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đặt biểu thức đó là X
ta có :
\(3X=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3X-X=1-\frac{1}{729}\)
\(\Rightarrow X=\frac{728}{729}.\frac{1}{2}=\frac{364}{729}\)
\(b,\)Đặt \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37\cdot38\cdot39}\)
\(B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{37.38\cdot38}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2B=\frac{1}{1.2}-\frac{1}{38.39}\)
\(\Rightarrow B=\frac{\left(\frac{1}{1.2}-\frac{1}{38.39}\right)}{2}=\frac{185}{741}\)
\(a)\) \(427-98=329\)
\(b)\) \(2\cdot19\cdot15+3\cdot43\cdot10+62\cdot80\)
\(=\left(2\cdot15\right)\cdot19+\left(3\cdot10\right)\cdot43+62\cdot80\)
\(=30\cdot19+30\cdot43+62\cdot80\)
\(=30\cdot\left(19+43\right)+62\cdot80\)
\(=30\cdot62+62\cdot80\)
\(=62\cdot\left(30+80\right)\)
\(=62\cdot110=6820\)
\(c)\) 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 \(M=\frac{364}{729}\)
a) Ta có: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}=1\text{-}\frac{1}{3}+\frac{1}{3}\text{-}\frac{1}{5}+...+\frac{1}{11}\text{-}\frac{1}{13}=1\text{-}\frac{1}{13}=\frac{12}{13}\)
Thay vào ta có:
\(\frac{12}{13}+x=\frac{24}{13}\Rightarrow x=\frac{24}{13}\text{-}\frac{12}{13}\Rightarrow x=\frac{12}{13}\)
a) \(\frac{1}{9}+3,25+5\frac{3}{16}+4\frac{1}{3}+2,8+0,5=\frac{1}{9}+\frac{13}{4}+\frac{83}{16}+\frac{13}{3}+\frac{14}{5}+\frac{1}{2}\)
\(=\frac{11651}{720}\)
B) \(2\frac{1}{3}+0,45+4,25+\frac{1}{81}+6\frac{8}{27}=\frac{7}{3}+\frac{9}{20}+\frac{17}{4}+\frac{1}{81}+\frac{170}{27}\)
\(=\frac{10807}{810}\)
C) \(1,25+2\frac{1}{4}+4\frac{2}{5}+0,3+2,14+4\frac{1}{8}=\frac{5}{4}+\frac{9}{4}+\frac{22}{5}+\frac{3}{10}+\frac{107}{50}+\frac{33}{8}\)
\(=\frac{2893}{200}\)
CHÚC BN HỌC TỐT!!!!!
Bài 1 mik học xong quên hết òi (mấy bài kia là hok biết luôn :V)
Bài 1.
b) \(\frac{5+55+555+5555}{9+99+999+9999}\)
= \(\frac{5\left(1+11+111+1111\right)}{9\left(1+11+111+1111\right)}=\frac{5}{9}\)
c) \(39,2\cdot27+39,2\cdot43+78,4\cdot15\)
= \(39,2\cdot27+39,2\cdot43+39,2\cdot2\cdot15\)
= \(39,2\left(27+43+30\right)=39,2\cdot100=3920\)
d) \(\frac{4}{17}\cdot\frac{3}{11}+\frac{8}{11}\cdot\frac{4}{17}-\frac{4}{17}\)
= \(\frac{4}{17}\left(\frac{3}{11}+\frac{8}{11}-1\right)=\frac{4}{17}\cdot0=0\)
Bài 2.
a) \(\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{57\cdot59}\)
= \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{57}-\frac{1}{59}\)
= \(\frac{1}{5}-\frac{1}{59}=\frac{54}{295}\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)-\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)
= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\)
= \(\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
c) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2012}\right)\)
= \(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2011}{2012}=\frac{1}{2012}\)
Câu a
\(S=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{2019-2017}{2017x2019}.\)
\(S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}=1-\frac{1}{2019}=\frac{2018}{2019}\)
Câu b
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^6}+\frac{1}{3^7}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^5}+\frac{1}{3^6}\)
\(2A=3A-A=1-\frac{1}{3^7}\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^7}\)