^4-16}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\frac{8a^5b^5}{a^8b^8-16^2}.\frac{a^4b^4+16}{a^4b^4}=\frac{8a^5b^5\left(a^4b^4+16\right)}{\left(a^4b^4-16\right)\left(a^4b^4+16\right).a^4b^4}\)
\(=\frac{8ab}{a^4b^4-16}\)
b) Khi \(\frac{a^2+4}{b^2+9}=\frac{a^2}{9}\)
=> (a2 + 4).9 = a2(b2 + 9)
=> 9a2 + 36 = a2b2 + 9a2
=> a2b2 = 36
=> (ab)2 = 36
=> \(\orbr{\begin{cases}ab=6\left(tm\right)\\ab=-6\left(tm\right)\end{cases}}\)
Khi ab = 6 => P = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.6}{6^4-16}=\frac{48}{1280}=\frac{3}{80}\)
Khi ab = -6 => P = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.\left(-6\right)}{\left(-6\right)^4-16}=-\frac{3}{80}\)
\(A=\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right).....\left(51^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)....\left(52^4+\frac{1}{4}\right)}\)
\(=\frac{\left(1+1+\frac{1}{2}\right)\left(1-1+\frac{1}{2}\right)....\left(11^2-11+\frac{1}{2}\right)}{\left(2+2^2+\frac{1}{2}\right)\left(2^2-2+\frac{1}{2}\right)....\left(12^2-12+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)....\left(11.12+\frac{1}{2}\right)}{\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)....\left(12.13+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{12.13+\frac{1}{2}}\)
\(=\frac{1}{313}\)
Chúc bạn học tốt !!!