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\(b,\)Đặt \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37\cdot38\cdot39}\)
\(B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{37.38\cdot38}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2B=\frac{1}{1.2}-\frac{1}{38.39}\)
\(\Rightarrow B=\frac{\left(\frac{1}{1.2}-\frac{1}{38.39}\right)}{2}=\frac{185}{741}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A= 1/3+1/9+1/27+1/81+1/243
A= 1/3+1/3^2+1/3^3+1/3^4+1/3^5
3A=1+1/3+1/3^2+1/3^3+1/3^4
3A-A=1+1/3+1/3^2+1/3^3+1/3^4-1/3-1/3^2-1/3^3-1/3^4-1/3^5
2A=1-1/3^5
2A=242/243
A=121/243
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đặt biểu thức đó là X
ta có :
\(3X=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3X-X=1-\frac{1}{729}\)
\(\Rightarrow X=\frac{728}{729}.\frac{1}{2}=\frac{364}{729}\)
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Muốn cho số có hai chữ số giống nhau và chia hết cho 2 thì số đó phải là một trong các số 22, 44, 66, 88. Bây giờ ta tìm trong những số này số mà chia cho 5 thì dư 3.
Đó là số 88.
Xem thêm tại: http://loigiaihay.com/bai-99-trang-39-sgk-toan-6-tap-1-c41a3896.html#ixzz4xczZ4dOb
Đặt biểu thức là \(A\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2015}}\)
\(3A=3\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2015}}\right)\)
\(3A=3+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2016}}\)
\(2A=3A-A=\left(3+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2016}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2015}}\right)\)
\(2A=2+\frac{1}{3}+\frac{1}{3^{2016}}\)
\(A=\left(2+\frac{1}{3}+\frac{1}{3^{2016}}\right):2\)
\(A=\frac{1}{2}\left(2+\frac{1}{3}+\frac{1}{3^{2016}}\right)\)
\(A=1+\frac{1}{6}+\frac{1}{3^{2016}.2}\)
\(A=\frac{7}{6}+\frac{1}{3^{2016}.2}\)
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\(a)\) \(427-98=329\)
\(b)\) \(2\cdot19\cdot15+3\cdot43\cdot10+62\cdot80\)
\(=\left(2\cdot15\right)\cdot19+\left(3\cdot10\right)\cdot43+62\cdot80\)
\(=30\cdot19+30\cdot43+62\cdot80\)
\(=30\cdot\left(19+43\right)+62\cdot80\)
\(=30\cdot62+62\cdot80\)
\(=62\cdot\left(30+80\right)\)
\(=62\cdot110=6820\)
\(c)\) Đặt \(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(\Rightarrow3M-M=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)\)
\(\Rightarrow2M=1-\frac{1}{3^6}\)
\(\Rightarrow M=\frac{728}{2\cdot729}=\frac{364}{729}\)
Vậy \(M=\frac{364}{729}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2009}}+\frac{1}{2^{2010}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2008}}+\frac{1}{2^{2009}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2008}}+\frac{1}{2^{2009}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2009}}+\frac{1}{2^{2010}}\right)\)
\(A=1-\frac{1}{2^{2010}}\)
\(A=\frac{2^{2010}-1}{2^{2010}}\)
Vậy \(A=\frac{2^{2010}-1}{2^{2010}}\)
Chúc bạn học tốt
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\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ =\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\\ =>3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\\ =>3A-A=2A=1-\dfrac{1}{3^5}\\ =>A=\dfrac{1-\dfrac{1}{3^5}}{2}=\dfrac{3^5-1}{2.3^5}\)
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\(\frac{\frac{2}{45}-\frac{4}{13}-\frac{1}{3}}{\frac{3}{13}-\frac{4}{15}+\frac{2}{3}}=\frac{\frac{-349}{585}}{\frac{41}{65}}=\frac{-349}{585}:\frac{41}{65}=\frac{-349}{585}\cdot\frac{65}{41}=-3\frac{52}{99}\)
\(=3^2.\frac{1}{3^5}.3^8.\frac{1}{3^3}\)
\(=9\)