\(\frac{2}{a+b\sqrt{5}}-\frac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\)

\(\Leftrightarrow\frac{2a-2b\sqrt{5}-3a-3b\sqrt{5}}{a^2-5b^2}=-9-20\sqrt{5}\)

\(\Leftrightarrow\frac{a+5b\sqrt{5}}{a^2-5b^2}=9+20\sqrt{5}\)

\(\Leftrightarrow\sqrt{5}\left(100b^2+5b-20a^2\right)=9a^2-a-45b^2\)

Ta nhận thây VT là sô vô tỷ còn VP là sô hữu tỷ.

\(\Rightarrow\hept{\begin{cases}100b^2+5b-20a^2=0\\9a^2-a-45b^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=0\\b=0\end{cases}\left(loai\right)}\)hoặc \(\hept{\begin{cases}a=9\\b=4\end{cases}\left(nhan\right)}\)

ĐK: \(a\ne\pm b\sqrt{5}\)(*)

\(\frac{2}{a+b\sqrt{5}}-\frac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\)

\(\Leftrightarrow2\left(a-b\sqrt{5}\right)-3\left(a+b\sqrt{5}\right)=-\left(9+20\sqrt{5}\right)\left(a+b\sqrt{5}\right)\left(a-b\sqrt{5}\right)\)

\(\Leftrightarrow9a^2-45b^2-a=\sqrt{5}\left(-20a^2+100b^2+5b\right)\)(*)

Ta thấy (*) có dạng \(A=B\sqrt{5}\)nếu \(B\ne0\)thì \(\sqrt{5}=\frac{A}{B}\in Z\)(vô lý) Vậy B=0 => A=0

Do đó: (*) \(\Leftrightarrow\hept{\begin{cases}9a^2-45b^2-a=0\\-20a^2+100b^2+5b=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}9a^2-45b^2-a=0\\-9a^2+45b^2+\frac{9}{4}b=0\end{cases}\Leftrightarrow\hept{\begin{cases}9a^2-45b^2-a=0\\a=\frac{9}{4}b\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}a=\frac{9}{4}b\\b^2-4b=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=9\\b=4\end{cases}}}\)hoặc \(\hept{\begin{cases}a=0\\b=0\end{cases}}\)( Loại vì không thoả mãn đk (*))

=> a=9;b=4.

\(\frac{a^4+b^4}{2}\ge ab^3+a^3b-a^2b^2\)

\(\Leftrightarrow a^4+b^4-2ab^3-2a^3b+2a^2b^2\ge0\)

\(\Leftrightarrow a^3\left(a-2b\right)-b^3\left(a-2b\right)+2a^2b^2\ge0\)

\(\Leftrightarrow\left(a-2b\right)\left(a-b\right)\left(a^2+ab+b^2\right)+2a^2b^2\ge0\left(1\right)\)

Do BĐT trên đối xứng,ko mất tính tổng quát giả sử \(a\le b\)

Khi đó \(\left(a-2b\right)\left(a-b\right)\left(a^2+2ab+b^2\right)\ge0\)

\(\Rightarrow\left(1\right)\ge0\left(true\right)\)

P/S:E ko bt chỗ giả sử có đúng ko nx:(((

\(\left(a-b\right)\left(a-2b\right)\left(a^2+ab+b^2\right)\ge0\) ạ.em viết nhầm:(((

a. ĐK \(\hept{\begin{cases}a\ge0\\a\ne4\\a\ne9\end{cases}}\)

P=\(\frac{2\sqrt{a}-9-\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)+\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

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

\(=\frac{a-\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}+1}{\sqrt{a}-3}\)

b. P = \(\frac{\sqrt{a}+1}{\sqrt{a}-3}=1+\frac{4}{\sqrt{a}-3}\)

P nguyên \(\sqrt{a}-3\inƯ\left(4\right)\Rightarrow\sqrt{a}-3\in\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow\sqrt{a}\in\left\{1;2;4;5;7\right\}\Rightarrow a\in\left\{1;4;16;25;49\right\}\)

c. \(P< 1\Rightarrow P-1< 0\Rightarrow\frac{\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}-3}< 0\Rightarrow\frac{4}{\sqrt{a}-3}< 0\)

\(\Rightarrow0\le a< 9\)và \(a\ne4\)