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1. \(\frac{14}{45}=\frac{1}{9}+\frac{1}{5}\)
2. \(\left(1-\frac{1}{12}\right).\left(1-\frac{1}{11}\right).\left(1-\frac{1}{10}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{8}\right)\)
\(=\frac{11}{12}.\frac{10}{11}.\frac{9}{10}.\frac{8}{9}.\frac{7}{8}\)
Triệt tử với mẫu:
\(=\frac{7}{12}\)
1.ket qua la 1/5+1/9
2.=11/12x10/11x9/10x8/9x7/8
=(11x10x9x8x7)/(12x11x10x9x8)
=7/12
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{3999.4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{3999}-\frac{1}{4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{3}+...+\frac{1}{3999}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}=1\)
Ta lại có:
\(B=\frac{\left(17+1\right)\left(\frac{17}{2}+1\right)...\left(\frac{17}{19}+1\right)}{\left(1+\frac{19}{17}\right)\left(1+\frac{19}{16}\right)...\left(1+19\right)}\)
\(=\frac{\frac{18}{1}.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{\frac{36}{17}.\frac{35}{16}.\frac{34}{15}...\frac{20}{1}}\)
\(=\frac{1.2.3...36}{1.2.3...36}=1\)
Từ đây ta suy ra được
\(A-B=1-1=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bđt \(\Leftrightarrow\)\(\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)\ge\left(\frac{10}{3}\right)^3abc\) (*)
đặt \(\left(\sqrt{ab};\sqrt{bc};\sqrt{ca}\right)=\left(x;y;z\right)\)\(\Rightarrow\)\(xyz\le\frac{1}{27}\)
(*) \(\Leftrightarrow\)\(\left(x^2+1\right)\left(y^2+1\right)\left(z^2+1\right)\ge\left(\frac{10}{3}\right)^3xyz\)
\(VT\ge\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\)
Có \(xy+1\ge10\sqrt[10]{\frac{xy}{9^9}}\)
Tương tự với \(yz+1\)\(;\)\(zx+1\)\(\Rightarrow\)\(VT\ge10^3\sqrt[10]{\frac{\left(xyz\right)^2}{9^{27}}}\)
Ta cần CM \(10^3\sqrt[10]{\frac{\left(xyz\right)^2}{9^{27}}}\ge\frac{10^3}{3^3}xyz\) đúng với \(xyz\le\frac{1}{27}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
Đặt \(P=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)\)
Vì a+b+c=1 nên
\(P=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)=abc+\frac{1}{abc}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\)
Từ BĐt Cosi cho 3 số dương ta có:
\(\frac{1}{3}=\frac{a+b+c}{3}\ge\sqrt[3]{abc}\Rightarrow abc\le\frac{1}{27}\)
đặt x=abc thì \(0< x\le\frac{1}{27}\)
do đó: \(x+\frac{1}{x}-27-\frac{1}{27}=\frac{\left(27-x\right)\left(1-27x\right)}{27x}\ge0\)
=> \(x+\frac{1}{x}=abc+\frac{1}{abc}\ge27+\frac{1}{27}=\frac{730}{27}\)
Mặt khác: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)
Nên \(P\ge\frac{730}{27}+10=\frac{1000}{27}=\left(\frac{10}{3}\right)^3\)
Dấu "=" xảy ra khi a=b=c\(=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
4.x+1/2.3+1/3.4+1/4.5+1/5.6=1
4.x+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6=1
4x+1-1/6=1
4x=1-5/6
4x=1/6
x=1/6:4
x=1/24
Chúc Học Tốt!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-\frac{1}{6}\right)x\left(1-\frac{1}{7}\right)x\left(1-\frac{1}{8}\right)x\left(1-\frac{1}{9}\right)x\left(1-\frac{1}{10}\right)\)
\(=\frac{5}{6}x\frac{6}{7}x\frac{7}{8}x\frac{8}{9}x\frac{9}{10}\)
\(=\frac{1}{2}\)
=(1/1-1/6)x(1/1-1/7)x(1/1-1/8)x(1/1-19)x(1/1-1/10)
=5/6x6/6x7/8x8/9x9/10
![](https://rs.olm.vn/images/avt/0.png?1311)
Cách giải
X = 8/9 / ( 1+1/3+1/6+1/10+1/15+1/21)
1+1/3=4/3
4/3+1/6=3/2
3/2+1/10=8/5
8/5+1/15=5/3
5/3+1/21=12/7
4/3+3/2=17/6
8/5+5/3=49/15
(17/6+49/15)+17/6=183/30+17/6=286/30
8/9:286/30=2288/270=1140/135=228/27=76/9
X = 76/9
![](https://rs.olm.vn/images/avt/0.png?1311)
a) $\frac{{11}}{{12}} - \frac{1}{3} + \frac{1}{4} = \frac{{11}}{{12}} - \frac{4}{{12}} + \frac{3}{{12}} = \frac{7}{{12}} + \frac{3}{{12}} = \frac{{10}}{{12}} = \frac{5}{6}$
b) $1 - \left( {\frac{1}{6} + \frac{1}{3}} \right) = 1 - \left( {\frac{1}{6} + \frac{2}{6}} \right) = 1 - \frac{1}{2} = \frac{1}{2}$
Co j k mik nha
Sr gui lon