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2 tháng 1 2021

CM: \(a=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}-\frac{\sqrt{2}}{8}\Rightarrow a+\frac{\sqrt{2}}{8}=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}\)

\(\Leftrightarrow\left(a+\frac{\sqrt{2}}{8}\right)^2=\left(\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}\right)^2\)\(\Leftrightarrow a^2+\frac{a\sqrt{2}}{4}+\frac{1}{32}=\frac{1}{4}\left(\sqrt{2}+\frac{1}{8}\right)\Leftrightarrow a^2+\frac{2\sqrt{a}}{4}+\frac{1}{32}=\frac{\sqrt{2}}{4}+\frac{1}{32}\)

\(\Leftrightarrow4a^2+\sqrt{2}a-\sqrt{2}=0\)

Theo trên: \(4a^2+\sqrt{2}a-\sqrt{2}=0\Rightarrow a^2=\frac{\sqrt{2}\left(1-a\right)}{4}\Rightarrow a^4=\frac{a^2-2a+1}{8}\)

\(\Rightarrow a^4+a+1=\frac{a^2-2a+1}{8}+a+1=\left(\frac{a+3}{2\sqrt{2}}\right)^2\)

\(B=a^2+\sqrt{a^4+a+1}=a^2+\frac{a+3}{2\sqrt{2}}=\frac{2\sqrt{2}a^2+a+3}{2\sqrt{2}}\)\(=\frac{4a^2+\sqrt{2}a+3\sqrt{2}}{4}=\frac{4\sqrt{2}}{4}=\sqrt{2}\)

6 tháng 8 2017

\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)

         \(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)

            \(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)

           =\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)

          \(=\frac{-8}{2}=-4\)

\(\Rightarrow A=-4\sqrt{2}\)

27 tháng 8 2016

Ta có: 

\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\frac{1-x}{\sqrt{2}}\right)^2\)

\(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(1-x\right)^2}{2}\)

\(P=\left(\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(x-1\right)^2}{2}\)

\(P=\left(\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2}{2}\)

\(P=\left(-\sqrt{x}\right)\left(\sqrt{x}-1\right)\)

\(P=\sqrt{x}-x\)

b) Để \(P>0\) thì \(\sqrt{x}-x>0\)

  • \(\sqrt{x}-x>0\)

   \(\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)

Suy ra: TH1\(\sqrt{x}< 0\) và \(1-\sqrt{x}< 0\) (Loại) vì \(\sqrt{x}\ge0\)

            TH2:\(\sqrt{x}>0\)  và \(1-\sqrt{x}>0\) (Nhận)

Ta có \(\sqrt{x}>0\) và \(1-\sqrt{x}>0\) để \(P>0\)

  • \(\sqrt{x}>0\) \(\Rightarrow x>0\)
  • \(1-\sqrt{x}>0\) \(\Rightarrow\sqrt{x}< 1\) \(\Rightarrow x< 1\)

Vậy để \(P>0\) thì \(0< x< 1\)

c)\(P=\sqrt{x}-x\)

\(P=-\left(x-\sqrt{x}\right)\)

\(P=-\left(\left(\sqrt{x}\right)^2-2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}-\frac{1}{4}\right)\)

\(P=-\left(\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\right)\)

\(P=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\)

Nên \(-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\) \(\Rightarrow x=\frac{1}{4}\)

Vậy GTLN của \(P\) là \(\frac{1}{4}\) khi \(x=\frac{1}{4}\)