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 = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.\left(-6\right)}{\left(-6\right)^4-16}=-\frac{3}{80}\)

mấy cái trên la a^2.b chứ không pải a tất cả mũ 2b

\(a^3-a^2b+ab^2-6b^3=0\)

\(\Leftrightarrow\left(a-2b\right)\left(a^2+ab+3b^2\right)=0\left(1\right)\)

Vì a>b>0 =>a²+ab+3b²>0 nên từ (1) ta có a=2b

Vậy biểu thức \(A=\frac{a^4-4b^4}{b^4-4a^4}=\frac{16b^4-4b^4}{b^4-64b^4}=\frac{12b^4}{-63b^4}=-\frac{4}{21}\)

= 4/ 2 ko

 \(a^3-a^2b+ab^2-6b^3=0\)

\(\Leftrightarrow\left(a^3-a^2b\right)+\left(a^2b-ab^2\right)+\left(3ab^2-6b^3\right)=0\)

\(\Leftrightarrow a^2\left(a-2b\right)+ab\left(a-2b\right)+3b^2\left(a-2b\right)=0\)          

\(\Leftrightarrow\left(a-2b\right)\left(a^2+ab+3b^2\right)=0\left(1\right)\)

Vì \(a>b>0\Rightarrow a^2+ab+3b^2>0\)nên từ (1) ta có \(a-2b=0\Leftrightarrow a=2b\)

Giá trị biểu thức \(P=\frac{a^4-4b^4}{b^4-4a^4}=\frac{16b^4-4b^4}{b^4-64b^4}=\frac{12b^4}{-63b^4}=-\frac{4}{21}\)

quy đồng mẫu số ta được

\(\frac{\left(a-b\right)^2}{a\left(a^2-b^2\right)}+\frac{\left(a+b\right)^2}{a\left(a^2-b^2\right)}=\frac{a\left(3a-b\right)}{a\left(a^2-b^2\right)}\)<=> (a-b)² +(a+b)² = a(3a-b) <=> a²- ab- 2b²= 0 <=> (a+ b)(a- 2b) = 0

<=> a=-b hoăc a =2b

với a= -b => P= \(\frac{-b^3+2b^3+2b^3}{-2b^3-b^3+2b^3}=-3\)

với a =2b => P= \(\frac{\left(2b\right)^3+2.\left(2b\right)^2b+2b^3}{2.\left(2b\right)^3+2b.b^2+2b^3}=\frac{3}{2}\)

\(A=\frac{9a^5-ab^4-18a^4b+2b^5}{3a^2b^2+ab^4-6a^2b^3-2b^5}\)

\(=\frac{a\left(9a^4-b^4\right)-2b\left(9a^4-b^4\right)}{ab^2\left(3a^2+b^2\right)-2b^3\left(3a^2+b^2\right)}\)

\(=\frac{\left(9a^4-b^4\right)\left(a-2b\right)}{\left(3a^2+b^2\right)\left(ab^2-2b^3\right)}\)

\(=\frac{\left(3a^2-b^2\right)\left(3a^2+b^2\right)\left(a-2b\right)}{\left(3a^2+b^2\right)b^2\left(a-2b\right)}\)

\(=\frac{3a^2-b^2}{b^2}\)

\(=3.\left(\frac{a}{b}\right)^2-1=3.\left(\frac{2}{3}\right)^2-1=\frac{1}{3}\)

A=\(\frac{\left(a-2\right)\left(a+2\right)}{b\left(a+2\right)-\left(a+2\right)}\)

=\(\frac{\left(a-2\right)\left(a+2\right)}{\left(a+2\right)\left(b-1\right)}\)

Vì a+2 khác 0 nên A = \(\frac{a-2}{b-1}\)

Thay số vào => A=\(\frac{3-2}{-\frac{3}{4}-1}\)=\(\frac{1}{-\frac{7}{4}}\) =-\(\frac{4}{7}\)

\(A=\frac{a^2-4}{ab+2b-a-2}\)

Thay \(a=3\) và \(b=\frac{-3}{4}\) vào A ta có :

\(A=\frac{3^2-4}{3.\frac{-3}{4}+2.\frac{-3}{4}-3-2}\)

A = \(\frac{9-4}{\frac{-9}{4}+\frac{-3}{2}-3-2}\)

A = \(\frac{5}{\frac{-23}{4}}\)

A = \(\frac{-20}{23}\)