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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: \(\dfrac{205}{107}>\dfrac{20}{23}>\dfrac{7}{10}>\dfrac{214}{315}>-\dfrac{5}{8}>-\dfrac{16}{19}\)
Giải:
a)
\(\dfrac{7}{48}=\dfrac{105}{720};\)
\(\dfrac{11}{72}=\dfrac{110}{720};\)
\(\dfrac{17}{120}=\dfrac{102}{720}\)
Vì \(102< 105< 110\)
\(\Leftrightarrow\dfrac{102}{720}< \dfrac{105}{720}< \dfrac{110}{720}\)
\(\Leftrightarrow\dfrac{17}{120}< \dfrac{7}{48}< \dfrac{11}{72}\)
Vậy ...
b) \(\dfrac{31}{49}=\dfrac{60140}{95060};\)
\(\dfrac{62}{97}=\dfrac{60760}{95060};\)
\(\dfrac{93}{140}=\dfrac{63147}{95060}\)
Vì \(60140< 60760< 63147\)
\(\Leftrightarrow\dfrac{60140}{95060}< \dfrac{60760}{95060}< \dfrac{63147}{95060}\)
\(\Leftrightarrow\dfrac{31}{49}< \dfrac{62}{97}< \dfrac{93}{140}\)
Vậy ...
a ) \(\dfrac{7}{48}\) = \(\dfrac{105}{720}\)
\(\dfrac{11}{72}\) = \(\dfrac{110}{720}\)
\(\dfrac{17}{120}\) = \(\dfrac{102}{720}\)
Vì 102 < 105 < 110
\(\Leftrightarrow\) \(\dfrac{102}{720}\) < \(\dfrac{105}{720}\) < \(\dfrac{110}{720}\)
\(\Leftrightarrow\) \(\dfrac{17}{120}\) < \(\dfrac{7}{48}\) < \(\dfrac{11}{72}\)
Vậy .....................
( k cho tớ nha . Tớ chỉ bt lm phần a )
a) Ta có: \(\dfrac{19}{33}=\dfrac{38}{66};\dfrac{6}{12}=\dfrac{1}{2}=\dfrac{33}{66};\dfrac{13}{22}=\dfrac{39}{66}\)
Mà \(\dfrac{33}{66}< \dfrac{38}{66}< \dfrac{39}{66}\Rightarrow\dfrac{6}{12}< \dfrac{19}{33}< \dfrac{13}{22}\)
Vậy các số hữu tỉ được sắp xếp theo thứ tự tăng dần là: \(\dfrac{6}{12};\dfrac{19}{33};\dfrac{13}{22}\)
b) Ta có:
\(\dfrac{-18}{12}=\dfrac{-3}{2}=\dfrac{-105}{70};\dfrac{-10}{7}=\dfrac{-100}{70};\dfrac{-8}{5}=\dfrac{-112}{70}\)
Mà \(\dfrac{-112}{70}< \dfrac{-105}{70}< \dfrac{-100}{70}\Rightarrow\dfrac{-8}{5}< \dfrac{-18}{12}< \dfrac{-10}{7}\)
Vậy các số hữu tỉ được sắp xếp theo thứ tự tăng dần là: \(\dfrac{-8}{5};\dfrac{-18}{12};\dfrac{-10}{7}\)
a. \(\dfrac{19}{33};\dfrac{6}{12};\dfrac{13}{22}\) ( \(MC=132\) )
Quy đồng : \(\dfrac{19}{33}=\dfrac{76}{132}\) ; \(\dfrac{6}{12}=\dfrac{66}{132}\) ; \(\dfrac{13}{22}=\dfrac{78}{132}\)
Vì \(\dfrac{66}{132}< \dfrac{76}{132}< \dfrac{78}{132}\) => \(\dfrac{6}{12}< \dfrac{19}{33}< \dfrac{13}{22}\)
b. \(\dfrac{-18}{12};\dfrac{-10}{7};\dfrac{-8}{5}\) ( \(MC=420\) )
Quy đồng : \(\dfrac{-18}{12}=\dfrac{-630}{420}\) ; \(\dfrac{-10}{7}=\dfrac{-600}{420}\) ; \(\dfrac{-8}{5}=\dfrac{-672}{420}\)
Vì : \(\dfrac{-672}{420}< \dfrac{-630}{420}< \dfrac{-600}{420}\) => \(\dfrac{-8}{5}< \dfrac{-18}{12}< \dfrac{-10}{7}\)
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
\(\dfrac{205}{107}\),\(\dfrac{20}{23}\),\(\dfrac{7}{10}\),\(\dfrac{214}{315}\),\(\dfrac{-5}{8}\),\(\dfrac{-16}{19}\)
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: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)
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
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
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 :\(\dfrac{1}{2}\);\(\dfrac{2}{3}\);\(\dfrac{3}{4}\);\(\dfrac{4}{5}\)
\(\Rightarrow\)\(\dfrac{7}{10}\) : \(\dfrac{7}{5}\) ; \(\dfrac{3}{2}\) :\(\dfrac{9}{4}\) ; \(\dfrac{6}{7}\) : \(\dfrac{8}{7}\) ; \(\dfrac{48}{55}\):\(\dfrac{12}{11}\).
Vậy theo thứ tự tăng dần lần lượt là