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a) \(\left(6\dfrac{1}{9}+3\dfrac{7}{11}\right)-1\dfrac{1}{9}\)
\(=\left(\left(6+3\right)+\left(\dfrac{1}{9}+\dfrac{7}{11}\right)\right)-\dfrac{10}{9}\)
\(=\left(9+\dfrac{74}{99}\right)-\dfrac{10}{9}\)
\(=9\dfrac{74}{99}-\dfrac{10}{9}\)
\(=\dfrac{965}{99}-\dfrac{10}{9}\)
\(=\dfrac{95}{11}\)
b) \(1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:5\%\)
\(=1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:\dfrac{1}{20}\)
\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+3\cdot20\)
\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+60\)
\(=\dfrac{1275}{22}\)
c) \(4\dfrac{3}{4}-0,37+\dfrac{1}{8}-1,28-2,5+3\cdot\dfrac{1}{2}\)
\(=\dfrac{19}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{3}{2}\)
\(=\dfrac{89}{40}\)
d) \(\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)
\(=\dfrac{5}{11}+\dfrac{10}{77}-5\cdot\dfrac{2}{11}\)
\(=\dfrac{5}{11}+\dfrac{10}{77}-\dfrac{10}{11}\)
\(=-\dfrac{25}{77}\)
e) \(\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
(câu này chờ mình một chút)
Câu e) anh rút thừa số chung là 2 cùng mũ số ra ngoài thì phân số thành số nguyên. Em nghĩ thế!
a )
\(\frac{2}{7}+\frac{7}{7}.\frac{14}{25}\)
\(=\frac{2}{7}+1.\frac{14}{25}=\frac{2}{7}+\frac{14}{25}\)
\(=\frac{50}{175}+\frac{98}{175}=\frac{148}{175}\)
b)
\(\frac{6}{7}+\frac{5}{7}:5-\frac{8}{9}\)
\(=\frac{6}{7}+\frac{5}{30}-\frac{8}{9}\)
\(=\frac{6}{7}+\frac{1}{6}-\frac{8}{9}\)
\(=\frac{36}{42}+\frac{7}{42}-\frac{8}{9}\)
\(=\frac{43}{42}-\frac{8}{9}=\frac{129}{126}-\frac{112}{126}=\frac{17}{126}\)
tk ủng hộ mk nha!!!!!!!!
a: \(=\dfrac{4\cdot2+4\cdot9}{55}+\dfrac{5}{6}=\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{49}{30}\)
b: \(=\dfrac{3}{2}\cdot\dfrac{3}{5}-\left(\dfrac{3}{7}+\dfrac{3}{20}\right)\cdot\dfrac{10}{3}\)
\(=\dfrac{9}{10}-\dfrac{81}{140}\cdot\dfrac{10}{3}\)
\(=\dfrac{9}{10}-\dfrac{27}{14}=\dfrac{-36}{35}\)
c: \(=15+\dfrac{3}{13}-3-\dfrac{4}{7}-8-\dfrac{3}{13}\)
\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)
d: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=5\)
a) \(6\dfrac{5}{7}-\left(1\dfrac{3}{4}+2\dfrac{5}{7}\right)\)
\(=6\dfrac{5}{7}-1\dfrac{3}{4}-2\dfrac{5}{7}\)
\(=\left(6\dfrac{5}{7}-2\dfrac{5}{7}\right)-1\dfrac{3}{4}\)
\(=4-1\dfrac{3}{4}\)
\(=3\dfrac{3}{4}\)
b) \(7\dfrac{5}{11}-\left(2\dfrac{3}{7}+3\dfrac{5}{11}\right)\)
\(=7\dfrac{5}{11}-2\dfrac{3}{7}-3\dfrac{5}{11}\)
\(=\left(7\dfrac{5}{11}-3\dfrac{5}{11}\right)-2\dfrac{3}{7}\)
\(=4-2\dfrac{3}{7}\)
\(=2\dfrac{3}{7}\)
a)
C1: \(2\dfrac{1}{4}+1\dfrac{1}{6}\)
\(=\dfrac{9}{4}+\dfrac{7}{6}\)
\(=\dfrac{27}{12}+\dfrac{14}{12}\)
\(=\dfrac{41}{12}=3\dfrac{5}{12}\)
C2: \(2\dfrac{1}{4}+1\dfrac{1}{6}\)
\(=2\dfrac{3}{12}+1\dfrac{2}{12}\)
\(=3\dfrac{5}{12}\)
Đáp án và hướng dẫn giải bài 109:
a) Cách 1
cách 2
b) Cách 1
cách 2
c, cách 1
cách 2
Các câu dễ tự làm nha:
\(D=\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(D=\dfrac{1}{99}-\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)\(D=-\dfrac{1}{100}-1\)
a: \(=\dfrac{32}{9}+\dfrac{13}{6}=\dfrac{32\cdot2+13\cdot3}{18}=\dfrac{64+39}{18}=\dfrac{103}{18}\)
b: \(=\dfrac{43}{8}-\dfrac{43}{6}=\dfrac{-43}{24}\)
c:\(=4-2-\dfrac{1}{6}=2-\dfrac{1}{6}=\dfrac{11}{6}\)
d: \(=5+\dfrac{2}{3}+7+\dfrac{1}{2}-3-\dfrac{1}{2}+1+\dfrac{2}{3}\)
\(=10+\dfrac{4}{3}=\dfrac{34}{3}\)
a. \(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}\)
= \(\left(\dfrac{5}{7}+\dfrac{-12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)\)
= \(\left(-1\right)+1=0\)
b. \(\left(7-2\dfrac{3}{5}\right).2\dfrac{4}{33}-2\dfrac{3}{5}:\dfrac{1}{2}+\dfrac{7}{5}\)
= \(\dfrac{22}{5}.2\dfrac{4}{33}-\dfrac{26}{5}+\dfrac{7}{5}\)
= \(\dfrac{28}{3}-\dfrac{33}{5}=\dfrac{41}{15}\)
c\(\dfrac{-7}{9}:\dfrac{8}{15}+\dfrac{-7}{9}.\dfrac{7}{15}+5\dfrac{7}{9}\)
= \(\dfrac{-56}{135}+\dfrac{-49}{135}+5\dfrac{7}{9}\)
= \(\dfrac{-7}{9}+5\dfrac{7}{9}=5\)
d. \(\left(16\dfrac{3}{8}-19\dfrac{3}{4}\right)-\left(12\dfrac{3}{8}-17\dfrac{3}{4}\right)\)
= \(16\dfrac{3}{8}-19\dfrac{3}{4}-12\dfrac{3}{8}+17\dfrac{3}{4}\)
= \(\left(16\dfrac{3}{8}-12\dfrac{3}{8}\right)-\left(19\dfrac{3}{4}-17\dfrac{3}{4}\right)\)
=\(4+2=6\)
a)\(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}\)
=\(\left(\dfrac{5}{7}-\dfrac{12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)=-1+1=0\)
b)\(\left(7-2\dfrac{3}{5}\right).2\dfrac{4}{33}-2\dfrac{3}{5}:\dfrac{1}{2}+\dfrac{7}{5}\)
=\(\dfrac{22}{5}.\dfrac{70}{33}-\dfrac{13}{5}.2+\dfrac{7}{5}\)
=\(\dfrac{28}{3}-\dfrac{26}{5}+\dfrac{7}{5}=\dfrac{83}{15}\)
Các câu sau tương tự
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
a) Cách 1:
\(3\dfrac{4}{9}+5\dfrac{1}{6}=\left(3+5\right)+\left(\dfrac{4}{9}+\dfrac{1}{6}\right)=8+\dfrac{11}{18}=8\dfrac{11}{18}\)
Cách 2:
\(3\dfrac{4}{9}+5\dfrac{1}{6}=\dfrac{31}{9}+\dfrac{31}{6}=\dfrac{155}{18}=8\dfrac{11}{18}\)
b) Cách 1:
\(8\dfrac{1}{14}-6\dfrac{3}{7}=\left(7-6\right)+\left(\dfrac{15}{14}-\dfrac{3}{7}\right)=1+\dfrac{9}{14}=1\dfrac{9}{14}\)
Cách 2:
\(8\dfrac{1}{14}-6\dfrac{3}{7}=\dfrac{113}{14}-\dfrac{45}{7}=\dfrac{23}{14}=1\dfrac{9}{14}\)
c) Cách 1:
\(7-3\dfrac{6}{7}=\left(6-3\right)+\left(1-\dfrac{6}{7}\right)=3+\dfrac{1}{7}=3\dfrac{1}{7}\)
Cách 2:
\(7-3\dfrac{6}{7}=\dfrac{49}{7}-\dfrac{27}{7}=\dfrac{22}{7}=3\dfrac{1}{7}\)