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a. \(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
\(=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)
b. \(D=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{97.100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
\(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-....-\frac{1}{66}\)
\(C=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)
\(D=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-....-\frac{1}{100}\right)\)
\(D=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
Câu 1:
\(-12+\left(16-11\right).4=-12+5.4=-12+20=8\)
Câu 2:
\(2\dfrac{3}{7}+1\dfrac{4}{7}=\dfrac{17}{7}+\dfrac{11}{7}=\dfrac{17+11}{7}=\dfrac{28}{7}=4\)
Câu 3:
\(\dfrac{-2}{3}.\dfrac{4}{5}+\dfrac{1}{5}:\dfrac{9}{11}=\dfrac{-8}{15}+\dfrac{11}{45}=\dfrac{-13}{45}\)
Câu 1:
\(-12+\left(16-11\right).4=-12+5.4=-12+20=8\)
Câu 2:
\(2\dfrac{3}{7}+1\dfrac{4}{7}=\dfrac{17}{7}+\dfrac{11}{7}=\dfrac{28}{7}=4\)
Câu 3:
\(\dfrac{-2}{3}.\dfrac{4}{5}+\dfrac{1}{5}:\dfrac{9}{11}=\dfrac{-8}{15}+\dfrac{1}{5}.\dfrac{11}{9}=\dfrac{-8}{15}+\dfrac{11}{45}=\dfrac{-13}{45}\)
Câu 1 :
a, 8.( -5 ).( -4 ).2
= [ 8.2 ].[( -5 ).(-4 ]
= 16.20
= 320
b, \(1\frac{3}{7}+\frac{-1}{3}+2\frac{4}{7}\)
\(=\frac{10}{7}+\frac{-1}{3}+\frac{18}{7}\)
\(=\frac{11}{3}\)
c, \(\frac{8}{5}.\frac{2}{3}+\frac{-5.5}{3.5}\)
\(=\frac{8}{3}+\frac{-5}{3}\)
\(=\frac{3}{3}=1\)
d, \(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.\left(-2\right)^2\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{16}.4\)
\(=\frac{55}{56}-\frac{3}{4}\)
\(=\frac{13}{56}\)
Câu 2 :
a, 2x + 10 = 16
2x = 16 + 10
2x = 26
x = 26 : 2
x = 13
b, \(x-\frac{1}{3}=\frac{5}{4}\)
\(x=\frac{5}{4}+\frac{1}{3}\)
\(x=\frac{19}{12}\)
c, \(2x+3\frac{1}{3}=7\frac{1}{3}\)
\(2x+\frac{10}{3}=\frac{22}{3}\)
\(2x=\frac{22}{3}-\frac{10}{3}\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
d, \(\left(\frac{2}{11}+\frac{1}{3}\right)x=\left(\frac{1}{7}-\frac{1}{8}\right).56\)
\(\frac{17}{33}x=1\)
\(x=1-\frac{17}{33}\)
\(x=\frac{16}{33}\)
Câu 1:
\(\dfrac{2}{5}-\dfrac{1}{4}+\dfrac{3}{10}=\dfrac{8}{20}-\dfrac{5}{20}+\dfrac{6}{20}=\dfrac{8-5+6}{20}=\dfrac{9}{20}\)
Câu 2:
\(\dfrac{-2}{5}:\left(1-\dfrac{1}{10}\right)=\dfrac{-2}{5}:\dfrac{9}{10}=\dfrac{-2}{5}.\dfrac{10}{9}=\dfrac{-2.10}{5.9}=\dfrac{-20}{45}=\dfrac{-4}{9}\)
Câu 3:
\(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}:\dfrac{5}{14}=\dfrac{7}{18}+\dfrac{1}{5}=\dfrac{53}{90}\)
Câu 4:
\(\dfrac{2}{7}.\dfrac{3}{11}+\dfrac{2}{7}.\dfrac{8}{11}\)
\(=\dfrac{2}{7}.\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=\dfrac{2}{7}.1\)
\(=\dfrac{2}{7}\)
Câu 1
\(\dfrac{2}{5}\)-\(\dfrac{1}{4}\)+\(\dfrac{3}{10}\)= \(\dfrac{8}{20}\)-\(\dfrac{5}{20}\)+\(\dfrac{6}{20}\)=\(\dfrac{3}{20}\)+\(\dfrac{6}{20}\)=\(\dfrac{9}{20}\)
Câu 2
-\(\dfrac{2}{5}\):(1-\(\dfrac{1}{10}\))= -\(\dfrac{2}{5}\):\(\dfrac{9}{10}\)=-\(\dfrac{2}{5}\).\(\dfrac{10}{9}\)=-\(\dfrac{4}{9}\)
Câu 3
\(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}:\dfrac{5}{14}\)= \(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}.\dfrac{14}{5}\)=\(\dfrac{7.4}{4.2.9}+\dfrac{1.14}{14.5}\)=\(\dfrac{7}{18}+\dfrac{1}{5}\)=\(\dfrac{35}{90}+\dfrac{18}{90}\)=\(\dfrac{53}{90}\)
Câu 4
\(\dfrac{2}{7}.\dfrac{3}{11}+\dfrac{2}{7}.\dfrac{8}{11}\)=\(\dfrac{2}{7}.\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)=\(\dfrac{2}{7}.1\)=\(\dfrac{2}{7}\)
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
a; \(\dfrac{3}{11}\) + \(\dfrac{5}{-9}\) + \(\dfrac{4}{11}\) - \(\dfrac{4}{9}\) + \(\dfrac{3}{17}\) + \(\dfrac{15}{11}\)
= (\(\dfrac{3}{11}\) + \(\dfrac{4}{11}\) + \(\dfrac{15}{11}\)) - (\(\dfrac{5}{9}\) + \(\dfrac{4}{9}\)) + \(\dfrac{3}{17}\)
= 2 - 1 + \(\dfrac{3}{17}\)
= 1 + \(\dfrac{3}{17}\)
= \(\dfrac{20}{17}\)
c; N = \(\dfrac{\dfrac{5}{7}-\dfrac{5}{9}-\dfrac{5}{11}}{\dfrac{15}{7}+\dfrac{15}{9}+\dfrac{15}{11}}\)
Phải là - \(\dfrac{5}{7}\) chỗ tử số mới đúng em nhé!
a) \(4^8\cdot4^4=\left(2^2\right)^8\cdot\left(2^2\right)^4=2^{16}\cdot2^8=2^{16+8}=2^{24}\)
b) \(5^{12}\cdot7-5^{11}\cdot10\)
\(=5^{11}\cdot\left(5\cdot7-10\right)=5^{11}\cdot\left(35-10\right)=5^{11}\cdot25\)
\(=5^{11}\cdot5^2=5^{11+2}=5^{13}\)
d) \(27^{16}:9^{10}\)
\(=\left(3^3\right)^{16}:\left(3^2\right)^{10}=3^{48}:3^{20}=3^{48-20}=3^{28}\)
e) \(125^3:25^4=\left(5^3\right)^3:\left(5^2\right)^4=5^9:5^8=5^{9-8}=5\)
f) \(24^4:3^4-32^{12}:16^{12}\)
\(=\left(24:4\right)^4-\left(32:16\right)^{12}\)
\(=6^4-2^{12}\)
\(=2^4\cdot\left(3^4-2^8\right)=2^4\cdot-175=-2800\)
a) 4.(13 - 16) - (3-5) .(-3)2
b) -2/5+ 7/11+ -11/10 + 7/-11
c) 1/2+2/3+1/6+-2/5
d) 34 .4-66/ 35.5+10.34
a) Ta có: \(4\cdot\left(13-16\right)-\left(3-5\right)\cdot\left(-3\right)^2\)
\(=4\cdot13-4\cdot16+2\cdot9\)
\(=52-64+18\)
=6
b) Ta có: \(-\dfrac{2}{5}+\dfrac{7}{11}+\dfrac{-11}{10}+\dfrac{7}{-11}\)
\(=-\left(\dfrac{2}{5}+\dfrac{11}{10}\right)+\left(\dfrac{7}{11}-\dfrac{7}{11}\right)\)
\(=-\left(\dfrac{4}{10}+\dfrac{11}{10}\right)\)
\(=\dfrac{-15}{10}=\dfrac{-3}{2}\)
\(3^3[(4^2-5):11]-16+2\cdot10^3\\=27[(16-5):11]-16+2\cdot1000\\=27(11:11)-16+2000\\=27-16+2000\\=11+2000\\=2011\)
3³.[(4² - 5) : 11] - 16 + 2.10³
= 27.[(16 - 5) : 11] - 16 + 2.1000
= 27 . (11 : 11) - 16 + 2000
= 27 . 1 + 1984
= 27 + 1984
= 2011