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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 : \(\dfrac{-672}{420}< \dfrac{-630}{420}< \dfrac{-600}{420}\) => \(\dfrac{-8}{5}< \dfrac{-18}{12}< \dfrac{-10}{7}\)
Bài 3:
Để A là số nguyên thì \(n-2+5⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{3;1;7;-3\right\}\)
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)
\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)
b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)
\(=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)
\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
Bài 1. Tính giá trị của biểu thức:
a, \(A=\dfrac{3}{11}.\dfrac{-7}{19}+\dfrac{17}{11}.\dfrac{-3}{19}+\dfrac{3}{19}.\dfrac{25}{11}.\)
\(=\dfrac{3}{19}.\dfrac{-7}{11}+\dfrac{-17}{11}.\dfrac{3}{19}+\dfrac{3}{19}.\dfrac{24}{11}.\)
\(=\dfrac{3}{19}\left(\dfrac{-7}{11}+\dfrac{-17}{11}+\dfrac{24}{11}\right).\)
\(=\dfrac{3}{19}.0\)
\(=0.\)
Vậy A = 0.
b, \(B=\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}.....\dfrac{99^2}{99.100}.\)
\(=\dfrac{3.3.4.4.....99.99}{\left(3.4.5.....99\right)\left(4.5.6.....100\right)}.\)
\(=\dfrac{\left(3.4.5.....99\right)\left(3.4.5.....99\right)}{\left(3.4.5.....99\right)\left(4.5.6.....100\right)}.\)
\(=\dfrac{1.3}{1.100}.\)
\(=\dfrac{3}{100}.\)
Vậy \(B=\dfrac{3}{100}.\)
Bài 2. So sánh:
\(A=\dfrac{10^{2015}+7}{10^{2016}+7}\) và \(B=\dfrac{10^{2016}+7}{10^{2017}+7}.\)
Giải:
Ta có:
\(10A=\dfrac{\left(10^{2015}+7\right)10}{10^{2016}+7}.\)
\(=\dfrac{10^{2016}+70}{10^{2016}+7}.\)
\(=\dfrac{\left(10^{2016}+7\right)+63}{10^{2016}+7}.\)
\(=1+\dfrac{63}{10^{2016}+7}._{\left(1\right).}\)
\(10B=\dfrac{\left(10^{2016}+7\right)10}{10^{2017}+7}.\)
\(=\dfrac{10^{2017}+70}{10^{2017}+7}.\)
\(=\dfrac{\left(10^{2017}+7\right)+63}{10^{2017}+7}.\)
\(=1+\dfrac{63}{10^{2017}+7}._{\left(2\right).}\)
Mà \(\dfrac{63}{10^{2016}+7}>\dfrac{63}{10^{2017}+7}._{\left(3\right).}\)
Từ (1), (2) và (3) suy ra: \(10A>10B.\).
\(\Rightarrow A>B.\)
Vậy A > B.
CHÚC BN HỌC GIỎI!!! ^ - ^
Đừng quên bình luận nếu bài mik sai nhé!!! Và nếu bài mik đúng thì nhớ tick mik nha!!!
a, \(A=\dfrac{3}{11}.\dfrac{-7}{19}+\dfrac{17}{11}.\dfrac{-3}{19}+\dfrac{3}{19}.\dfrac{25}{11}.\)
\(=\dfrac{3}{19}.\dfrac{-7}{11}+\dfrac{-17}{11}.\dfrac{3}{19}+\dfrac{3}{19}.\dfrac{25}{11}.\)
\(=\dfrac{3}{19}\left(\dfrac{-7}{11}+\dfrac{-17}{11}+\dfrac{25}{11}\right).\)
\(=\dfrac{3}{19}.\dfrac{1}{11}.\)
\(=\dfrac{3}{209}.\)
Vậy \(A=\dfrac{3}{209}.\)
Do phần a có 1 chút nhầm lẫn của mik nên bài mik bị sai nhé, xin lỗi bn!!!
CHÚC BN HỌC GIỎI!!!
a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
a) \(\dfrac{-2}{3};\dfrac{-3}{4};\dfrac{-7}{8};\dfrac{-18}{19};\dfrac{-27}{28}\)
b) \(\dfrac{1998}{1997};\dfrac{1}{173};0;\dfrac{-15}{31};\dfrac{-27}{53}\)
giải chi tiết ra vì sao nó như thế giúp mình nhé :)