Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
\(1,3.\dfrac{15}{39}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
\(=\dfrac{13}{10}.\dfrac{15}{39}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(=\dfrac{1}{2}-\dfrac{2}{3}=-\dfrac{1}{6}\)
\(4\dfrac{1}{3}\cdot\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\\ \dfrac{13}{3}\cdot\dfrac{-1}{3}\le x\le\dfrac{2}{3}\cdot\dfrac{-11}{12}\\ \dfrac{-13}{9}\le x\le\dfrac{-11}{18}\\ \dfrac{-26}{18}\le x\le\dfrac{-11}{18}\\ \Rightarrow x=-1\)
a, \(x+\dfrac{2}{3}=0,2\)
\(\Rightarrow x+\dfrac{2}{3}=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{-7}{10}\)
b, \(\dfrac{17}{7}-\dfrac{6}{5}x=\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{17}{7}-\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{-51}{28}\)
\(\Rightarrow x=\dfrac{-51}{28}:\dfrac{6}{5}\)
\(\Rightarrow x=\dfrac{-85}{56}\)
Câu 1:
\(B=\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)
\(\Rightarrow B=\dfrac{1}{7}\left(\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{1}{15.28}\right)\)
\(\Rightarrow B=\dfrac{1}{7}\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\right)\)
\(\Rightarrow B=\dfrac{1}{7}\left(\dfrac{1}{2}-\dfrac{1}{28}\right)\)
\(\Rightarrow B=\dfrac{1}{7}.\dfrac{13}{28}\)
\(\Rightarrow B=\dfrac{13}{196}\)
Vậy \(B=\dfrac{13}{196}\)
Câu 2:
\(D=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{20}}\)
\(\Rightarrow2D=1+\dfrac{1}{2}+...+\dfrac{1}{2^{19}}\)
\(\Rightarrow2D-D=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^{19}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{20}}\right)\)
\(\Rightarrow D=1-\dfrac{1}{2^{20}}< 1\)
\(\Rightarrow D< 1\left(đpcm\right)\)
Vậy...