Tính \(1\frac{2}{3}\)của \(3\frac{2}{9}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}}:\frac{3}{{ - 4}}} \right).\frac{4}{5} = \left( {\frac{2}{5}.\frac{{ - 4}}{3}} \right).\frac{4}{5}\\ = \frac{{ - 8}}{{15}}.\frac{4}{5} = \frac{{ - 32}}{{75}}\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}}:\left( {\frac{7}{{ - 5}}.\frac{{ - 3}}{2}} \right) = \frac{3}{4}:\frac{{ - 21}}{{ - 10}}\\ = \frac{3}{4}.\frac{{10}}{{21}} = \frac{{30}}{{84}} = \frac{5}{14}\end{array}\)
c)
\(\begin{array}{l}\frac{{ - 1}}{9}.\frac{{ - 3}}{5} + \frac{5}{{ - 6}}.\frac{{ - 3}}{5} + \frac{5}{2}.\frac{{ - 3}}{5}.\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 1}}{9} + \frac{5}{{ - 6}} + \frac{5}{2}} \right)\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 15}}{{18}} + \frac{{45}}{{18}}} \right)\\ = \frac{{ - 3}}{5}.\frac{{28}}{{18}}\\ = \frac{{ - 3}}{5}.\frac{{14}}{9}\\ = \frac{{ - 14}}{{15}}\end{array}\)
\(\frac{24\cdot47-23}{24+47\cdot23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
\(=\frac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\frac{3\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}\)
\(=\frac{24^2+24\cdot23-23}{24+24\cdot23+23^2}\cdot\frac{3}{9}\) \(=\frac{24^2+23\cdot\left(24-1\right)}{\left(23+1\right)\cdot24\cdot23^2}\cdot\frac{1}{3}=1\cdot\frac{1}{3}=\frac{1}{3}\)
\(\left[9-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
có 9 số 1 có 9 số hạng
\(=\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=\left[\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=1\)
tử số cộng thêm 1 vào mỗi phân số ở tử
là ra ngay nha bạn
k mình nha
=182.\(\orbr{\begin{cases}1.\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)\\2.\left(\frac{1}{2}+\frac{1}{9}+\frac{1}{27}\right)\end{cases}}:\frac{4.\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1.\left(\frac{1}{3}+\frac{1}{49}-\frac{1}{343}\right)}:\frac{91}{80} \)
=.\(182.\left(\frac{1}{2}:\frac{4}{1}\right).\frac{91}{80}\)
=\(182.\frac{1}{8}.\frac{91}{80}\)
=.\(182.\frac{91}{640}\)
=\(\frac{8281}{320}\)
\(=182.\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2.\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4.\left(1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}}\right]:\frac{919191}{808080}\)
\(=182.\frac{1}{8}.\frac{808080}{919191}=\frac{182}{8}.\frac{80}{91}=20\)
\(2014:\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1\frac{2}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{1\frac{1}{6}+0,875-0,7}{\frac{1}{3}+0,25-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{2}{6}+\frac{2}{8}-\frac{2}{10}}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}{2\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7}{2}\right)=2014\)
(A) \(\frac{2}{3} + \frac{{ - 4}}{6} = \frac{4}{6} + \frac{{ - 4}}{6} = 0\) => A sai
(B) \(\frac{2}{3}.\frac{{ - 1}}{5} = \frac{{ - 2}}{{15}}\) mà \(\frac{{3 - 2}}{5} = \frac{1}{5}\) => B sai
(C) \(\frac{2}{3} - \frac{3}{5} = \frac{{10}}{{15}} - \frac{9}{{15}} = \frac{1}{{15}}\) => C đúng
(D) \(\frac{3}{5}:\frac{3}{{ - 5}} = \frac{3}{5}.\frac{{ - 5}}{3} = \frac{{ - 15}}{{15}} = - 1\) => D sai
=> Chọn C.
\(\left(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}\right)x\frac{9}{15}:\frac{7}{12}=\frac{1}{4}x\frac{9}{15}:\frac{7}{12}=\frac{3}{20}:\frac{7}{12}=\frac{9}{35}\)
ai hminh tích lại nhà
\(1\frac{2}{3}\)=\(\frac{5}{3}\)
\(3\frac{2}{9}\)=\(\frac{29}{9}\)thế này à mik ko bít (mik hơi nhu,thông cảm)
\(1\frac{2}{3}\) của \(3\frac{2}{9}\)
=>\(\frac{5}{3}\)và \(\frac{29}{9}\)
tính laf cộng hay trừ
HT