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Ta có : \(A=\frac{10^{2016}-1}{10^{2017}-11}\)
\(\Leftrightarrow10.A=\frac{10.\left(10^{2016}-1\right)}{10^{2017}-11}=\frac{10^{2017}-10}{10^{2017}-11}\)
\(=\frac{10^{2017}-11+1}{10^{2017}-11}=1+\frac{1}{10^{2017}-11}\)
Lại có : \(B=\frac{10^{2016}+1}{10^{2017}+9}\)
\(\Leftrightarrow10.B=\frac{10\left(10^{2016}+1\right)}{10^{2017}+9}=\frac{10^{2017}+10}{10^{2017}+9}\)
\(=\frac{10^{2017}+9+1}{10^{2017}+9}=1+\frac{1}{10^{2017}+9}\)
Do : \(10^{2017}-11< 10^{2017}+9\) \(\Rightarrow\frac{1}{10^{2017}-11}>\frac{1}{10^{2017}+9}\)
\(\Rightarrow1+\frac{1}{10^{2017}-11}>1+\frac{1}{10^{2017}+9}\)
hay \(A>B\)
Vậy : \(A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Với n > 0 Ta có:
\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)
\(=\sqrt{n+1}+\sqrt{n}\)
\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)
\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)
\(\sqrt{16}+\sqrt{9}=3+4=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)7/23<11/28
b)2014/2015+2015/2016>2014+2015/2015+2016
c) A= gì vậy
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\(\frac{1+\sqrt{3}}{\sqrt{3}-1}=\frac{\left(1+\sqrt{3}\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=2+\sqrt{3}\)
\(\frac{2}{\sqrt{2}-1}=\frac{2\sqrt{2}+2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=2\sqrt{2}+2=\sqrt{8}+2\)
\(\Rightarrow\frac{2}{\sqrt{2}-1}>\frac{1+\sqrt{3}}{\sqrt{3}-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Từ \(\frac{a}{1+a}+\frac{2b}{1+b}+\frac{3c}{1+c}+\frac{5d}{1+d}\le1\)
\(\Rightarrow1-\frac{a}{1+a}+2-\frac{2b}{1+b}+3-\frac{3c}{1+c}+5-\frac{5d}{1+d}\ge10\)
\(\Rightarrow\frac{1}{1+a}+\frac{2}{1+b}+\frac{3}{1+c}+\frac{5}{1+d}\ge10\)
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a+1}\ge\)\(\frac{2b}{1+b}+\frac{3c}{1+c}+\frac{5d}{1+d}\ge10\sqrt[10]{\frac{b^2c^3d^5}{\left(1+b\right)^2\left(1+c\right)^3\left(1+d\right)^5}}\)
Và \(\frac{1}{1+b}\ge\)\(\frac{a}{1+a}+\frac{b}{b+1}+\frac{3c}{c+1}+\frac{5d}{d+1}\)
\(\ge10\sqrt[10]{\frac{abc^3d^5}{\left(1+a\right)\left(1+b\right)\left(1+c\right)^3\left(1+d\right)^5}}\)
Và \(\frac{1}{1+c}\ge\frac{a}{1+a}+\frac{2b}{b+1}+\frac{2c}{c+1}+\frac{5d}{d+1}\)
\(\ge10\sqrt[10]{\frac{ab^2c^2d^5}{\left(1+a\right)\left(1+b\right)^2\left(1+c\right)^2\left(1+d\right)^5}}\)
Và \(\frac{1}{1+d}\ge\frac{a}{a+1}+\frac{2b}{b+1}+\frac{3c}{c+1}+\frac{4d}{d+1}\)
\(\ge10\sqrt[10]{\frac{ab^2c^3d^4}{\left(1+a\right)\left(1+b\right)^2\left(1+c\right)^3\left(1+d\right)^4}}\)
Nhân theo vế 4 BĐT có: \(\frac{1}{\left(1+a\right)\left(1+b\right)^2\left(1+c\right)^3\left(1+d\right)^5}\)
\(\ge10^{1+2+3+5}\sqrt[10]{\frac{a^{2+3+5}b^{2+2+6+10}c^{3+6+6+15}d^{5+10+15+20}}{\left(1+a\right)^{10}\left(1+b\right)^{20}\left(1+c\right)^{30}\left(1+d\right)^{50}}}\)
Tương đương với \(ab^2c^3d^5\le\frac{1}{10^{11}}\) (ĐPCM)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Có: \(2>1>0\)
\(\Rightarrow\sqrt{2}>1\Rightarrow1+\sqrt{2}>1+1\\ \Leftrightarrow1+\sqrt{2}>2\)
b) Có: \(0< \sqrt{3}< 3\)
\(\Rightarrow3+1>\sqrt{3}+1\\ \Rightarrow4>\sqrt{3}+1\)
c) Có: \(0< \sqrt{11}< \sqrt{25}\left(0< 11< 25\right)\)
\(\Rightarrow\sqrt{11}< 5\\ \Rightarrow-2\sqrt{11}>-2.5=-10\left(-2< 0\right)\)
d) Có: \(0< \sqrt{11}< \sqrt{16}=4\left(do.0< 11< 16\right)\)
\(\Rightarrow3\sqrt{11}< 3.4\\ \Leftrightarrow3\sqrt{11}< 12\)
a: 2=1+1<1+căn 2
b: 4=1+3>1+căn 3
c: -2căn 11=-căn 44
-10=-căn 100
mà 44<100
nên -2 căn 11>-10
d: 12=3*4=3*căn 16>3*căn 11