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\(\dfrac{3}{-2}=\dfrac{-9}{6};\dfrac{-1}{-7}=\dfrac{1}{7}\)
Sắp xếp:
\(\dfrac{-1}{-7};\dfrac{0}{8};\dfrac{-7}{6};\dfrac{3}{-2}\)
1) \(\dfrac{3}{2}\) : \(\dfrac{9}{4}\) =\(\dfrac{3}{2}\)x\(\dfrac{4}{9}\)=\(\dfrac{12}{18}\)=\(\dfrac{2}{3}\)
2)\(\dfrac{48}{55}\) : \(\dfrac{12}{11}\)= \(\dfrac{48}{55}\) x\(\dfrac{11}{12}\)= \(\)\(\dfrac{528}{660}\)=\(\dfrac{4}{5}\)
3)\(\dfrac{7}{10}\) : \(\dfrac{7}{5}\)=\(\dfrac{7}{10}\) x \(\dfrac{5}{7}\)= \(\dfrac{35}{70}\)=\(\dfrac{1}{2}\)
4)\(\dfrac{6}{7}\) : \(\dfrac{8}{7}\) = \(\dfrac{6}{7}\) x \(\dfrac{7}{8}\) =\(\dfrac{42}{56}\)=\(\dfrac{3}{4}\)
Sắp xếp các thương theo thứ tự tăng dần :\(\dfrac{1}{2}\);\(\dfrac{2}{3}\);\(\dfrac{3}{4}\);\(\dfrac{4}{5}\)
\(\Rightarrow\)\(\dfrac{7}{10}\) : \(\dfrac{7}{5}\) ; \(\dfrac{3}{2}\) :\(\dfrac{9}{4}\) ; \(\dfrac{6}{7}\) : \(\dfrac{8}{7}\) ; \(\dfrac{48}{55}\):\(\dfrac{12}{11}\).
a: \(\dfrac{-5}{6}=\dfrac{-20}{24};\dfrac{7}{8}=\dfrac{21}{24};\dfrac{7}{24}=\dfrac{7}{24};\dfrac{-3}{4}=-\dfrac{18}{24};\dfrac{2}{3}=\dfrac{16}{24}\)
Do đó: \(\dfrac{-5}{6}< -\dfrac{3}{4}< \dfrac{7}{24}< \dfrac{2}{3}< \dfrac{7}{8}\)
\(\dfrac{7}{8}=\dfrac{119}{136};\dfrac{16}{17}=\dfrac{128}{136}\)
mà 119<128
nên 7/8<16/17
DO đó: -5/6<-3/4<7/24<2/3<7/8<16/17
b: \(\dfrac{-5}{8}=\dfrac{-95}{8\cdot19};\dfrac{-16}{19}=\dfrac{-128}{19\cdot8}\)
Do đó: -5/8>-16/19
\(\dfrac{7}{10}=0.7;\dfrac{20}{23}\simeq0.87;\dfrac{214}{315}\simeq0.68;\dfrac{205}{107}>1\)
Do đó: \(\dfrac{205}{107}>\dfrac{20}{23}>\dfrac{7}{10}>\dfrac{214}{315}>-\dfrac{5}{8}>-\dfrac{16}{19}\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2007^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{2006.2007}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2006}-\dfrac{1}{2007}\)
\(=\dfrac{1}{4}-\dfrac{1}{2007}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2007^2}< \dfrac{1}{4}\left(đpcm\right)\)
Vậy...
a) \(\dfrac{1}{6};\dfrac{1}{3};\dfrac{1}{2};...\)
\(\Rightarrow\dfrac{1}{6};\dfrac{2}{6};\dfrac{3}{6};...\)
Dãy có quy luật tăng dần lên 1 đơn vị ở tử số
\(\Rightarrow\) Số tiếp theo của dãy là: \(\dfrac{4}{6}\)
b) \(\dfrac{1}{8};\dfrac{5}{24};\dfrac{7}{24};...\)
\(\Rightarrow\dfrac{3}{24};\dfrac{5}{24};\dfrac{7}{24};...\)
Dãy có quy luật tăng dần lên 2 đơn vị ở tử số
\(\Rightarrow\) Số tiếp theo của dãy là: \(\dfrac{9}{24}\)
c) \(\dfrac{1}{5};\dfrac{1}{4};\dfrac{1}{3};...\)
\(\dfrac{4}{20};\dfrac{5}{20};\dfrac{6}{20};...\)
Dãy có quy luật tăng dần lên 1 đơn vị ở tử số
\(\Rightarrow\) Số tiếp theo của dãy là: \(\dfrac{7}{20}\)
d) \(\dfrac{4}{15};\dfrac{3}{10};\dfrac{1}{3};...\)
\(\Rightarrow\dfrac{8}{30};\dfrac{9}{30};\dfrac{11}{30};...\)
Dãy có quy luật tăng dần lên 1 đơn vị ở tử số
\(\Rightarrow\) Số tiếp theo của dãy là: \(\dfrac{12}{30}\)
1.a) 0,7 = \(\dfrac{7}{10}\) ; 0,94=\(\dfrac{94}{100}=\dfrac{47}{50}\) ; 2,7\(=\dfrac{27}{10}\) ; 4,567\(=\dfrac{4567}{1000}\)
b) \(\dfrac{1}{4}=0,25\) ; \(\dfrac{7}{5}=1,4\) ; \(\dfrac{16}{25}=0,64\) ; \(\dfrac{3}{2}=1,5\)
2.a) 0,6 = 60% ; 0,48 = 48% ; 6,25 = 625%
b) 7% = 0,07 ; 37% = 0,37 ; 785% = 7,85
3.a) \(\dfrac{1}{4}\)giờ = 0,25 giờ
\(\dfrac{3}{2}\)phút = 1,5 phút
\(\dfrac{2}{5}\)giờ = 0,4 giờ
b) \(\dfrac{3}{4}\)kg = 0,75kg
\(\dfrac{7}{10}\)m = 0,7m
\(\dfrac{3}{5}\)km = 0,6km
4.a) 7,305 ; 7,35 ; 7,6 ; 7,602
b) 62,3 ; 61,98 ; 54,7 ; 54,68
5.
0,3< 0,31 ; 0,32 ; 0,33 <0,4
6.
\(\dfrac{7}{10}=0,7\) ; \(\dfrac{7}{100}=0,07\) ; \(6\dfrac{38}{100}=6,38\) ; \(\dfrac{2014}{1000}=2,014\)
\(\dfrac{3}{2}=1,5\) ; \(\dfrac{2}{5}=0,4\) ; \(\dfrac{5}{8}=0,625\) ; \(1\dfrac{1}{4}=1.25\)
bài này trong sách giáo khoa lớp 5 đúng ko dù bây h anh lớp 6
a,
= \(\dfrac{20}{100}.\dfrac{15}{36}-\)(\(\dfrac{2}{5}+\dfrac{2}{3}\)):\(\dfrac{6}{5}\)
= \(\dfrac{20.15}{100.36}\)- ( \(\dfrac{6}{15}+\dfrac{10}{15}\)): \(\dfrac{6}{5}\)
= \(\dfrac{1.5}{5.12}-\dfrac{16}{15}:\dfrac{6}{5}\)
= \(\dfrac{1.1}{1.12}-\dfrac{16.5}{10.6}\)
= \(\dfrac{1}{12}-\dfrac{16.1}{2.6}\)
= \(\dfrac{1}{12}\) - \(\dfrac{16}{12}\)
= \(\dfrac{1}{12}+\dfrac{\left(-16\right)}{12}\)
= \(\dfrac{15}{12}\) = \(\dfrac{5}{4}\)
b,
= \(\dfrac{6}{7}+\dfrac{5}{4}+\dfrac{\left(-4\right)}{7}+\dfrac{25}{4}\)
= (\(\dfrac{6}{7}+\dfrac{\left(-4\right)}{7}\)) + (\(\dfrac{5}{4}+\dfrac{25}{4}\))
= \(\dfrac{2}{7}+\dfrac{30}{4}\)
= \(\dfrac{2}{7}+\dfrac{15}{2}\)
= \(\dfrac{4}{14}+\dfrac{105}{14}\)
= \(\dfrac{109}{14}\)
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:
\(\dfrac{4}{12\cdot14}+\dfrac{4}{14\cdot16}+\dfrac{4}{16\cdot18}+...+\dfrac{4}{58\cdot60}\)
\(=2\left(\dfrac{1}{12\cdot14}+\dfrac{1}{14\cdot16}+\dfrac{1}{16\cdot18}+...+\dfrac{1}{58\cdot60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{18}+...+\dfrac{1}{58}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{5}{60}-\dfrac{1}{60}\right)\)
\(=2\cdot\dfrac{1}{15}\)
\(=\dfrac{2}{15}\)
Các câu dễ tự làm nha:
\(D=\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(D=\dfrac{1}{99}-\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)\(D=-\dfrac{1}{100}-1\)
Quy đồng mẫu số:
\(\dfrac{15}{30}\);\(\dfrac{20}{30}\);\(\dfrac{6}{30}\);\(\dfrac{10}{30}\);\(\dfrac{35}{30}\)
Sắp xếp :
\(\dfrac{6}{30}\);\(\dfrac{10}{30}\);\(\dfrac{15}{30}\);\(\dfrac{20}{30}\);\(\dfrac{35}{30}\)