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\(\frac{2}{a+b\sqrt{5}}-\frac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\)
\(\Leftrightarrow\frac{2a-2b\sqrt{5}-3a-3b\sqrt{5}}{a^2-5b^2}=-9-20\sqrt{5}\)
\(\Leftrightarrow\frac{a+5b\sqrt{5}}{a^2-5b^2}=9+20\sqrt{5}\)
\(\Leftrightarrow\sqrt{5}\left(100b^2+5b-20a^2\right)=9a^2-a-45b^2\)
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 \(\hept{\begin{cases}a=9\\b=4\end{cases}\left(nhan\right)}\)
\(x=9\Rightarrow\sqrt{x}=3\Rightarrow A=\frac{3+2}{3-5}=\frac{5}{-2}=-\frac{5}{2}\\ \)
\(B=\frac{3}{\sqrt{x}+5}+\frac{20-2\sqrt{x}}{x-25}=\frac{3.\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}+\frac{20-2\sqrt{x}}{\left(x+\sqrt{5}\right).\left(x-\sqrt{5}\right)}\)
\(=\frac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}+5}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}=\frac{1}{\sqrt{x}-5}\)
\(A=B.\left|x-4\right|\Leftrightarrow\left|x-4\right|=A:B=\frac{\sqrt{x}+2}{\sqrt{x}-5}:\frac{1}{\sqrt{x}-5}=\sqrt{x}+2\)
\(\Rightarrow\left(x-4\right)^2=\left(\sqrt{x}+2\right)^2\Leftrightarrow x^2-8x+16=x+4\sqrt{x}+4\)
\(\Leftrightarrow x^2-9x-4\sqrt{x}+12=0\Leftrightarrow x.\left(x-9\right)-4.\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow x.\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)-4.\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(x\sqrt{x}+3x-4\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(\left(x\sqrt{x}-x\right)+\left(4x-4\right)\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(x.\left(\sqrt{x}-1\right)+4.\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(\sqrt{x}-1\right).\left(x+4\sqrt{x}+4\right)=0\Leftrightarrow\left(\sqrt{x}-3\right).\left(\sqrt{x}-1\right).\left(\sqrt{x}+2\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}-3=0\\\sqrt{x}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}}\)(Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\left(\sqrt{x}+2\right)^2\ge4>0\))
\(\frac{2}{a+b\sqrt{5}}-\frac{3}{a-b\sqrt{5}}=\frac{2\left(a-b\sqrt{5}\right)-3\left(a+b\sqrt{5}\right)}{\left(a+b\sqrt{5}\right)\left(a-b\sqrt{5}\right)}\)\(=\frac{-a-5b\sqrt{5}}{a^2-5b^2}=\frac{-a}{a^2-5b^2}+\frac{-5b\sqrt{5}}{a^2-5b^2}\).
Suy ra:
\(\hept{\begin{cases}\frac{-a}{a^2-5b^2}=-9\\-\frac{5b}{a^2-5b^2}=-20\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\frac{a}{b}=\frac{9}{4}\\\frac{a}{a^2-5b^2}=-9\end{cases}}\)
\(\frac{a}{b}=\frac{9}{4}\Leftrightarrow\frac{a}{9}=\frac{b}{4}=k\)\(\Rightarrow\hept{\begin{cases}a=9k\\b=4k\end{cases}}\).
Suy ra \(\frac{a}{a^2-5b^2}=\frac{9k}{81k^2-5.16k^2}=\frac{9}{k}=-9\).
Suy ra \(k=-1\).
Vậy \(\hept{\begin{cases}a=9k\\b=4k\end{cases}\Leftrightarrow\hept{\begin{cases}a=-9\\b=-4\end{cases}}}\).
a;b có là số nguyên ko