Trước hết , ta cần chứng minh \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\)(*) (Bạn tự chứng minh)

Đặt \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)

\(\Rightarrow2A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{79}+\sqrt{80}}\)

\(>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)

Áp dụng (*) :\(\Rightarrow2A>\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{5}-\sqrt{4}\right)+...+\left(\sqrt{80}-\sqrt{79}\right)+\left(\sqrt{81}-\sqrt{80}\right)\)

\(\Rightarrow2A>\sqrt{81}-1=8\Rightarrow A>4\)(đpcm)

\(a,A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+...+\frac{\sqrt{99}-\sqrt{100}}{99-100}\)

\(=\frac{1-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{99}-\sqrt{100}}{-1}\)

\(=\frac{1-\sqrt{100}}{-1}=9\)

\(b,B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+..+\frac{1}{\sqrt{99}}\)

\(=\frac{2}{\sqrt{1}+\sqrt{1}}+\frac{2}{\sqrt{2}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{99}}>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)\(\Rightarrow B>2\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{99}+\sqrt{100}}\right)\)

\(\Rightarrow B>2\left(\frac{\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{99}-\sqrt{100}}{-1}\right)\)

\(\Rightarrow B>2\left(\frac{1-\sqrt{100}}{-1}\right)\)

\(\Rightarrow B>2.9=18\left(ĐPCM\right)\)

\(\frac{1}{\sqrt{1}+\sqrt{2}}+....\frac{1}{\sqrt{79}+\sqrt{80}}>\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\) (40 số)

................................................................\(>\frac{40}{10}=4\) 

=>đpcm

hc tốt

ko chắc lắm :)

dhasuxbhfc;CX

lôn dit me con cac

Tổng quát ta có: Với \(n\inℕ\)ta có:

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

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

Với \(n=2\)\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2}\)

Với \(n=3\)\(\Rightarrow\frac{1}{\sqrt{3}+\sqrt{4}}=\sqrt{4}-\sqrt{3}\)

...........................

Với \(n=79\)\(\Rightarrow\frac{1}{\sqrt{79}+\sqrt{80}}=\sqrt{80}-\sqrt{79}\)

\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+.....+\frac{1}{\sqrt{79}+\sqrt{80}}\)

\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+......+\sqrt{80}-\sqrt{79}\)

\(=\sqrt{80}-\sqrt{2}=\sqrt{40.2}-\sqrt{2}=\sqrt{40}.\sqrt{2}-\sqrt{2}\)

\(=\sqrt{2}.\left(\sqrt{40}-1\right)>\sqrt{2}.\left(\sqrt{36}-1\right)\)

\(=\sqrt{2}.\left(6-1\right)=5\sqrt{2}>4\)( đpcm )

2/

a/ \(\sqrt{a}+\frac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\frac{1}{\sqrt{a}}}=2\), dấu "=" khi \(a=1\)

b/ \(a+b+\frac{1}{2}=a+\frac{1}{4}+b+\frac{1}{4}\ge2\sqrt{a.\frac{1}{4}}+2\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}\)

Dấu "=" khi \(a=b=\frac{1}{4}\)

c/ Có lẽ bạn viết đề nhầm, nếu đề đúng thế này thì mình ko biết làm

Còn đề như vậy: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\) thì làm như sau:

\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\) ; \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\); \(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\)

Cộng vế với vế ta được:

\(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\ge\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{yz}}+\frac{2}{\sqrt{xz}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)

Dấu "=" khi \(x=y=z\)

d/ \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\)

\(=\frac{7+4\sqrt{3}}{3-4}-\frac{7-4\sqrt{3}}{3-4}=-7-4\sqrt{3}+7-4\sqrt{3}=-8\sqrt{3}\)

e/ \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\frac{\left(a-b\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{ab}}\)

\(=\frac{a^2-b^2}{\sqrt{ab}}-\left(a-b\right)\) (bạn chép đề sai)

@Akai Haruma Cô giúp em với ạ!!!

? Cosi thôi câu 1 2 phần II