19/25 . (3/7 . 25/19) - 3/7 . (19/25 - 7/3)
trình bày giúp mình nhé :)
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=19/25.3/7+19/25.25/19-3/7.19/25-3/7.7/3
=19/25.3/7+1-3/7.19/25-1
=19/25.3/7-19/25+1-1
=0
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
Xin phép đc ghi lại đề như sau \(C=\frac{19}{25}\left(\frac{3}{7}+\frac{25}{19}\right)-\frac{3}{7}\left(\frac{19}{25}-\frac{7}{3}\right)\)\(=\frac{19}{25}.\frac{3}{7}+\frac{19}{25}.\frac{25}{19}-\frac{3}{7}.\frac{19}{25}+\frac{3}{7}.\frac{7}{3}=\frac{19}{25}.\frac{3}{7}-\frac{3}{7}.\frac{19}{25}+\frac{19}{25}.\frac{25}{19}+\frac{3}{7}.\frac{7}{3}=0+1+1=2\)
\(A=\frac{19}{25}.\left(\frac{3}{7}+\frac{25}{19}\right)-\frac{3}{7}.\left(\frac{19}{25}-\frac{7}{3}\right)\)
=> \(A=\frac{19}{25}.\frac{232}{133}-\frac{3}{7}.\frac{232}{75}\)
=> \(A=\frac{232}{175}-\frac{232}{175}\)
=> \(A=0\)
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