Tìm MIN của \(P=\frac{81x^2+18225x+1}{9x}-\frac{6\sqrt{x}+8}{x+1},x>0\)
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\(A=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{6\sqrt{x}}{3\sqrt{x}+1}\)
\(A=\left[\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right].\frac{3\sqrt{x}+1}{6\sqrt{x}}\)
\(A=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{6\sqrt{x}}\)
\(A=\frac{3\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}.\frac{1}{6\sqrt{x}}\)
\(A=\frac{\sqrt{x}+1}{6\sqrt{x}-2}\)
\(A=\frac{5}{6}\Leftrightarrow\frac{\sqrt{x}+1}{6\sqrt{x}-2}=\frac{5}{6}\)
\(\Leftrightarrow6\sqrt{x}+6=30\sqrt{x}-10\)
\(\Leftrightarrow24\sqrt{x}=16\)
\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\Leftrightarrow x=\frac{4}{9}\)
\(A=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]\div\frac{6\sqrt{x}}{3\sqrt{x}+1}\)
\(A=\left[\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]\times\frac{3\sqrt{x}+1}{6\sqrt{x}}\)
\(A=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}\times\frac{1}{6\sqrt{x}}\)
\(A=\frac{3\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}\times\frac{1}{6\sqrt{x}}\)
\(A=\frac{\sqrt{x}+1}{6\sqrt{x}-2}\)
\(A=\frac{5}{6}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{6\sqrt{x}-2}=\frac{5}{6}\)
\(\Leftrightarrow6\sqrt{x}+6=30\sqrt{x}-10\)
\(\Leftrightarrow24\sqrt{x}=16\)
\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{4}{9}\)
con 6 tách trong căn thành nhân tử nhân 2 vế cho 2 rồi tách thành hđt