=\(\left(13\frac{9}{11}.\frac{49}{38}-5\frac{2}{11}.\frac{49}{38}\right).\left(\frac{38}{49}.\frac{11}{5}\right)\)

= \(\left(13\frac{9}{11}-5\frac{2}{11}\right).\frac{49}{38}.\frac{38}{49}.\frac{11}{5}\) = \(\left(13+\frac{9}{11}-5-\frac{2}{11}\right).\frac{11}{5}\)

=  \(\left(8+\frac{7}{11}\right).\frac{11}{5}\)= \(\frac{95}{11}.\frac{11}{5}=19\)

         \(\left(13\frac{9}{11}:\frac{38}{49}-5\frac{2}{11}:\frac{38}{49}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\)     

    \(=\left(\frac{152}{11}:\frac{38}{49}-\frac{57}{11}:\frac{38}{49}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\) 

    \(=\left(\frac{196}{11}-\frac{147}{22}\right):\frac{245}{418}=\frac{245}{22}:\frac{245}{418}=19\)

49\(\frac{8}{23}\) - (5\(\frac{7}{32}\) + 14\(\frac{8}{23}\))

= 49\(\frac{8}{23}\) - 5\(\frac{7}{32}\) - 14\(\frac{8}{23}\)

= (49\(\frac{8}{23}\) - 14\(\frac{8}{23}\)) - 5\(\frac{7}{32}\)

= 35 - 5 - \(\frac{7}{32}\)

= 30 - \(\frac{7}{32}\)

= \(\frac{960}{32}\) - \(\frac{7}{32}\)

= 953/32

Câu b:

71\(\frac{38}{45}\) - (43\(\frac{8}{45}\) - 1\(\frac{17}{57}\))

= 71\(\frac{38}{45}\) - 43\(\frac{8}{45}\) - 1\(\frac{17}{57}\)

= 28\(\frac{30}{45}\) - 1\(\frac{17}{57}\)

= 28\(\frac23\) - 1\(\frac{17}{57}\)

= \(\frac{86}{3}\) - \(\frac{74}{57}\)

= \(\frac{1634}{57}\) - \(\frac{74}{57}\)

= \(\frac{1560}{57}\)

= \(\frac{520}{19}\)

Mình làm thử, không biết có sai không, nếu sai thì cho mình đáp án rút kinh nghiệm nhé:

     \(\left(13\frac{9}{11}:\frac{38}{49}-5\frac{2}{11}:\frac{38}{49}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\)   

\(=\left(\frac{152}{11}:\frac{38}{49}-\frac{57}{11}:\frac{38}{49}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\)

\(=\left(\frac{196}{11}-\frac{147}{22}\right):\frac{245}{418}\)

\(=\frac{245}{22}:\frac{245}{418}=19\)

\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)

\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)

\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)

\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)

\(A=\left[35-0\right]-5\frac{7}{32}\)

\(A=35-5\frac{7}{32}\)

\(A=\frac{953}{32}\)

\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)

\(B=71\frac{38}{45}-\frac{36377}{855}\)

\(B=\frac{1670}{57}\)

\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)

\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\frac{153}{14}:\frac{4}{5}\)

\(C=\frac{765}{56}\)

\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)

\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0-\frac{1}{4}\)

\(D=-\frac{1}{4}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)

\(\)\(E=\frac{22}{45}\)

CHUC BAN HOC TOT >.<

Oihgh

a)\(3\frac{1}{2}-2\frac{7}{8}=3\frac{4}{8}-2\frac{7}{8}=2\frac{12}{8}-2\frac{7}{8}=\left(2-2\right)+\left(\frac{12}{8}-\frac{7}{8}\right)=\frac{5}{8}\)

b)\(10-2\frac{38}{39}=9\frac{39}{39}-2\frac{38}{39}=\left(9-2\right)+\left(\frac{39}{39}-\frac{38}{39}\right)=7+\frac{1}{39}=7\frac{1}{39}\)

c)\(3\frac{1}{4}.2\frac{6}{13}=\frac{13}{4}.\frac{32}{13}=8\)

a) =7/8

b)=274/39

c)=8

a) \(\frac{x+1}{3}=\frac{x-2}{4}\)

=> (x+1).4 = (x - 2) . 3

=> 4x + 4 = 3x - 6

=> 4x - 3x = - 6 - 4

=> x = - 10

b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)

\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)

\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)

\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0

\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)

Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0

=> x = -1

c) Xem lại đề

Xin ỗi bạn nha! Đoạn x+3 sửa lại thành x+32 nha bạn !!!

M=5/3*8+5/8*13+5/13*18+...+5/33*38

M=1/3-1/8+1/8-1/13+1/13-1/18+...+1/33-1/38

M=1/3-1/38

Đến đây, còn lại tự giải nhé. Phép trừ phân số thôi mà.

a)

\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)

= \(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)

= \(\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)

= \(\frac{1}{3}+\left(-1\right)+1\)

= \(\frac{1}{3}+0=\frac{1}{3}\)

b)

\(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)

= \(\frac{38}{45}-\frac{8}{45}+\frac{17}{51}+\frac{3}{11}\)

= \(\left(\frac{38}{45}-\frac{8}{45}\right)+\left(\frac{17}{51}+\frac{3}{11}\right)\)

= \(\frac{2}{3}+\frac{20}{33}\)

= \(\frac{14}{11}\)

21)

\(\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{8}\right).\left(1+\dfrac{1}{15}\right).....\left(1+\dfrac{1}{9999}\right)\\ =\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{10000}{9999}\\ =\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{100.100}{99.101}\\ =\dfrac{2.3.4.....100}{1.2.3.....99}.\dfrac{2.3.4.....100}{3.4.5.....101}\\ =100.\dfrac{2}{101}\\ =\dfrac{200}{101}\)