ick mình

Phải làm đúng mới k

\(\frac{3}{7}.\frac{9}{26}-\frac{1}{14}.\frac{1}{13}+\frac{6}{7}\)

=\(\frac{3}{7}.\frac{9}{26}-\frac{1}{26}.\frac{1}{7}+\frac{6}{7}\)

=\(\frac{1}{7}.\frac{27}{26}-\frac{1}{26}.\frac{1}{7}+6.\frac{1}{7}\)

=\(\frac{1}{7}.\left(\frac{27}{26}-\frac{1}{26}+6\right)\)

=\(\frac{1}{7}.7\)

=1

C1:3/7 x 9/26 - 1/14 x 1/13 - 1/7                                       

   =27/182 - 1/182 -1/7

   =26/182 - 1/7

   =0                                                                        

bằng 0 nha

bằng 0 đó

\(\frac{3}{7}\times\frac{9}{26}-\frac{1}{14}\times\frac{1}{13}\)

\(=\frac{27}{182}-\frac{1}{182}\)

\(=\frac{1}{7}\)

27/182 - 1/182

1/7

Muốn cho số có hai chữ số giống nhau và chia hết cho 2 thì số đó phải là một trong các số 22, 44, 66, 88. Bây giờ ta tìm trong những số này số mà chia cho 5 thì dư 3.

Đó là số 88.

Xem thêm tại: http://loigiaihay.com/bai-99-trang-39-sgk-toan-6-tap-1-c41a3896.html#ixzz4xczZ4dOb

C = \(\frac{3}{7}\cdot\frac{9}{26}-\frac{1}{14}\cdot\frac{1}{13}-\frac{1}{7}\)

C = \(\frac{1}{7}\cdot3\cdot\frac{9}{26}-\frac{1}{7}\cdot\frac{1}{2}\cdot\frac{1}{13}-\frac{1}{7}\)

C = \(\frac{1}{7}\cdot\frac{27}{26}-\frac{1}{7}\cdot\frac{1}{26}-\frac{1}{7}\)

C = \(\frac{1}{7}\left(\frac{27}{26}-\frac{1}{26}-1\right)\)

C = \(\frac{1}{7}\left(1-1\right)\)

C = 0

Chúc bn học tốt

0

Ta có : 

\(P=\frac{\frac{6}{8}+\frac{6}{10}+\frac{6}{14}+\frac{6}{26}}{\frac{11}{4}+\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)

\(\Rightarrow P=\frac{\frac{3}{4}+\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)

\(\Rightarrow P=\frac{3\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)

\(\Rightarrow P=\frac{3}{11}\)

Vậy \(P=\frac{3}{11}\)

\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}=\frac{3}{11}\)

đề bài của bn sai nên mk sửa luôn nha

\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}-\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

\(A=\frac{155-5\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{403-13\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}-\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{7}{91}+\frac{2}{10}-\frac{3}{10}}\)

\(A=\frac{155-5}{403-13}-\frac{3\left(\frac{1}{5}+\frac{1}{13}\right)-\frac{9}{10}}{\frac{7}{91}+\left(-\frac{1}{10}\right)}\)

\(A=\frac{5}{13}-\frac{\left(-\frac{9}{130}\right)}{\left(-\frac{3}{130}\right)}=\frac{5}{13}-\frac{\frac{9}{130}}{\frac{3}{130}}\)

\(A=\frac{5}{13}-\frac{9}{130}\cdot\frac{130}{3}\)

\(A=\frac{5}{13}-3=-\frac{34}{13}\)

\(B=\frac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\)

\(B=\frac{30\cdot\left(2^2\right)^7\cdot3^{29}-5\cdot\left(2\cdot7\right)^5\cdot2^{12}}{54\cdot\left(2\cdot3\right)^{14}\cdot\left(3^2\right)^7-12\cdot\left(2^3\right)^5\cdot7^5}\)

\(B=\frac{30\cdot2^{14}\cdot3^{29}-5\cdot2^5\cdot7^5\cdot2^{12}}{54\cdot2^{14}\cdot3^{14}\cdot3^{14}-12\cdot2^{15}\cdot7^5}\)

\(B=\frac{30\cdot3^{29}-5\cdot2^{17}\cdot7^5}{54\cdot3^{28}-12\cdot2^{15}\cdot7^5}=\frac{30\cdot3-5\cdot2^2}{54-12}=\frac{5}{3}\)

= 1/3 - 1/3 + 5/7 - 5/7 - 7/9 + 7/9 +9/11 - 9/11 -11/13 + 11/13 +13/15

= 0 + 0 - 0 + 0 -0 + 13/15

= 0 + 13/15

= 13/15

\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)

\(=\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{3}{5}-\frac{3}{5}\right)+\left(\frac{5}{7}-\frac{5}{7}\right)+\left(\frac{7}{9}-\frac{7}{9}\right)+\left(\frac{9}{11}-\frac{9}{11}\right)+\left(\frac{11}{13}-\frac{11}{13}\right)+\frac{13}{15}\)

\(=0+0+0+0+0+0+\frac{13}{15}\)

\(=\frac{13}{15}\)