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\(A=\frac{7^{2011}+1}{7^{2013}+1}\)
\(7^2.A=\frac{7^{2013}+49}{7^{2013}+1}=\frac{7^{2013}+1+48}{7^{2013}+1}=\)\(\frac{7^{2013}+1}{7^{2013}+1}+\frac{48}{7^{2013}+1}=1\frac{48}{7^{2013}+1}\)
\(B=\frac{7^{2013}+1}{7^{2015}+1}\)
\(7^2.B=\)\(=\frac{7^{2015}+49}{7^{2015}+1}=\)\(\frac{7^{2015}+1+48}{7^{2015}+1}=\)\(\frac{7^{2015}+1}{7^{2015}+1}+\frac{48}{7^{2015}+1}=1\frac{48}{7^{2015}+1}\)
\(Vì\) \(1\frac{48}{7^{2013}+1}>1\frac{48}{7^{2013}+1}\)\(\Rightarrow7^2.A>7^2.B\)\(\Rightarrow A>B\)
\(Vậy\) \(A>B\)
Bài 2 nè
ta xét B trước:
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..\)\(.....+\frac{1}{2015}-\frac{1}{2016}\)
=\(\left(\frac{1}{1}+\frac{1}{3}+....+\frac{1}{2015}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{2016}\right)\)
\(=\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}\right)-\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
vậy A:B\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)\(:\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
\(=1\)
Gần đúng được không
\(A=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)cách làm \(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\) tách hết ra bạn thấy cái giữa tự triệt tiêu nhau
\(B=\frac{5}{1.3}+...+\frac{5}{23.25}\) { nếu đúng là \(\frac{5}{23.5}\) thì làm đến \(\frac{5}{21.23}\) rồi cộng lẻ cái cuối
\(\frac{2B}{5}=\frac{2}{1.3}+...+\frac{1}{23.25}=1-\frac{1}{25}\) cách làm giống (a)
\(\frac{2}{5}B=1-\frac{1}{25}=\frac{24}{25}\Rightarrow B=\frac{24}{25}.\frac{5}{2}=\frac{12}{5}\)
A=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
=\(\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{19}-\frac{1}{20}\right)\)
=\(\frac{1}{2}-\frac{1}{20}=\frac{10-1}{20}=\frac{9}{20}\)
áp dụng công thức nhân chéo của phân số (A/B=C/D = AxD=BxC)
a ) \(\Rightarrow3\times x=1\times\left(-18\right)\)
\(\Rightarrow x=-18\div3\)
\(\Rightarrow x=-6\)
KL x=-6
b) \(\Rightarrow x\times\left(-3\right)=\left(-2\right)\times\left(-15\right)\)
\(\Rightarrow x=30\div\left(-3\right)\)
\(\Rightarrow x=-6\)
\(\Rightarrow\left(-6\right)\times y=200\times15\)
\(\Rightarrow y=3000\div\left(-6\right)\)
\(\Rightarrow y=-500\)
KL x=-6 ; y = -500
.........................
= \(\frac{1}{2}\). ( \(\frac{2}{1.3}\) + \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) ... + \(\frac{2}{x.\left(x+2\right)}\) )
= \(\frac{1}{2}\) . ( 1 - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{7}\) + ... + \(\frac{1}{x}\)- \(\frac{1}{x+2}\) )
= ................
Bạn tự làm tiếp nhé ! Chúc bạn học tốt :)
Ta có:\(\frac{2019}{1}\)+\(\frac{2018}{2}\)+...+\(\frac{2}{2018}\)+\(\frac{1}{2019}\)
= (1+1+1+...+1) +\(\frac{2018}{2}\)+...+\(\frac{2}{2018}\)+\(\frac{1}{2019}\)
=\((\)1+\(\frac{2018}{2}\)\()\)+...+\((\)1+\(\frac{2}{2018}\)\()\)+\((\)1+\(\frac{1}{2019}\)\()\) +1
=\(\frac{2020}{2}\) +...+\(\frac{2020}{2018}\)+\(\frac{2020}{2019}\)+\(\frac{2020}{2020}\)
=2020\(\times\)\((\)\(\frac{1}{2}\)+...+\(\frac{1}{2018}\)+\(\frac{1}{2019}\)+\(\frac{1}{2020}\)\()\)
\(\)\(\Rightarrow\)G=2020\(\times\)\((\)\(\frac{1}{2}\)+...+\(\frac{1}{2018}\)+\(\frac{1}{2019}\)+\(\frac{1}{2020}\)\()\)\(\div\)\((\)\(\frac{1}{2}\)+\(\frac{1}{3}\)+...+\(\frac{1}{2019}\)+\(\frac{1}{2020}\)\()\)
G=2020
Vậy G=2020
n=\(\frac{2}{3}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\times\frac{98}{99}\)
n=\(\frac{196}{297}\)
Câu \(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{2}{99.100}\)Bạn viết \(\frac{3}{99.100}=\frac{2}{99.100}\)mik sửa lại nhé.
\(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.100}\)
\(M=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{100-99}{99.100}\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)
\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{97.99}\)
\(N=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{99-97}{97.99}\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{3}{2}.\frac{98}{99}=\frac{49}{33}\)
Ta thấy : \(\frac{297}{200}>\frac{49}{33}\Rightarrow M>N\)