ko biet

2, a, \(a+\dfrac{1}{a}\ge2\)

\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)

\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)

\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)

vậy...................

Câu 1:

\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)

\(=\sqrt{4+5}=3\)

\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

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

\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

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

Lời giải:

a) Ta thấy: \(a+b-2\sqrt{ab}=(\sqrt{a}-\sqrt{b})^2\geq 0, \forall a,b>0\)

\(\Rightarrow a+b\geq 2\sqrt{ab}>0\Rightarrow \frac{1}{a+b}\le \frac{1}{2\sqrt{ab}}\).

Vì $a> b$ nên dấu bằng không xảy ra . Tức \(\frac{1}{a+b}< \frac{1}{2\sqrt{ab}}\)

Ta có đpcm

b)

Áp dụng kết quả phần a:

\(\frac{1}{3}=\frac{1}{1+2}< \frac{1}{2\sqrt{2.1}}\)

\(\frac{1}{5}=\frac{1}{3+2}< \frac{1}{2\sqrt{2.3}}\)

\(\frac{1}{7}=\frac{1}{4+3}< \frac{1}{2\sqrt{4.3}}\)

.....

\(\frac{1}{4021}=\frac{1}{2011+2010}< \frac{1}{2\sqrt{2011.2010}}\)

Do đó:

\(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}\)

\(< \frac{\sqrt{2}-\sqrt{1}}{2\sqrt{2.1}}+\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3.2}}+\frac{\sqrt{4}-\sqrt{3}}{2\sqrt{4.3}}+....+\frac{\sqrt{2011}-\sqrt{2010}}{2\sqrt{2011.2010}}\)

\(=\frac{1}{2}-\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}-\frac{1}{2\sqrt{3}}+...+\frac{1}{2\sqrt{2010}}-\frac{1}{2\sqrt{2011}}\)

\(=\frac{1}{2}-\frac{1}{2\sqrt{2011}}< \frac{1}{2}\) (đpcm)

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

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

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

b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)

\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)

\(=\dfrac{a}{a-b}\)

khúc \(\dfrac{a}{a-b}\) sai nhé

\(=\dfrac{a-b}{a-b}=1\)

a: \(a+\dfrac{1}{a}\ge2\sqrt{a\cdot\dfrac{1}{a}}=2\)

b: \(\Leftrightarrow\dfrac{a^2+a+1+1}{\sqrt{a^2+a+1}}>=2\)

=>\(\sqrt{a^2+a+1}+\dfrac{1}{\sqrt{a^2+a+1}}>=2\)(1)

\(\sqrt{a^2+a+1}+\dfrac{1}{\sqrt{a^2+a+1}}>=2\sqrt{\sqrt{a^2+a+1}\cdot\dfrac{1}{\sqrt{a^2+a+1}}}=2\)

nên (1) đúng