Ta có \(AB=\dfrac{2}{5}CD\)

\(\Rightarrow S_{ABD}=\dfrac{2}{5}S_{BCD}\)

\(S_{ABCD}=S_{ABD}+S_{BCD}=S_{ABD}+\dfrac{2}{5}S_{ABD}=\left(1+\dfrac{2}{5}\right)S_{ABD}=\dfrac{7}{5}S_{ABD}\\ \Rightarrow S_{ABCD}=\dfrac{7}{5}S_{ABD}\)

\(S_{ABD}\) và \(S_{ABCD}\) mà ?? 

trên ghi 0,32 mà sao dưới ghi 0,024 vậy 

\(\dfrac{1}{5}< x< \dfrac{12}{5}\\ \Rightarrow1< 5x< 12\\ \Rightarrow\left\{{}\begin{matrix}1< 5x\\5x< 12\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x< \dfrac{12}{5}\end{matrix}\right.\\ \Rightarrow x=\left\{\dfrac{2}{5};\dfrac{3}{5};\dfrac{4}{5};\dfrac{5}{5};\dfrac{6}{5};\dfrac{7}{5};\dfrac{8}{5};\dfrac{9}{5};\dfrac{10}{5};\dfrac{11}{5}\right\}\)

Có 10 số

a, \(45;450;504;540\)

b,  \(450;540;490;590\)

54 x 133 không phải 113

và 113 x 100 =11300 

\(=113\times\left(45+1\right)+54\times133\\ =113\times46+54\times133\\ =5198+7182\\ =12380\)

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{10-1}{10}=\dfrac{9}{10}\)