\(4\dfrac{1}{2}\) + \(\dfrac{2}{3...">
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16 tháng 4 2017

a) \(\dfrac{4}{7}+\dfrac{5}{6}:5-0.375\cdot\left(-2\right)^2\)

\(=\dfrac{4}{7}+\dfrac{5}{6}\cdot\dfrac{1}{5}-\dfrac{3}{8}\cdot4\\ =\dfrac{4}{7}+\dfrac{1}{6}-\dfrac{3}{2}\\ =-\dfrac{16}{21}\)

b) \(\dfrac{1}{4}+\dfrac{3}{4}\cdot\left(-\dfrac{1}{2}+\dfrac{2}{3}\right)\)

\(=\dfrac{1}{4}+\dfrac{3}{4}\cdot\dfrac{1}{6}\\ =\dfrac{1}{4}+\dfrac{1}{8}\\ =\dfrac{3}{8}\)

17 tháng 4 2017

Giải bài 89 trang 43 SGK Toán 6 Tập 2 | Giải toán lớp 6

1 tháng 5 2018

Giải bà i 89 trang 43 SGK Toán 6 Tập 2 | Giải toán lá»p 6

20 tháng 7 2018

a) \(\dfrac{3}{4}+\dfrac{3}{5}-\dfrac{18}{60}\) ( MTC: 60)

= \(\dfrac{3.15}{4.15}+\dfrac{3.12}{5.12}-\dfrac{18}{60}\)

= \(\dfrac{45}{60}+\dfrac{36}{60}-\dfrac{18}{60}\)

= \(\dfrac{45+36-18}{60}\)=\(\dfrac{63}{60}=\dfrac{21}{20}\)

b)\(\dfrac{17}{8}-\dfrac{11}{6}-\dfrac{2}{9}\) (MTC:72)

=\(\dfrac{17.9}{8.9}-\dfrac{11.12}{6.12}-\dfrac{2.8}{9.8}\)

= \(\dfrac{153}{72}-\dfrac{132}{72}-\dfrac{16}{72}\)

=\(\dfrac{153-132-16}{72}\)

=\(\dfrac{5}{72}\)

c)\(\dfrac{23}{29}+\dfrac{5}{11}+\dfrac{17}{11}\) (MTC:319)

= \(\dfrac{23.11}{29.11}+\dfrac{5.29}{11.29}+\dfrac{17.29}{11.29}\)

=\(\dfrac{253}{319}+\dfrac{145}{319}+\dfrac{493}{319}\)

=\(\dfrac{253+145+493}{319}\)=\(\dfrac{891}{319}=\dfrac{81}{29}\)

20 tháng 7 2018

c) \(\dfrac{20}{45}+\dfrac{14}{35}+\dfrac{32}{44}\)

= \(\dfrac{4}{9}+\dfrac{2}{5}+\dfrac{8}{11}\)(Rút gọn b/thức)(MTC:495)

=\(\dfrac{4.55}{9.55}+\dfrac{2.99}{5.99}+\dfrac{8.45}{11.45}\)

=\(\dfrac{220}{495}+\dfrac{198}{495}+\dfrac{360}{495}\)

=\(\dfrac{220+198+360}{495}\)=\(\dfrac{778}{495}\)

e)\(17\dfrac{25}{27}+3\dfrac{7}{2}\)

= \(\dfrac{484}{27}+\dfrac{13}{2}\) (MTC:54)

=\(\dfrac{484.2}{27.2}+\dfrac{13.27}{2.27}\)

\(=\dfrac{968}{54}+\dfrac{351}{54}\)

=\(\dfrac{968+351}{54}=\dfrac{1319}{54}\)

5 tháng 4 2017

1) \(-\dfrac{5}{9}+1\dfrac{5}{9}\cdot\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =-\dfrac{5}{9}+\dfrac{14}{9}\cdot\dfrac{7}{20}\cdot\dfrac{1}{49}\\ =-\dfrac{5}{9}+\dfrac{1}{90}=\dfrac{-49}{90}\)

2) \(1\dfrac{13}{15}\cdot0,75-\left(\dfrac{104}{195}+25\%\right)\cdot\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}\cdot\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right)\cdot\dfrac{24}{47}-\dfrac{51}{13}\cdot\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}\cdot\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =-\dfrac{4}{13}\)

3) \(1\dfrac{13}{15}\cdot\left(0,5\right)^2\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{5}-\dfrac{47}{60}\cdot\dfrac{24}{47}\\ =\dfrac{7}{5}-\dfrac{2}{5}\\ =1\)

Tìm x : 1) \(60\%x+0,4x+x:3=2\\ \Leftrightarrow\dfrac{3}{5}x+\dfrac{2}{5}x+\dfrac{1}{3}x=2\\ \Leftrightarrow\dfrac{4}{3}x=2\\ \Leftrightarrow x=\dfrac{3}{2}\)

5 tháng 4 2017

Nốt nè bn

\(-2x-\dfrac{-3}{5}:\left(0,5\right)^2=-1\dfrac{1}{4}\\ \Leftrightarrow-2x+\dfrac{3}{5}:\dfrac{1}{4}=-\dfrac{5}{4}\\ \Leftrightarrow-2x+\dfrac{12}{5}=-\dfrac{5}{4}\\ \Leftrightarrow-2x=-\dfrac{73}{20}\\ x=-\dfrac{73}{40}\)

\(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\\ \Leftrightarrow\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\\ \Leftrightarrow\dfrac{2}{3}-x=\dfrac{3}{20}\\ \Leftrightarrow x=\dfrac{2}{3}-\dfrac{3}{20}\\ \Leftrightarrow x=\dfrac{31}{60}\)

20 tháng 4 2017

a) \(\dfrac{8}{5}-\dfrac{9}{5}=\dfrac{8-9}{5}=\dfrac{-1}{5}\)

b) \(\dfrac{5}{2}+\dfrac{2}{3}=\dfrac{15}{6}+\dfrac{4}{6}=\dfrac{15+4}{6}=\dfrac{19}{6}\)

c) \(\dfrac{-5}{9}\cdot\dfrac{2}{11}=\dfrac{-5\cdot2}{9\cdot11}=\dfrac{-10}{99}\)

d) \(\dfrac{-2}{9}:\dfrac{1}{3}=\dfrac{-2}{9}\cdot3=\dfrac{-2}{3}\)

e) \(\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{5}{12}=\dfrac{9}{24}-\dfrac{6}{24}+\dfrac{10}{24}=\dfrac{9-6+10}{24}=\dfrac{13}{24}\)

f) \(\dfrac{-4}{3}\cdot\dfrac{5}{4}:\dfrac{7}{3}=\dfrac{-4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{3}{7}=\dfrac{-4\cdot5\cdot3}{3\cdot4\cdot7}=\dfrac{-5}{7}\)

a: \(=\left(\dfrac{37}{9}+\dfrac{13}{4}\right)\cdot\dfrac{9}{4}+\dfrac{11}{4}\)

\(=\dfrac{265}{36}\cdot\dfrac{9}{4}+\dfrac{11}{4}=\dfrac{309}{16}\)

b: \(=1+\dfrac{9-8}{10}:\dfrac{19}{6}\)

\(=1+\dfrac{1}{10}\cdot\dfrac{6}{19}=1+\dfrac{3}{95}=\dfrac{98}{95}\)

30 tháng 4 2017

Bài 1:

a) \(\dfrac{2}{5}\cdot x-\dfrac{1}{4}=\dfrac{1}{10}\)

\(\dfrac{2}{5}\cdot x=\dfrac{1}{10}+\dfrac{1}{4}\)

\(\dfrac{2}{5}\cdot x=\dfrac{7}{20}\)

\(x=\dfrac{7}{20}:\dfrac{2}{5}\)

\(x=\dfrac{7}{8}\)

Vậy \(x=\dfrac{7}{8}\).

b) \(\dfrac{3}{5}=\dfrac{24}{x}\)

\(x=\dfrac{5\cdot24}{3}\)

\(x=40\)

Vậy \(x=40\).

c) \(\left(2x-3\right)^2=16\)

\(\left(2x-3\right)^2=4^2\)

\(\circledast\)TH1: \(2x-3=4\\ 2x=4+3\\ 2x=7\\ x=\dfrac{7}{2}\)

\(\circledast\)TH2: \(2x-3=-4\\ 2x=-4+3\\ 2x=-1\\ x=\dfrac{-1}{2}\)

Vậy \(x\in\left\{\dfrac{7}{2};\dfrac{-1}{2}\right\}\).

Bài 2:

a) \(25\%-4\dfrac{2}{5}+0.3:\dfrac{6}{5}\)

\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}:\dfrac{6}{5}\)

\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}\cdot\dfrac{5}{6}\)

\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{1}{4}\)

\(=\dfrac{5}{20}-\dfrac{88}{20}+\dfrac{5}{20}\)

\(=\dfrac{5-88+5}{20}\)

\(=\dfrac{78}{20}=\dfrac{39}{10}\)

b) \(\left(\dfrac{1}{6}-\dfrac{1}{5^2}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)

\(=\left(\dfrac{1}{6}-\dfrac{1}{25}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)

\(=\left(\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)

\(=\left(\dfrac{5}{30}-\dfrac{6}{30}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)

\(=\left(\dfrac{5-6+1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)

\(=0\cdot\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)

\(=0\)

Bài 3:

a) \(\dfrac{4}{19}\cdot\dfrac{-3}{7}+\dfrac{-3}{7}\cdot\dfrac{15}{19}\)

\(=\dfrac{-3}{7}\left(\dfrac{4}{19}+\dfrac{15}{19}\right)\)

\(=\dfrac{-3}{7}\cdot1\)

\(=\dfrac{-3}{7}\)

b) \(7\dfrac{5}{9}-\left(2\dfrac{3}{4}+3\dfrac{5}{9}\right)\)

\(=\dfrac{68}{9}-\dfrac{11}{4}-\dfrac{32}{9}\)

\(=\dfrac{68}{9}-\dfrac{32}{9}-\dfrac{11}{4}\)

\(=4-\dfrac{11}{4}\)

\(=\dfrac{16}{4}-\dfrac{11}{4}\)

\(\dfrac{5}{4}\)

Bài 4:

\(\dfrac{4}{12\cdot14}+\dfrac{4}{14\cdot16}+\dfrac{4}{16\cdot18}+...+\dfrac{4}{58\cdot60}\)

\(=2\left(\dfrac{1}{12\cdot14}+\dfrac{1}{14\cdot16}+\dfrac{1}{16\cdot18}+...+\dfrac{1}{58\cdot60}\right)\)

\(=2\left(\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{18}+...+\dfrac{1}{58}-\dfrac{1}{60}\right)\)

\(=2\left(\dfrac{1}{12}-\dfrac{1}{60}\right)\)

\(=2\left(\dfrac{5}{60}-\dfrac{1}{60}\right)\)

\(=2\cdot\dfrac{1}{15}\)

\(=\dfrac{2}{15}\)

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: \(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{45}{4}\cdot\dfrac{2}{15}\)

\(=\dfrac{77}{12}\cdot\dfrac{4}{11}+\dfrac{3}{2}\)

\(=\dfrac{7}{3}+\dfrac{3}{2}=\dfrac{23}{6}\)

b: \(=\left(\dfrac{3}{5}+\dfrac{415}{1000}-\dfrac{3}{200}\right)\cdot\dfrac{8}{3}\cdot\dfrac{1}{4}\)

\(=\dfrac{600+415-15}{1000}\cdot\dfrac{2}{3}=\dfrac{2}{3}\)

c: \(=\dfrac{28}{15}\cdot\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right)\cdot\dfrac{3}{7}\)

\(=\dfrac{7}{5}-\dfrac{3}{4}\cdot\dfrac{3}{7}=\dfrac{7}{5}-\dfrac{9}{28}=\dfrac{151}{140}\)

2 tháng 4 2017

\(1\dfrac{13}{15}-\left(0.5\right)^2.3+37.5\%:\left(\dfrac{8}{15}-1\dfrac{9}{60}\right)\)

\(=\dfrac{28}{15}.\dfrac{3}{4}+\dfrac{3}{8}:-\dfrac{37}{60}\)

\(=\dfrac{7}{5}+\left(-\dfrac{45}{74}\right)\)

\(=\dfrac{293}{370}\)

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2 tháng 4 2017

\(=\dfrac{28}{15}.0,25.3+0,375:\left(-\dfrac{37}{60}\right)\)

\(=1,4-\dfrac{45}{74}\)

=\(\dfrac{293}{370}\)