Bài này dễ,ông không chịu làm thì có ^_^:

Ta có:\(B=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+....+\left(\frac{1}{2^{2014}+1}+....+\frac{1}{2^{2015}}\right)+\frac{1}{2^{2015}+1}+...+\frac{1}{2^{2016}-1}\)

\(>1+\frac{1}{2}+2.\frac{1}{2^2}+2^2.\frac{1}{2^3}+........+2^{2014}.\frac{1}{2^{2015}}\)

\(=1+\frac{1}{2}+\frac{1}{2}+.........+\frac{1}{2}\)  (có 2015 phân số  \(\frac{1}{2}\))

\(=1+2014.\frac{1}{2}+\frac{1}{2}=1008+\frac{1}{2}>1008\)

Lời giải:

$B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}$

$2B=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2015}}$

Trừ theo vế:

$2B-B=1-\frac{1}{2^{2016}}$

$B=1-\frac{1}{2^{2016}}< 1$ (đpcm)

b)\(\left(2016.1017+2017.2018\right).\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)

\(\left(2016.2017+2017.2018\right)\left(1+\frac{1}{3}-\frac{4}{3}\right)\)

\(\left(2016.2017+2017.2018\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)

\(\left(2016.2017+2017.2018\right).0\)

\(=0\)

a) \(1001.789+456.128.128-789+912.436\)

\(=\left(1001.789-789\right)+\left(456.2.64.128+912.436\right)\)

\(=789.1000+912.4\left(2048+109\right)\)

\(=789000+912.4.2157\)

\(=8657736\)

c)\(5\frac{9}{10}+\frac{3}{2}-\left(2\frac{1}{3}.4\frac{1}{2}-2.2\frac{1}{3}\right):\frac{7}{4}\)

\(=\frac{59}{10}+\frac{3}{2}-\left(\frac{7}{3}.\frac{9}{2}-2.\frac{7}{3}\right):\frac{7}{4}\)

\(=\frac{59}{10}+\frac{3}{2}-\left[\frac{7}{3}\left(\frac{9}{2}-2\right)\right]:\frac{7}{4}\)

\(=\frac{59}{10}+\frac{3}{2}-\left(\frac{7}{3}.\frac{5}{2}\right):\frac{7}{4}\)

\(=59+\frac{3}{2}-\frac{35}{6}.\frac{4}{7}\)

\(=\frac{59}{10}+\frac{3}{2}-\frac{10}{3}\)

\(=\frac{177+45-100}{30}=\frac{122}{30}=\frac{61}{15}\)

Chứng minh rổng quát, Nếu:

\(A=\frac{1}{a^{2.k}}-\frac{1}{a^{2.\left(k+1\right)}}+\frac{1}{a^{2.\left(k+2\right)}}-\frac{1}{a^{2.\left(k+3\right)}}+...+\frac{1}{a^{2.\left(k+n\right)}}-\frac{1}{a^{2.\left(k+n+1\right)}}\) (a;b \(\in\) N*)

\(a^{2.k}.A=1-\frac{1}{a^{2.k}}+\frac{1}{a^{2.\left(k+1\right)}}-\frac{1}{a^{2.\left(k+2\right)}}+...+\frac{1}{a^{2.\left(k+n-1\right)}}-\frac{1}{a^{2.\left(k+n\right)}}\)

\(a^{2.k}.A+A=\left(1-\frac{1}{a^{2.k}}+\frac{1}{a^{2.\left(k+1\right)}}-\frac{1}{a^{2.\left(k+2\right)}}+..+\frac{1}{a^{2.\left(k+n-1\right)}}-\frac{1}{a^{2.\left(k+n\right)}}\right)-\left(\frac{1}{a^{2.k}}-\frac{1}{a^{2.\left(k+1\right)}}+\frac{1}{a^{2.\left(k+2\right)}}-\frac{1}{a^{2.\left(k+3\right)}}+..+\frac{1}{a^{2.\left(k+n\right)}}-\frac{1}{a^{2.\left(k+n+1\right)}}\right)\)

\(A.\left(a^{2.k}+1\right)=1-\frac{1}{a^{2.\left(k+n+1\right)}}< 1\)

\(A< \frac{1}{a^{2.k}+1}\)

Áp dụng vào bài toán dễ thấy a = 3; k = 1

Như vậy, \(A< \frac{1}{3^{2.1}+1}=\frac{1}{3^2+1}=\frac{1}{9+1}=\frac{1}{10}=0,1\left(đpcm\right)\)

\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{2014}}-\frac{1}{3^{2016}}\)

\(\Rightarrow9A=1-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+...+\frac{1}{3^{2012}}-\frac{1}{3^{2014}}\)

\(\Rightarrow10A=1-\frac{1}{3^{2016}}\)

\(\Rightarrow A=\frac{1-\frac{1}{3^{2016}}}{10}\)

Vì 0,1 = \(\frac{1}{10}\) nên \(\frac{1-\frac{1}{3^{2016}}}{10}< \frac{1}{10}\) hay A < 0,1

Để bài toán trông quen thuộc hơn:

Đặt a =x; \(\frac{1}{b}=y\) thì bài toán trở thành:

Cho x, y > 0 thỏa mãn x + y =1. CMR: \(\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge\frac{25}{2}\).

-------------------------------------------------------------------------

Áp dụng BĐT Cauchy-Schwarz dạng Engel:

\(VT\ge\frac{1}{2}\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2=\frac{25}{2}^{\left(đpcm\right)}\)

P/s: Is it true?

Xí, hôm qua buồn ngủ quá làm thiếu:V

\(VT\ge\frac{1}{2}\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(x+y+\frac{4}{x+y}\right)^2=\frac{25}{2}\)(đpcm)

Sử dụng một vài bất đẳng thức đơn giản:

\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{\left(a+\frac{1}{a}+b+\frac{1}{b}\right)}{2}\)

\(=\frac{\left(1+b+\frac{1}{b}\right)^2}{2}=\frac{\left(1+\frac{ab+1}{a}\right)^2}{2}\)

\(=\frac{\left(1+\frac{b}{a}\right)^2}{2}\)(1)

(Dấu "=" khi \(a+\frac{1}{a}=b+\frac{1}{b}\)và \(a+\frac{1}{b}=1\))

Ta có: \(\left(a+\frac{1}{b}\right)^2\ge4\frac{a}{b}\Leftrightarrow1\ge4\frac{a}{b}\Leftrightarrow\frac{b}{a}\ge4\)

(Dấu "=" khi \(a=\frac{1}{b}\)và \(a+\frac{1}{b}=1\)(2)

Từ (1) và (2) suy ra \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{\left(1+4\right)^2}{2}=\frac{25}{2}\)

Đẳng thức xảy ra khi \(a=\frac{1}{2};b=2\)

1/ \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\)

\(\Leftrightarrow\frac{a+b+c}{abc}=0\)(đúng)

Vậy ta có ĐPCM

2/ \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2005}+\sqrt{2006}}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2006}-\sqrt{2005}\)

\(=\sqrt{2006}-1\)

b/ Ta có

\(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}>\frac{2}{\sqrt{n}+\sqrt{n+1}}\)

\(=2\left(\sqrt{n+1}-\sqrt{n}\right)\)

Áp dụng vài bài toán ta có

\(B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{2005}}\)

\(>2.\sqrt{2}-2.\sqrt{1}+2.\sqrt{3}-2.\sqrt{2}+...+2.\sqrt{2006}-2.\sqrt{2005}\)

\(=2.\sqrt{2006}-2=2\left(\sqrt{2006}-1\right)\)