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G=\(\dfrac{\left(-2\right)}{3}+\dfrac{\left(-5\right)}{7}+\dfrac{2}{3}+\dfrac{\left(-2\right)}{7}\)
\(\Rightarrow G=\dfrac{\left(-2\right)}{3}+\dfrac{2}{3}+\dfrac{\left(-5\right)}{7}+\dfrac{\left(-2\right)}{7}\)
\(\Rightarrow G=\dfrac{\left(-2\right)+2}{3}+\dfrac{\left(-5\right)+\left(-2\right)}{7}\)
\(\Rightarrow G=0+\dfrac{-7}{7}\)
\(\Rightarrow G=-1\)
\(H=\dfrac{\left(-5\right)}{7}\cdot\dfrac{2}{11}+\dfrac{\left(-5\right)}{7}\cdot\dfrac{9}{11}\)
\(\Rightarrow H=\dfrac{\left(-5\right)}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)\)\(\Rightarrow H=\dfrac{\left(-5\right)}{7}\cdot\left(\dfrac{2+9}{11}\right)\)
\(\Rightarrow H=\dfrac{\left(-5\right)}{7}\cdot1\)
\(\Rightarrow H=\dfrac{-5}{7}\)
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
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK
11: \(=\dfrac{-5}{7}+\dfrac{5}{67}+\dfrac{13}{30}+\dfrac{1}{2}-\dfrac{11}{6}+\dfrac{17}{14}+\dfrac{2}{5}\)
\(=\left(\dfrac{-5}{7}+\dfrac{1}{2}+\dfrac{17}{14}\right)+\left(\dfrac{13}{30}-\dfrac{11}{6}+\dfrac{2}{5}\right)+\dfrac{5}{67}\)
\(=\dfrac{-10+7+17}{14}+\dfrac{13-55+12}{30}+\dfrac{5}{67}\)
\(=1-1+\dfrac{5}{67}=\dfrac{5}{67}\)
12: \(=\dfrac{-1}{4}\cdot\dfrac{152}{11}-\dfrac{1}{4}\cdot\dfrac{68}{11}\)
\(=\dfrac{-1}{4}\left(\dfrac{152}{11}+\dfrac{68}{11}\right)=-\dfrac{1}{4}\cdot20=-5\)
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\\ =\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\\ =\dfrac{1}{2}+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\dfrac{4}{5}\\ =\dfrac{1}{2}+0+\dfrac{4}{5}\\ =\dfrac{1}{2}+\dfrac{4}{5}\\ =\dfrac{5}{10}+\dfrac{8}{10}\\ =\dfrac{13}{10}\)
\(\dfrac{-3}{7}+\dfrac{3}{4}:\dfrac{3}{14}\\ =\dfrac{-3}{7}+\dfrac{3}{4}\cdot\dfrac{14}{3}\\ =\dfrac{-3}{7}+\dfrac{7}{2}\\ =\dfrac{-6}{14}+\dfrac{49}{14}\\ =\dfrac{43}{14}\)
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
Vậy......
\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)
\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)
\(=\dfrac{5}{8}:\dfrac{15}{16}.\)
\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)
Vậy......
c, (làm tương tự câu b).
~ Học tốt!!! ~
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: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)
b: \(\dfrac{\left(-4\right)^2}{5}=\dfrac{16}{5}\)
\(B=\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)
c: \(C=\left(6-2.8\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\)
\(=5\cdot2-\dfrac{32}{5}=10-\dfrac{32}{5}=\dfrac{18}{5}\)
d: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)
\(=\dfrac{-5}{12}+\dfrac{5}{12}+\dfrac{6}{11}+\dfrac{5}{11}+\dfrac{7}{18}=1+\dfrac{7}{18}=\dfrac{25}{18}\)