1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)

2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)

= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)

Vậy ......

hok tốt

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:

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

\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)

\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)

\(x=1\div\left(-\frac{3}{5}\right)\)

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

\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)

\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)

\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)

\(-\frac{4}{21}\cdot x=\frac{3}{5}\)

\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)

\(x=-\frac{63}{20}\)

\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)

\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)

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

\(x=1\div\frac{5}{7}\)

\(x=\frac{7}{5}\)

\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)

\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)

\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)

\(\frac{17}{33}\cdot x=\frac{9}{14}\)

\(x=\frac{9}{14}\div\frac{17}{33}\)

\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)

a)\(\frac{3}{5}\times x=1\)                                        

\(x=1\div\frac{3}{5}\)

\(x=\frac{5}{3}\)

Mình chỉ làm duoc câu a thôi 

\(=\frac{11}{-5}\cdot\frac{-9}{11}\cdot\frac{15}{-14}\cdot\frac{2}{5}+-\frac{2}{77}\cdot\frac{5}{-3}\)
\(=\frac{9}{5}\cdot-\frac{15}{14}\cdot\frac{2}{5}+\frac{10}{231}\)
\(=-\frac{841}{1155}\)

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

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

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

\(=-49\)

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

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

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

\(A=-0,25x1\)

\(A=-0,25\)

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

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

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

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

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

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

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

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

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

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

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

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

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

\(D=-49\)

giúp em với : Akai Haruma

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

tính giá trị biểu thức chứ còn cái gì nữa

a, \(A=\frac{22}{27}\)

b,\(B=\frac{1}{57}\)

C,\(C=\frac{1}{50}\)

d, \(D=0\)

a) x = 99/20

b) x = 7

c) x = 2

( chỉ lm đc đến đó thui nk )