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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(A=\dfrac{19}{25}\left(\dfrac{3}{7}+\dfrac{25}{19}\right)-\dfrac{3}{7}\left(\dfrac{19}{25}-\dfrac{7}{3}\right)\\ A=\dfrac{19}{25}.\dfrac{3}{7}+\dfrac{19}{25}.\dfrac{25}{19}-\dfrac{3}{7}.\dfrac{19}{25}+\dfrac{3}{7}.\dfrac{7}{3}\\ A=\dfrac{19}{25}.\dfrac{3}{7}-\dfrac{19}{25}.\dfrac{3}{7}+\dfrac{25}{9}.\dfrac{19}{25}+\dfrac{3}{7}.\dfrac{7}{3}\\ A=0+1+1\\ A=2\)
Vậy A= 2
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![](https://rs.olm.vn/images/avt/0.png?1311)
19/25 . 3/7 . 25/19 - 3/7 . 19/25 . 7/3
( 19/25 . 25/19 ) . 3/7 - ( 3/7 . 7/3 ) . 19/25
1 . 3/7 - 1 . /19/25
3/7 - 19/25
-58/175
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![](https://rs.olm.vn/images/avt/0.png?1311)
giúp mình bài này với
bạn nam đọc 1 quyển sách trong 3 ngày . ngày 1 đọc 1 phần 3 số trang . ngày thứ 2 đọc 2 phần 5 số trang còn lại .ngày 3 đọc 36 trang còn lại. hỏi quyển sách có bao nhiêu trang
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\(\dfrac{26}{19}.\dfrac{4}{25}+\dfrac{7}{5}.\dfrac{3}{5}-\dfrac{4}{25}.\dfrac{7}{19}=\dfrac{104}{475}+\dfrac{7}{5}.\dfrac{3}{5}-\dfrac{4}{25}.\dfrac{7}{19}=\dfrac{104}{475}+\dfrac{21}{25}-\dfrac{4}{25}.\dfrac{7}{19}=\dfrac{503}{475}-\dfrac{4}{25}.\dfrac{7}{19}=\dfrac{503}{475}-\dfrac{28}{475}=\dfrac{475}{475}=1\)
\(\dfrac{26}{19}\cdot\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{3}{5}-\dfrac{4}{25}\cdot\dfrac{7}{19}=\dfrac{4}{25}\cdot\left(\dfrac{26}{19}-\dfrac{7}{19}\right)+\dfrac{7}{5}\cdot\dfrac{3}{5}=\dfrac{4}{25}\cdot1+\dfrac{21}{25}=\dfrac{4}{25}+\dfrac{21}{25}=\dfrac{25}{25}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{7}{-25}+\dfrac{-18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
\(=\left(\dfrac{-7}{25}+\dfrac{-18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
\(=\left(-1\right)+1+\dfrac{5}{7}\)
\(=0+\dfrac{5}{7}=\dfrac{5}{7}\)
b, \(\dfrac{7}{19}.\dfrac{8}{11}+\dfrac{7}{19}.\dfrac{3}{11}+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\)
\(=\dfrac{7}{19}.1+\dfrac{12}{19}\)
\(=\dfrac{7}{19}+\dfrac{12}{19}=1\)
a. \(\dfrac{7}{-25}+\dfrac{-18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
\(=\left(\dfrac{-7}{25}+\dfrac{-18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
\(=-1+1+\dfrac{5}{7}\)
\(=\dfrac{5}{7}\)
b. \(\dfrac{7}{19}\cdot\dfrac{8}{11}+\dfrac{7}{19}\cdot\dfrac{3}{11}+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\cdot1+\dfrac{12}{19}\)
\(=\dfrac{7}{19}+\dfrac{12}{19}\)
\(=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
e gõ latex hoặc vt r chụp lên nhé như vầy ko nhìn đc cái đề nx e
a) \(2\dfrac{3}{4}-\left(\dfrac{1}{7}+1\dfrac{3}{4}\right)\)
\(=\dfrac{11}{4}-\dfrac{1}{7}-\dfrac{7}{4}=\left(\dfrac{11}{4}-\dfrac{7}{4}\right)-\dfrac{1}{7}=1-\dfrac{1}{7}=\dfrac{6}{7}\)
b) \(\dfrac{7}{-25}+\dfrac{-18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
\(=\left(\dfrac{-7}{25}+\dfrac{-18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}=-1+1+\dfrac{5}{7}=\dfrac{5}{7}\)
c) \(\dfrac{-1}{4}\left(12\dfrac{3}{4}-7,75\right)-25\%.3\dfrac{1}{2}\)
\(=\dfrac{-1}{4}\left(12\dfrac{3}{4}-7\dfrac{3}{4}\right)-\dfrac{1}{4}.\dfrac{7}{2}=\dfrac{-1}{4}.5-\dfrac{7}{8}\\ =-\dfrac{5}{4}-\dfrac{7}{8}=-\dfrac{17}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{7}{-25}+\dfrac{18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
=\(\left(\dfrac{-7}{25}+\dfrac{18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
= \(\dfrac{11}{25} +1+\dfrac{5}{7}\)
= \(1\dfrac{11}{25}+\dfrac{5}{7}\)
= \(2\dfrac{27}{175}\)
b) \(-2+\dfrac{15}{19}+\dfrac{-15}{17}+\dfrac{15}{23}+\dfrac{4}{19}\)
=\(-2+\left(\dfrac{ }{ }\right)\)
\(A=\frac{19}{25}.\left(\frac{3}{7}+\frac{25}{19}\right)-\frac{3}{7}.\left(\frac{19}{25}-\frac{7}{3}\right)\)
=> \(A=\frac{19}{25}.\frac{232}{133}-\frac{3}{7}.\frac{232}{75}\)
=> \(A=\frac{232}{175}-\frac{232}{175}\)
=> \(A=0\)
A =0 nha