B1      :

\(\frac{0,1\left(6\right)+0,\left(3\right)}{0,\left(3\right)+1,1\left(6\right)}\) . x = 0,(2)

=\(\frac{0,5}{1,5}\).x=0,(2)

x=0,(2):\(\frac{0,5}{1,5}\)

x=0,(6)=\(\frac{2}{3}\)

b2:

 [12,(1) - 2,3(6)] : 4,(21)

=9,7(4):4,(21)

=\(\frac{9,7\left(4\right)}{4,\left(21\right)}\)

di ban

\(A=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right)\left(63.1,2-21.3,6+1\right)}{1-2+3-4+....+99-100}\)

\(=\frac{\frac{100\left(100+1\right)}{2}\left(\frac{3+2-6}{12}\right)\left[63\left(1,2-1,2\right)+1\right]}{\left(1-2\right)+\left(3-4\right)+....+\left(99-100\right)}\)

\(=\frac{5050.\left(-\frac{1}{12}\right).1}{-1+\left(-1\right)+\left(-1\right)+...+\left(-1\right)}\)

\(=\frac{2525.\left(-\frac{1}{6}\right)}{-50}=\frac{101}{12}\)

101/12 bạn nha

CHÚC BẠN HỌC GIỎI

232/301

Gợi ý cho bn: Tách chúng ra nhé, rồi rút gọn số có ở trên cả tử và mẫu.

\(=\dfrac{2^{12}.3^{15}.7^{30}-31.2^{12}.3^{12}.3^2.7^{28}}{2^{11}.3^{14}.7^{14}.7^{15}+2^{12}.3^{15}.7^{30}}=\)

\(=\dfrac{2^{12}.3^{15}.7^{30}-31.2^{12}.3^{14}.7^{28}}{2^{11}.3^{14}.7^{29}+2^{12}.3^{15}.7^{30}}=\)

\(=\dfrac{2^{11}.3^{14}.7^{28}\left(2.3.7^2-31.2\right)}{2^{11}.3^{14}.7^{28}\left(7+2.3\right)}=\dfrac{2.3.7^2-31.2}{7+2.3}=\dfrac{232}{13}\)

a) \(10,\left(3\right)+0,\left(4\right)-8,\left(6\right)\)

\(=\frac{31}{3}+\frac{4}{9}-\frac{26}{3}\)

\(=\left(\frac{31}{3}-\frac{26}{3}\right)+\frac{4}{9}=\frac{5}{3}+\frac{4}{9}=\frac{15}{9}+\frac{4}{9}=\frac{19}{9}\)

b) \(\left[12,\left(1\right)-2,3\left(6\right)\right]:4,\left(21\right)\)

\(=\left[\frac{109}{9}-\frac{71}{30}\right]:\frac{139}{33}\)

\(=-\frac{52}{45}:\frac{139}{33}=-\frac{52}{45}\cdot\frac{33}{139}=-\frac{572}{2085}\)(số xấu quá)

c) \(3\frac{1}{2}\cdot\frac{4}{49}-\left[2,\left(4\right)\cdot2\frac{5}{11}\right]:\frac{-42}{53}\)

\(=\frac{7}{2}\cdot\frac{4}{49}-\left[\frac{22}{9}\cdot\frac{27}{11}\right]\cdot\frac{-53}{42}\)

\(=\frac{2}{7}-6\cdot\left(-\frac{53}{42}\right)=\frac{2}{7}-\left(-\frac{53}{7}\right)=\frac{2}{7}+\frac{53}{7}=\frac{55}{7}\)

qwertyuiopasdfgggggghjkllzxcvbnmm,.//234567890-=`

\(5\frac{4}{7}\): [ x : 1,3 + 8,4 . \(\frac{6}{7}\). ( 6 - \(\frac{\left(2,3+5\div6,25\right)\times7}{8\times0,-125+6,9}\)) ] = \(1\frac{1}{14}\)

\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - \(\frac{\left(2,3+0,8\right).7}{0,1+6,9}\)) ] = \(\frac{15}{14}\)

\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - \(\frac{3,1.7}{7}\)) ] = \(\frac{15}{14}\)

\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - 3,1 ) ] = \(\frac{15}{14}\)

x : 1,3 + \(\frac{36}{5}\). 2,9 = \(\frac{39}{7}\): \(\frac{15}{14}\)

x : 1,3 + 20,88 = 5,2

x : 1,3 = - 15,68

x = - 15,68 . 1,3

x = - 20,384

ta có 

\(5\frac{4}{7}:\left\{x:1,3+8,4.\frac{6}{7}.\left[6-\frac{\left(2,3+5:6,25\right).7}{8.0,0125+6,9}\right]\right\}=1\frac{1}{14}\)

\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.\left[6-\frac{\left(2,3+0,8\right).7}{0,1+6,9}\right]\right\}=\frac{15}{14}\)

\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.\left[6-\frac{3,1.7}{7}\right]\right\}=\frac{15}{14}\)

\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.2,9\right\}=\frac{15}{14}\Leftrightarrow\left\{x:1,3+7,2.2,9\right\}=\frac{39}{7}:\frac{15}{14}\)

\(\Leftrightarrow x:1,3+20,88=5,2\Leftrightarrow x:1,3=-15,68\Leftrightarrow x=-20,384\)

B = \(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right)...\left(\frac{1}{100}-1\right)\)

B = \(\frac{-3}{4}.\frac{-8}{9}...\frac{-99}{100}\)

B = \(-\left(\frac{3}{4}.\frac{8}{9}...\frac{99}{100}\right)\)

B = \(-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{9.11}{10.10}\right)\)

B = \(-\left(\frac{1.2...9}{2.3...10}.\frac{3.4...11}{2.3...10}\right)\)

B = \(-\left(\frac{1}{10}.\frac{11}{2}\right)\)

B = \(\frac{-11}{20}\)

Vì  \(\frac{11}{20}>\frac{11}{21}\)nên \(\frac{-11}{20}< \frac{-11}{21}\)

Vậy \(B< \frac{-11}{21}\)

Hồ Ngọc Minh Châu Võ cho mình hỏi nhưng bài kia mỗi bài 1 dòng hay là cả một bài vậy bạn