giúp minh với mink đang cần gấp vì hôm nay kt 1 tiết rồi

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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

C=2

Xin phép đc ghi lại đề như sau \(C=\frac{19}{25}\left(\frac{3}{7}+\frac{25}{19}\right)-\frac{3}{7}\left(\frac{19}{25}-\frac{7}{3}\right)\)\(=\frac{19}{25}.\frac{3}{7}+\frac{19}{25}.\frac{25}{19}-\frac{3}{7}.\frac{19}{25}+\frac{3}{7}.\frac{7}{3}=\frac{19}{25}.\frac{3}{7}-\frac{3}{7}.\frac{19}{25}+\frac{19}{25}.\frac{25}{19}+\frac{3}{7}.\frac{7}{3}=0+1+1=2\)

\(A=\frac{19}{25}.\left(\frac{3}{7}+\frac{25}{19}\right)-\frac{3}{7}.\left(\frac{19}{25}-\frac{7}{3}\right)\)

=> \(A=\frac{19}{25}.\frac{232}{133}-\frac{3}{7}.\frac{232}{75}\)

=> \(A=\frac{232}{175}-\frac{232}{175}\)

=> \(A=0\)

A =0 nha

A=0 nha

Đáp án là : 2

=2 nha

Thank bạn