K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
V
2
7 tháng 4 2020
c) \(\frac{3\cdot7-3\cdot19}{3\cdot4}=\frac{3\left(7-19\right)}{3\cdot4}=\frac{3\cdot\left(-12\right)}{3\cdot4}=\frac{3\cdot4\cdot\left(-3\right)}{3\cdot4}=-3\)
Vậy \(\frac{3\cdot7-3\cdot19}{3\cdot4}=-3\)
d,\(\frac{2^3\cdot9^4-2^4\cdot3^7}{2^4\cdot3^7}=\frac{2^3\cdot\left(3^2\right)^4-2^4\cdot3^7}{2^4\cdot3^7}=\frac{2^3\cdot3^8-2^4\cdot3^7}{2^2\cdot3^7}=\frac{2^3\cdot3^7\left(3-2\right)}{2^2\cdot3^7}=2\)
VM
21 tháng 7 2018
Bài 1
a.\(\frac{-3}{4}\)-y:\(\frac{1}{5}\)=\(\frac{9}{28}\)
y:\(\frac{1}{5}\)=\(\frac{-15}{14}\)
y= \(\frac{-3}{14}\)
b.5x + 5x+2=650
5x . 1 + 5x + 52=650
5x(1+25)=650
5x.26=650
5x=25
x=2
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
4 tháng 8 2023
\(S=5+5^2+5^3+5^4+...+5^{2022}\\ =\left(5+5^2\right)+5^2.\left(5+5^2\right)+...+5^{2020}.\left(5+5^2\right)\\ =30+30.5^2+...+30.5^{2020}\\ =30.\left(1+5^2+...+5^{2020}\right)⋮30\)
ND
Nguyễn Đức Trí
VIP
4 tháng 8 2023
\(S=5+5^2+5^3+...+5^{2022}\)
\(\Rightarrow S=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{2000}\left(5+5^2\right)\)
\(\Rightarrow S=20+5^2.20+...+5^{2000}.20\)
\(\Rightarrow S=20\left(1+5^2+...+5^{2000}\right)⋮20\)
\(\Rightarrow dpcm\)
PT
7 tháng 5 2017
Giải:
a) \(\left(-\dfrac{5}{28}+1,75+\dfrac{8}{35}\right):\left(\dfrac{-39}{20}\right)\)
\(=\left(-\dfrac{5}{28}+\dfrac{7}{4}+\dfrac{8}{35}\right):\left(\dfrac{-39}{20}\right)\)
\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-\dfrac{39}{20}\right)\)
\(=\dfrac{9}{5}:\left(-\dfrac{39}{20}\right)\)
\(=\dfrac{9.\left(-20\right)}{5.39}\)
\(=\dfrac{3.\left(-4\right)}{1.13}\)
\(=\dfrac{-12}{13}\)
b) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{22}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{22}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
c) \(70,5-528:\dfrac{15}{2}\)
\(=70,5-528.\dfrac{2}{15}\)
\(=70,5-\dfrac{1056}{15}\)
\(=70,5-70,4\)
\(=0,1\)
2 tháng 5 2017
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{5}{12}\)
\(x\cdot\left(-\dfrac{5}{6}\right)=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\left(-\dfrac{5}{6}\right)\)
\(x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\).
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3\cdot7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{53}{10}-\dfrac{2}{5}\)
\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{57}{10}\)
\(3x-3\cdot7=-\dfrac{57}{10}:\dfrac{3}{5}\)
\(3x-3\cdot7=-\dfrac{19}{2}\)
\(3x-21=-\dfrac{19}{2}\)
\(3x=-\dfrac{19}{2}+21\)
\(3x=\dfrac{23}{2}\)
\(x=\)\(\dfrac{23}{2}:3\)
\(x=\dfrac{23}{6}\)
Vậy \(x=\dfrac{23}{6}\).
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)+\dfrac{5}{3}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=\dfrac{23}{27}-\dfrac{5}{3}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=-\dfrac{22}{27}\)
\(2+\dfrac{3}{4x}=\dfrac{7}{9}:-\dfrac{22}{27}\)
\(2+\dfrac{3}{4x}=-\dfrac{21}{22}\)
\(\dfrac{3}{4x}=-\dfrac{21}{22}-2\)
\(\dfrac{3}{4x}=-\dfrac{65}{22}\)
\(4x=\dfrac{3\cdot22}{-65}\)
\(4x=-\dfrac{66}{65}\)
\(x=-\dfrac{66}{65}:4\)
\(x=-\dfrac{33}{130}\)
Vậy \(x=-\dfrac{33}{130}\).
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}:-\dfrac{2}{3}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\).
e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}\)
\(\left|x\right|=\dfrac{29}{12}\)
\(x=\dfrac{29}{12}\) hoặc \(=-\dfrac{29}{12}\)
Vậy \(x\in\left\{\dfrac{29}{12};-\dfrac{29}{12}\right\}\).
H
0
\(3^2+2^4.3^3.19+3^5=9+16.27.19+243=9+8208+243=8460\)
32 + 24 . 33 .19 + 35
= 9 + 16 . 27 . 19 + 243
= 9 + 8208 + 243
= 8460
#Học tốt!!!
~NTTH~