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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: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)
\(=\frac{42}{43}\)
\(a,\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)
\(=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(=\frac{3\left[\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right]}{11\left[\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right]}=\frac{3}{11}\)
Câu b tương tự
#)Giải :
b)\(\frac{5}{9}:\left(\frac{1}{3}+\frac{1}{4}\right)+\frac{5}{9}:\left(\frac{1}{9}+\frac{2}{3}\right)\)
\(=\frac{5}{9}:\frac{1}{3}+\frac{5}{9}:\frac{1}{4}+\frac{5}{9}:\frac{1}{9}+\frac{5}{9}:\frac{2}{3}\)
\(=\frac{5}{9}:\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{9}+\frac{2}{3}\right)\)
\(=\frac{5}{9}:\frac{49}{36}\)
\(=\frac{20}{49}\)
a, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}.\frac{12}{7}\)
\(=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)
\(=\frac{-5}{7}.1+\frac{12}{7}=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)
\(1\frac{13}{15}.0,75-\left(\frac{104}{195}+25\%\right).\frac{24}{47}-3\frac{13}{12}:3\)
= \(\frac{28}{15}.\frac{3}{4}-\left(\frac{104}{195}+\frac{1}{4}\right).\frac{24}{47}-\frac{49}{12}.\frac{1}{3}\)
= \(\frac{4.3.7}{4.3.5}-\frac{47}{60}.\frac{24}{47}-\frac{49}{36}\)
= \(\frac{7}{5}-\frac{2}{5}-\frac{49}{36}\)
= \(1-\frac{49}{36}\)
= \(-\frac{13}{36}\)
Chúc bạn làm bài tốt
\(1\frac{13}{15}.0,75-\left(\frac{104}{195}+25\%\right).\frac{24}{27}-3\frac{13}{12}:3\)
=\(\frac{28}{15}.\frac{3}{4}-\left(\frac{8}{15}+\frac{1}{4}\right).\frac{24}{47}-\frac{49}{12}.\frac{1}{3}\)
=\(\frac{7}{5}-\frac{47}{60}.\frac{24}{47}-\frac{49}{36}\)
=\(\frac{7}{5}-\frac{2}{5}-\frac{49}{36}\)
=\(\frac{-13}{36}\)
a)\(=\frac{-5}{2}:\left(\frac{3}{4}-\frac{2}{4}\right)\)
\(=\frac{-5}{2}:1=\frac{-5}{2}\)
b)\(=\frac{146}{13}-\left(\frac{18}{7}+\frac{68}{13}\right)\)
\(=\frac{146}{13}-\frac{18}{7}-\frac{68}{13}\)
\(=\frac{146}{13}-\frac{68}{13}-\frac{18}{7}\)
\(=\frac{78}{13}-\frac{18}{7}=6-\frac{18}{7}=\frac{42}{7}-\frac{18}{7}=\frac{24}{7}\)
c)\(=\left(\frac{-5}{24}+\frac{18}{24}+\frac{14}{24}\right):\left(\frac{-9}{4}\right)^2\)
\(=\frac{27}{24}:\frac{\left(-9\right)^2}{4^2}\)
\(=\frac{27\times16}{24\times81}=\frac{2}{9}\)
Còn mình nè
Lý 7đ ; Sinh 9đ ; Sử 9đ ; Địa 8,5đ ; Gdcd 9đ ; Công nghệ 8đ
Chả biết có đc HSG không nữa
a) B = \(\frac{1}{2}.\frac{4}{3}.20.\frac{7}{35}.0,75\)
B = \(\frac{1.1.2.1.1}{1.1.1.1.1}\)
B= 2
b) \(\frac{1}{3}x=16\frac{1}{4}-13\frac{1}{4}\)
= > \(\frac{1}{3}x=3\)
= > x = 9
c)\(\frac{2}{3}x=\frac{-1}{4}-\frac{1}{3}\)
=> \(\frac{2}{3}x=-\frac{7}{12}\)
=> x = \(\frac{-7}{12}:\frac{2}{3}\)
=> x = \(\frac{-7}{8}\)