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Dựng tam giác vuông cân ABC có \(AB=AC=1\); \(BC=\sqrt{2}\)

Dựng phân giác BD của góc B \(\Rightarrow\widehat{ABD}=\frac{45}{2}=22,5^0\)

Theo t/c phân giác: \(\frac{AD}{AB}=\frac{CD}{BC}\Rightarrow CD=\sqrt{2}AD\)

Mà \(AD+CD=AB\Rightarrow AD+\sqrt{2}AD=1\Rightarrow AD=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\)

\(BD=\sqrt{AB^2+BD^2}=\sqrt{1+\left(\sqrt{2}-1\right)^2}=\sqrt{4-2\sqrt{2}}\)

\(\Rightarrow sin22,5^0=sin\widehat{ABD}=\frac{AD}{BD}=\frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}}\)

\(A=\left(\dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}-\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(\dfrac{\sqrt{3}-1}{3\sqrt{2}-\sqrt{6}}\right)\)

\(=\dfrac{5+2\sqrt{6}-5+2\sqrt{6}}{-1}\cdot\dfrac{1}{\sqrt{6}}\)

=-4

=5+2√6−5+2√6−1⋅1√6=5+26−5+26−1⋅16

=-4

Ta có: \(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-3x+8\sqrt{x}-5-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

\(\Leftrightarrow A-\dfrac{2}{3}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{2}{3}\)

\(\Leftrightarrow A-\dfrac{2}{3}=\dfrac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow A-\dfrac{2}{3}=\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\le0\)

\(\Leftrightarrow A\le\dfrac{2}{3}\)

\(P=\left(\dfrac{x-2+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)