Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{20}\right)\)
\(=\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)\cdot\left(\dfrac{5}{20}-\dfrac{4}{20}-\dfrac{1}{20}\right)\)
\(=0\cdot\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)=0\)
b) Ta có: \(\dfrac{12}{5}\cdot\left(\dfrac{10}{3}-\dfrac{5}{12}\right)\)
\(=\dfrac{12}{5}\cdot\left(\dfrac{40}{12}-\dfrac{5}{12}\right)\)
\(=\dfrac{12}{5}\cdot\dfrac{35}{12}\)
=7
\(1,Y=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ Y=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ Y=13\left(1+3^3+...+3^{96}\right)⋮13\\ 2,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{2019}\right)\\ A=4\left(1+3^2+...+3^{2019}\right)⋮4\\ 3,\Leftrightarrow2\left(x+4\right)=60\Leftrightarrow x+4=30\Leftrightarrow x=36\)
+) \(176+483+24+117\)
\(=\left(176+24\right)+\left(483+117\right)\)
\(=200+600\)
\(=800\)
+) \(239+518+761+482\)
\(=\left(239+761\right)+\left(518+482\right)\)
\(=1000+1000\)
\(=2000\)
+) \(32+33+34+...+78+79+80\)
Số số hạng của dãy là :
\(\left(80-32\right):1+1=49\) (số)
Tổng của dãy là :
\(\frac{\left(80+32\right)\times49}{2}=2744\)
+) \(5\times125\times2\times4\)
\(=\left(5\times2\right)\times\left(125\times4\right)\)
\(=10\times500\)
\(=5000\)
+) \(25\times50\times4\times20\)
\(=\left(25\times4\right)\times\left(50\times20\right)\)
\(=100\times1000\)
\(=100000\)
+) \(17\times32+43\times17+17\times25\)
\(=17\times\left(32+43+25\right)\)
\(=17\times100\)
\(=1700\)
+) \(24\times19+29\times24+18\times24+24\times33+24\)
\(=24\times\left(19+29+18+33+1\right)\)
\(=24\times100\)
\(=2400\)
_Chúc bạn học tốt_
c: \(=\dfrac{4}{9}\left(\dfrac{5}{7}+\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{7}{5}\right)=\dfrac{4}{9}\cdot\left(1-1\right)=0\)
d: \(=\dfrac{4-3-1}{12}\cdot\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)=0\cdot\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)=0\)
Câu 11:
(\(\dfrac{11}{4}\). \(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\).\(\dfrac{11}{4}\)).\(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(\(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\)). \(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(-1).\(\dfrac{8}{33}\)
= - \(\dfrac{2}{3}\)
Bài 2
a: =>x=-23-7=-30
b: =>75:x=15
hay x=5
c: =>2(x+4)=80
=>x+4=40
hay x=36
d: \(\Leftrightarrow2^x\cdot11=88\)
hay x=3
Bài 3:
Có thể chia được nhiều nhất 12 tổ vì UCLN(180;132)=12
Khi đó, mỗi tổ có 15 nam và 11 nữ
(617/191+29/33-115/117)*(1/4-1/5-1/20)
= (617/191+29/33-115/117)*(5/20-4/20-1/20)
=(617/191+29/33-115/117)*0
=0
\(\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{117}\right)\times\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{20}\right)\\ =\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{117}\right)\times0\\ =0\)