Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{6}{7}+\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}.\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}\left(\dfrac{2}{7}+\dfrac{5}{7}\right).\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}.1.\)
\(A=\dfrac{6}{7}+\dfrac{1}{7}=1.\)
Vậy \(A=1.\)
\(B=\dfrac{40}{9}.\dfrac{13}{3}-\dfrac{4}{3}.\dfrac{40}{9}.\)
\(B=\dfrac{4}{9}.\dfrac{13}{3}-\dfrac{4}{9}.\dfrac{40}{3}.\)
\(B=\dfrac{4}{9}\left(\dfrac{13}{3}-\dfrac{40}{3}\right).\)
\(B=\dfrac{4}{9}.\left(-9\right).\)
\(B=-4.\)
Vậy \(B=-4.\)
ta có
\(2.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(5.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{11}\right)\)
_______________________ X ________________________
\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)
= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)
13: \(=\dfrac{4}{9}\cdot\left(-7\right)+\left(6+\dfrac{5}{9}\right)\cdot\left(-7\right)\)
\(=\left(-7\right)\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)=\left(-7\right)\cdot7=-49\)
14: \(=\left(\dfrac{-3}{4}+\dfrac{5}{13}\right)\cdot\dfrac{7}{2}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right)\cdot\dfrac{7}{2}\)
\(=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)\)
\(=\dfrac{7}{2}\left(-3-\dfrac{3}{13}\right)=\dfrac{7}{2}\cdot\dfrac{-42}{13}=\dfrac{-147}{13}\)
\(D=\dfrac{1}{2}+\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{1}{6}+\dfrac{-3}{35}+\dfrac{1}{3}+\dfrac{1}{41}\)
\(D=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{-3}{35}\right)+\dfrac{1}{41}\)
\(D=1+-1+\dfrac{1}{41}\)
\(D=0+\dfrac{1}{41}\)
\(D=\dfrac{1}{41}\)
\(C=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{-3}{4}+\dfrac{-1}{36}+\dfrac{-2}{9}\right)+\dfrac{1}{57}\)
\(=\dfrac{5+9+1}{15}+\dfrac{-27-1-8}{36}+\dfrac{1}{57}\)
=1/57
\(E=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}=\dfrac{1}{127}\)
a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)
\(\Rightarrow-4< x< \dfrac{-3}{10}\)
\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)
\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)
b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)
\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)
\(\Rightarrow x=\varnothing\)
c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)
\(\Rightarrow x\in\left\{1;2\right\}\)
+) Với \(x=1\)
\(\Rightarrow y\in\left\{1;2\right\}\)
+) Với \(x=2\)
\(\Rightarrow y=2\)
Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).
1) \(\dfrac{5}{6}-\dfrac{6}{7}+\dfrac{7}{8}-\dfrac{8}{9}+\dfrac{10}{9}-\dfrac{5}{6}+\dfrac{6}{7}-\dfrac{7}{8}+\dfrac{8}{9}\)
\(=\left(\dfrac{5}{6}-\dfrac{5}{6}\right)-\left(\dfrac{6}{7}+\dfrac{6}{7}\right)+\left(\dfrac{7}{8}-\dfrac{7}{8}\right)-\left(\dfrac{8}{9}+\dfrac{8}{9}\right)+\dfrac{10}{9}\)
\(=0-0+0-0+\dfrac{10}{9}\)
\(=\dfrac{10}{9}\)
2) \(\dfrac{1}{13}+\dfrac{16}{7}+\dfrac{3}{105}-\dfrac{9}{7}-\dfrac{-12}{13}\)
\(=\left(\dfrac{1}{13}-\left(-\dfrac{12}{13}\right)\right)+\left(\dfrac{16}{7}-\dfrac{9}{7}\right)+\dfrac{3}{105}\)
\(=1+1+\dfrac{3}{105}\)
\(=\dfrac{213}{105}=\dfrac{71}{35}\)
a) \(6\dfrac{5}{7}-\left(1\dfrac{3}{4}+2\dfrac{5}{7}\right)\)
\(=6\dfrac{5}{7}-1\dfrac{3}{4}-2\dfrac{5}{7}\)
\(=\left(6\dfrac{5}{7}-2\dfrac{5}{7}\right)-1\dfrac{3}{4}\)
\(=4-1\dfrac{3}{4}\)
\(=3\dfrac{3}{4}\)
b) \(7\dfrac{5}{11}-\left(2\dfrac{3}{7}+3\dfrac{5}{11}\right)\)
\(=7\dfrac{5}{11}-2\dfrac{3}{7}-3\dfrac{5}{11}\)
\(=\left(7\dfrac{5}{11}-3\dfrac{5}{11}\right)-2\dfrac{3}{7}\)
\(=4-2\dfrac{3}{7}\)
\(=2\dfrac{3}{7}\)
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: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=5\)
Gọi kết quả của phép tính là A
=> \(\dfrac{4}{7.25}\) + \(\dfrac{5}{25.10}\) + \(\dfrac{6}{10.36}\) + \(\dfrac{9}{36.15}\) = A
<=> \(\dfrac{4}{175}\) + \(\dfrac{5}{250}\) + \(\dfrac{6}{360}\) + \(\dfrac{9}{540}\) = A
<=> (\(\dfrac{4}{175}\) + \(\dfrac{1}{50}\)) + (\(\dfrac{1}{60}\) + \(\dfrac{1}{60}\)) = A
<=> (\(\dfrac{200}{8750}\) + \(\dfrac{175}{8750}\)) + \(\dfrac{2}{60}\) = A
<=> \(\dfrac{375}{8750}\) + \(\dfrac{1}{30}\) = A
<=> \(\dfrac{3}{70}\) + \(\dfrac{1}{30}\) = A
<=> \(\dfrac{90}{2100}\) + \(\dfrac{70}{2100}\) = A
<=> \(\dfrac{160}{2100}\)
Lỡ gửi trả lời còn nữa nè :
<=> \(\dfrac{160}{2100}\) = A
<=> A = \(\dfrac{8}{105}\)
Vậy tổng của biểu thức là \(\dfrac{8}{105}\)