\(A=21\frac{4}{11}-\left(1\frac{3}{5}+7\frac{4}{11}\right)\)

\(A=\frac{235}{11}-\left(\frac{8}{5}+\frac{81}{11}\right)\)

\(A=\left(\frac{235}{11}-\frac{81}{11}\right)+\frac{8}{5}\)

\(A=\frac{154}{11}+\frac{8}{5}\)

\(\Rightarrow A=\frac{78}{5}\)

\(B=\left(7\frac{8}{9}+2\frac{3}{13}\right)-\left(4\frac{8}{9}-7\frac{10}{13}\right)\)

\(B=\left(\frac{71}{9}+\frac{29}{13}\right)-\left(\frac{44}{9}-\frac{101}{13}\right)\)

\(B=\left(\frac{71}{9}-\frac{44}{9}\right)+\left(\frac{29}{13}-\frac{101}{13}\right)\)

\(B=\frac{27}{9}+\frac{-72}{13}\)

\(B=3+\frac{-72}{13}\)

\(\Rightarrow B=\frac{-33}{13}\)

P/s: Hoq chắc :v

\(C=\frac{-3}{5}+\frac{2}{7}+\frac{-3}{7}.\frac{3}{5}+\frac{-3}{7}\)

\(C=\left(\frac{-3}{5}+\frac{3}{5}\right).\left(\frac{2}{7}+\frac{-3}{7}+\frac{-3}{7}\right)\)

\(C=0.\frac{-4}{7}\)

\(C=0\)

a) = -3/7 . 5/11 + -3/7 . 6/11 + 9/7

   = -3/7. ( 5/11 + 6/11 ) + 9/7

  = -3/7. 1 + 9/7

  = -3/7 + 9/7

  = 6/7

b) = 4/13 + 9/13 + -11/5 + 6/5 - 3/4

    = 13/13 + -5/5 - 3/4

    = 1 + (-1) - 3/4

    = 0 - 3/4

    = -3/4

c) = -19/17. 4/7 + 19/17. -3/7 + 19/17

    = 19/17. -4/7 + 19/17. -3/7 + 19/17.1

    = 19/17.( -4/7 + -3/7 + 19/17

    = 19/17. -7/7 + 19/17

    = 19/17. (-1) + 19/17

    = -19/17 + 19/17

    = 0

tk mk nha,thanks

a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)

\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)

b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)

c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)

\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)

\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)

e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)

f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)

a) \(-\frac{1}{4}.13\frac{9}{11}-0,25.6\frac{2}{11}\)

\(=-\frac{1}{4}.13\frac{9}{11}-\frac{1}{4}.6\frac{2}{11}\)

\(=-\frac{1}{4}\left(13\frac{9}{11}+6\frac{2}{11}\right)\)

\(=-\frac{1}{4}.20\)

\(=-5\)

b) \(B=\frac{-5}{6}.\frac{4}{19}+\frac{-7}{12}.\frac{4}{19}-\frac{40}{57}\)

\(=\frac{4}{19}\left(\frac{-5}{6}+\frac{-7}{12}\right)-\frac{40}{57}\)

\(=\frac{4}{19}.\frac{-17}{12}-\frac{40}{57}\)

\(=\frac{-17}{57}-\frac{40}{57}\)

\(=-1\)

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

\(=\frac{3}{7}.\frac{9}{26}-\frac{1}{2}.\frac{1}{7}.\frac{1}{13}-\frac{1}{7}\)

\(=\frac{1}{7}\left(3.\frac{9}{26}-\frac{1}{2}.\frac{1}{13}-1\right)\)

\(=\frac{1}{7}.0\)

\(=0\)

d) \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

\(=\left(\frac{4}{9}+6\frac{5}{9}\right):\left(-\frac{1}{7}\right)\)

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

\(=-49\)

\(A=-\frac{1}{4}x3\frac{9}{11}-0,25x6\frac{2}{11}\)

\(A=-0,25x3\frac{9}{11}-0,25x6\frac{2}{11}\)

\(A=-0,25x\left(\frac{9}{11}+\frac{2}{11}\right)\)

\(A=-0,25x1\)

\(A=-0,25\)

\(B=-\frac{5}{6}x\frac{4}{19}+-\frac{7}{12}x\frac{4}{19}-\frac{40}{57}\)

\(B=-\frac{5}{6}x\frac{4}{19}+-\frac{7}{12}x\frac{4}{19}-\frac{4}{19}x190\)

\(B=\frac{4}{19}x\left(-\frac{5}{6}-\frac{7}{12}-190\right)\)

\(B=\frac{4}{19}x\left(-\frac{2297}{12}\right)\)

\(B=\frac{-2297}{57}\)

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

\(C=\frac{1}{7}x3x\frac{9}{26}-\frac{1}{7}x\frac{1}{2}x\frac{1}{13}-\frac{1}{7}\)

\(C=\frac{1}{7}x\left(3x\frac{9}{26}-\frac{1}{2}x\frac{1}{13}-1\right)\)

\(C=\frac{1}{7}x0\)

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

\(D=\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

\(D=\left(\frac{4}{9}+6\frac{5}{9}\right):\left(-\frac{1}{7}\right)\)

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

\(D=-49\)

\(a,15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\\ =15\frac{3}{13}-3\frac{4}{7}-8\frac{3}{13}\\ =7\frac{3}{13}-3\frac{4}{7}\\ =\frac{94}{13}-\frac{25}{7}\\ =\frac{94\cdot7-25\cdot13}{13\cdot7}\\ =\frac{333}{91}\)

\(b,\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\\ =7\frac{4}{9}-3\frac{4}{9}+4\frac{7}{11}\\ =4\frac{4}{9}+4\frac{7}{11}\\ =\frac{40}{9}+\frac{51}{11}\\ =\frac{40\cdot11+51\cdot9}{11\cdot9}\\ =\frac{899}{99}\)

\(c,=-\frac{7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\\ =-\frac{7}{9}\cdot1+\frac{52}{9}\\ =\frac{-7+52}{9}\\ =\frac{45}{9}=5\)

\(d,=\frac{50}{100}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{75}{100}\\ =\frac{50\cdot2\cdot2\cdot2\cdot5\cdot7\cdot25\cdot3}{50\cdot2\cdot3\cdot5\cdot7\cdot25\cdot2\cdot2}=1\)

\(e,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\\ =1-\frac{1}{43}\\ =\frac{42}{43}\)

a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)

\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)

\(=4-\frac{4}{7}=\frac{24}{7}\)

b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)

\(=7+\frac{4}{9}+4+\frac{7}{11}-3-\frac{4}{9}\)

\(=8+\frac{7}{11}=\frac{95}{11}\)

c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)

\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)

\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)

\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)

d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)

\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)

\(=5\cdot\frac{7}{35}=1\)

e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)

\(=\frac{42}{43}\)

Để nhân các phân số này, ta chỉ cần nhân tử số với nhau và mẫu số với nhau:

\[
\frac{1}{3} \times \frac{2}{5} \times \frac{3}{7} \times \frac{4}{9} \times \frac{5}{11} \times \frac{6}{15} \times \frac{7}{15} \times \frac{8}{15} \times \frac{9}{19} \times \frac{10}{21} \times \frac{11}{32} \times \frac{12}{25} \times \left( \frac{126}{252} - 4 \right)
\]

Sau đó, ta thực hiện các phép tính:

1. Nhân tử số:
\[1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \times 11 \times 12 \times 126 = 997920\]

2. Nhân mẫu số:
\[3 \times 5 \times 7 \times 9 \times 11 \times 15 \times 15 \times 15 \times 19 \times 21 \times 32 \times 25 \times 252 = 7621237680\]

Kết quả là:
\[\frac{997920}{7621237680}\]

Bây giờ, ta có thể rút gọn phân số này bằng cách chia tử số và mẫu số cho 160:

\[ \frac{997920}{7621237680} = \frac{997920 ÷ 160}{7621237680 ÷ 160} = \frac{6237}{47695230} \]