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a, ĐK :a >= 3
\(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{\left(a-3\right)\left(a+3\right)}+6\sqrt{\left(a-3\right)\left(a+3\right)}=0\)
\(\Leftrightarrow\sqrt{a-3}\left(5-\frac{14}{3}-\sqrt{a+3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{a-3}=0\\\sqrt{a+3}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{2}{9}\left(loai\right)\end{cases}}\)
b, \(ĐK:x\ge-\frac{1}{2}\)
\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\frac{4}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\sqrt{2x+1}=3\)
\(\Leftrightarrow x=4\left(tm\right)\)
a) đk: \(a\ge3\)
pt \(\Leftrightarrow25\frac{\sqrt{a-3}}{\sqrt{25}}-7\frac{\sqrt{4\left(a-3\right)}}{\sqrt{9}}-7\sqrt{a^2-9}+18\frac{\sqrt{9\left(a^2-9\right)}}{\sqrt{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{7.2}{3}\sqrt{a-3}-7\sqrt{a^2-9}+\frac{18.3}{9}\sqrt{a^2-9}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}=\sqrt{a^2-9}\)
\(\Leftrightarrow\frac{1}{9}\left(a-3\right)=a^2-9\)
\(\Leftrightarrow a^2-\frac{1}{9}a-\frac{26}{3}=0\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{26}{9}\left(loại\right)\end{cases}}\)
a) ĐKXĐ: \(\hept{\begin{cases}x-9\ne0\\\sqrt{x}\ge0\\\sqrt{x}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ge0\\x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne9\\x>0\end{cases}}}\)
\(A=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{x+\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}+3}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{x-9}\)
b) \(x=\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(\Leftrightarrow x=\sqrt{4+4\sqrt{2}+2}-\sqrt{2+2\sqrt{2}+1}\)
\(\Leftrightarrow x=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(\Leftrightarrow x=\left|2+\sqrt{2}\right|-\left|\sqrt{2}+1\right|\)
\(\Leftrightarrow x=2+\sqrt{2}-\sqrt{2}-1=1\left(TM\right)\)
Vậy với x= 1 thì giá trị của biểu thức \(A=\frac{\left(1+1\right)\left(1-3\right)}{1-9}=\frac{2.\left(-2\right)}{-8}=\frac{-4}{-8}=\frac{1}{2}\)
c)
Ta có :
\(\frac{x-9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\)
+) \(\frac{1}{A}\)nguyên
\(\Leftrightarrow1+\frac{2}{\sqrt{x}+1}\)nguyên
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\)
\(\Leftrightarrow x=1\)
Vậy ..............