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Xét bài toán :
So sánh \(\frac{a}{b}\)và \(\frac{a+m}{b+m}\)( a>b , m>0)
Có \(\frac{a}{b}=\frac{a\left(b+m\right)}{b\left(b+m\right)}=\frac{ab+am}{b\left(b+m\right)}\)
\(\frac{a+m}{b+m}=\frac{b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+bm}{b\left(b+m\right)}\)
Mà a>b => am > bm => \(\frac{ab+am}{b\left(b+m\right)}>\frac{ab+bm}{b\left(b+m\right)}\)hay \(\frac{a}{b}>\frac{a+m}{b+m}\)
Áp dụng : \(A=\frac{3^{2017}+5}{3^{2015}+5}>\frac{3^{2017}+5+4}{3^{2015}+5+4}=\frac{3^{2017}+9}{3^{2015}+9}=\frac{3^2\left(3^{2017}+9\right)}{3^2\left(3^{2015}+9\right)}\)
\(=\frac{3^{2015}+1}{3^{2013}+1}=B\)
=> A > B
\(A=\frac{99.100.101}{3}=333300\)
\(B=\frac{2015.2016.2017.2018}{4}-\frac{6.7.8.9}{4}=4133639960604\)
\(C=\frac{3^{51}-1}{3}+1\)
3A= 1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)
3A= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100
3a= 99.100.101
a, 1/2+1/2.(-3/5)
=1/2[1+(-3/5)]
=1/2.2/5
=1/5
b,1/3-(1/3.2/3-3/9)
=1/3-(2/9-3/9)
=1/3+1/9
=4/9
c,(-2 và 1/3+1/(5/4-7/3)
=(-2/3)+(-12/13)
=-62/39
d,2015/2017.3/4+3/4.2/2017+1/4
=3/4(2015/2017+2/2017)+1/4
=3/4+1/4
=1
\(A=\frac{3}{1\cdot3}+\frac{3}{2\cdot5}+\frac{3}{5\cdot7}+...+\frac{3}{2015\cdot2017}\)
\(\frac{2}{3}A=\frac{2}{3}\left(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{3}{2015\cdot2017}\right)\)
\(\frac{2}{3}A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2015\cdot2017}\)
\(\frac{2}{3}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(\frac{2}{3}A=1-\frac{1}{2017}\)
\(\frac{2}{3}A=\frac{2016}{2017}\)
\(A=\frac{2016}{2017}:\frac{2}{3}\)
\(A=\frac{2016}{2017}\cdot\frac{3}{2}\)
\(A=1,499256321\)
A x 2/3=2/3 x(3/1x3+3/3x5+...+3/2015x2017)
A x 2/3=2/1x3+2/3x5+...+2/2015x2017
A x 2/3=1-1/3+1/3-1/5+...+1/2015-1/2017
A x 2/3=1-1/2017
A x 2/3=2016/2017
A = 3024/2017