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ĐKXĐ: ...
\(P=\left(\frac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(P=\left(\frac{x+\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{\left(\sqrt{x}-3\right)}{\sqrt{x}}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)
\(x=5+\sqrt{2}-4-\sqrt{2}=1\)
\(\Rightarrow P=\frac{1+1}{1+3}=\frac{1}{2}\)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}+3}=1-\frac{2}{\sqrt{x}+3}\)
Do \(\sqrt{x}>0\Rightarrow\sqrt{x}+3>3\Rightarrow\frac{2}{\sqrt{x}+3}< \frac{2}{3}\)
\(\Rightarrow P>1-\frac{2}{3}=\frac{1}{3}\) (đpcm)
a.
\(B=\left(\frac{x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\\ =\left(\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\\ =\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}}\\ =\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
b. Ta có :
\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\\ =\sqrt{25+2\cdot5\cdot\sqrt{2}+2}-\sqrt{16+2\cdot4\cdot\sqrt{2}+2}\\ =\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\\ =5+\sqrt{2}-4-\sqrt{2}=1\)
\(B=\frac{\sqrt{x}+1}{\sqrt{x}+3}=\frac{1+1}{1+3}=\frac{2}{4}=\frac{1}{2}\)
c. Giả sử B>\(\frac{1}{3}\), ta có
\(B=\frac{\sqrt{x}+1}{\sqrt{x}+3}>\frac{1}{3}\\ \Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}+3}-\frac{1}{3}>0\\ \Leftrightarrow\\\frac{3\left(\sqrt{x}+1\right)-\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow\frac{2\sqrt{x}}{3\left(\sqrt{x}+3\right)}>0\left(luondungvoix>0\right)\)
Vậy.........
\(P=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
ĐKXĐ:\(x\ge0;x\ne9\)
\(=\left(\frac{x+3}{x-9}+\frac{1\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x+3}\right)}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\left(\frac{x+3+\sqrt{x}-3}{x-9}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{x+\sqrt{x}}{x-9}.\frac{\sqrt{x-3}}{\sqrt{x}}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b)
\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(=\sqrt{5^2+2.5\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{4^2+2.4\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)
\(=5+\sqrt{2}-4-\sqrt{2}\)
\(=1\)
Thay x=1 vào P ta có:
\(P=\frac{\sqrt{1}+1}{\sqrt{1}-3}\)
\(=\frac{2}{-2}=-1\)
huhu cảm ơn cậu nhiều lắm