giúp với ạ

giải dc nhưng mà hoi lâu

a)\(\dfrac{5}{7}+\dfrac{4}{9}=\dfrac{45}{63}+\dfrac{28}{63}=\dfrac{73}{63}\) ; \(\dfrac{9}{11}+\dfrac{3}{8}=\dfrac{72}{88}+\dfrac{33}{88}=\dfrac{105}{88}\)

\(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\); \(\dfrac{16}{25}-\dfrac{2}{5}=\dfrac{16}{25}-\dfrac{10}{25}=\dfrac{6}{25}\)

Giúp mik với mn ơi

1.\(\left(-\frac{6}{5}+\frac{6}{16}-\frac{6}{23}\right):\left(\frac{9}{5}-\frac{9}{16}+\frac{9}{23}\right)\)

\(=6\left(-\frac{1}{5}+\frac{1}{16}-\frac{1}{23}\right):\left(-9\right)\left(\frac{-1}{5}+\frac{1}{16}-\frac{1}{23}\right)\)

\(=6:\left(-9\right)=-\frac{2}{3}\)

2. \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{0.5-\frac{1}{3}+\frac{1}{4}}{-\frac{3}{2}+1-\frac{3}{4}}\)

\(=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{-3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)

\(=\frac{3}{5}-\frac{1}{3}\)

\(=\frac{9}{13}-\frac{5}{15}=\frac{4}{15}\)

a)\(\frac{11^4.6-11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5+9^8.7}\)

\(=1.6:\frac{9^8.3-9^8.9}{9^8.\left(5+7\right)}\)

\(=6:\frac{9^8.\left(3-9\right)}{9^8.12}\)

\(=6:\frac{9^8.\left(-6\right)}{9^8.12}\)

\(=6:\left(-\frac{6}{12}\right)\)

\(=6:\left(-\frac{1}{2}\right)\)

\(=-12\)

b) 3/5 : ( -1/5-1/6)+3/5:(-1/3-16/15) ( mình chuyển về ps luôn )

=3/5: (-11/30) + 3/5 : (-7/5) 

=3/5:[-11/30+(-7/5)]

=3/5:53/30

=18/53

c) (1/2-13/14):5/7-(-2/21+1/7):5/7

= -3/7:5/7-1/21:5/7

=(-3/7-1/21):5/7

=-10/21:5/7

=-2/3

câu b vá c mình làm tắt nha. chúc bạn học tốt

\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{2}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+\frac{10}{4}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)

\(\frac{A}{B}=10\)

\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{2}{8}+\frac{1}{9}\)

Tách 9=1+1+...+1 ( có 9 số 1)

\(\Rightarrow A=1+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{2}{8}+1\right)+\left(\frac{1}{9}+1\right)\)

\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{8}+\frac{10}{9}\)

\(A=10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)

\(\Rightarrow A:B=\frac{10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=10\) ( vì \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\ne0\) )

Vậy \(A:B=10\)

3- 1/5 + 1/7 + 1/9 

2.3 - 2/5 + 2/7 + 2/9

= 3- 1/5 + 1/7 + 1/9

2.( 3- 1/5 + 1/7 + 1/9)

=1/2

\(N=\frac{1}{6}.\frac{5}{9}+\frac{5}{9}.\frac{2}{7}+\frac{1}{9}.\frac{5}{7}-\frac{5}{9}.\frac{3}{7}\)

\(N=\frac{1}{6}.\left(\frac{5}{9}+\frac{1}{9}\right).\left(\frac{2}{7}+\frac{5}{7}-\frac{3}{7}\right)\)

\(N=\frac{1}{6}.\frac{6}{9}.\frac{4}{7}\)

\(N=\frac{1}{9}.\frac{4}{7}\)

\(N=\frac{4}{63}\)

\(N=\frac{1}{6}.\frac{5}{9}+\frac{5}{9}.\frac{2}{7}+\frac{1}{9}.\frac{5}{7}-\frac{5}{9}.\frac{3}{7}\)

\(N=\frac{1}{6}.\frac{5}{9}+\frac{5}{9}.\frac{2}{7}+\frac{5}{9}.\frac{1}{7}-\frac{5}{9}.\frac{3}{7}\)

\(N=\frac{5}{9}.\left(\frac{1}{6}+\frac{2}{7}+\frac{1}{7}-\frac{3}{7}\right)\)

\(N=\frac{5}{9}.\frac{1}{6}\)

\(N=\frac{5}{54}\)

17/5×1/2×10/17×-1/8

17/10×-10/136

-170/1360

-1/8

5/54+10/63+5/63+15/63

5/54+15/63+15/63

5/54+30/63

315/3402+1620/3402

1935/3402

\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)

\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)

\(=\frac{7}{12}+\frac{31}{12}\)

\(=\frac{38}{12}=\frac{19}{6}\)

\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)

\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)

\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)

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

\(=\frac{-5}{9}\cdot\frac{2}{13}\)

\(=\frac{-10}{117}\)

\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)

\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)

\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)

\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)

\(=\frac{-8}{5}-\frac{7}{5}\)

\(=-3\)

\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)

\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)

\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)

\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)

\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)

\(=\frac{4}{5}\cdot\frac{13}{7}\)

\(=\frac{52}{35}\)