- 1 + 3 + 5 + ... + 23
- 2 + 4 + 6 + ... +24
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
$(5^{1986}-5^{1985}):5^{1985}=5^{1985}(5-1):5^{1985}=5-1=4$
2.
\((7^{846}+7^{847}):7^{846}=7^{846}(1+7):7^{846}=1+7=8\)
3.
\((9^{2018}-3^{4036}):6^{2006}=[(3^2)^{2018}-3^{4036}]:6^{2006}\)
$=(3^{4036}-3^{4036}):6^{2006}=0:6^{2006}=0$
4.
$(7^{80}.8^{70}-56^{70}):56^{70}$
$=[7^{10}(7.8)^{70}-56^{70}]:56^{70}$
$=[7^{10}.56^{70}-56^{70}]:56^{70}$
$=56^{70}(7^{10}-1):56^{70}=7^{10}-1$
5.
$4^{4016}:(4^{4017}-4^{4016})=4^{4016}:[4^{4016}(4-1)]$
$=4^{4016}:4^{4016}:3=1:3=\frac{1}{3}$
6.
$(12^{206}.2^{207}-24^{206}):24^{206}$
$=(12^{206}.2^{206}.2-24^{206}):24^{206}$
$=[(12.2)^{206}.2-24^{206}]:24^{206}$
$=(24^{206}.2-24^{206}):24^{206}$
$=24^{206}(2-1):24^{206}=2-1=1$
7.
$(5^2-24)^{8946}+4^{30}:2^{60}=1^{8946}+(2^2)^{30}:2^{60}$
$=1+2^{60}:2^{60}=1+1=2$
8.
$(37.8^{1007}-7.2^{3021}):8^{1007}=[37.8^{1007}-7.(2^3)^{1007}]:8^{1007}$
$=[37.8^{1007}-7.8^{1007}]:8^{1007}$
$=8^{1007}(37-7):8^{1007}=37-7=30$
\(\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{4}{12}+\dfrac{9}{12}+\dfrac{6}{12}=\dfrac{19}{12}\)
\(\dfrac{6}{8}+\dfrac{2}{4}+\dfrac{6}{24}=\dfrac{3}{4}+\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{6}{4}=\dfrac{3}{2}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{5}{6}=\dfrac{10}{6}=\dfrac{5}{3}\)
\(\dfrac{3}{5}+\dfrac{3}{2}+2=\dfrac{6}{10}+\dfrac{15}{10}+\dfrac{20}{10}=\dfrac{41}{10}\)
\(\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{13}{12}+\dfrac{1}{2}=\dfrac{19}{12}\\ \dfrac{6}{8}+\dfrac{2}{4}+\dfrac{6}{24}=\dfrac{5}{4}+\dfrac{6}{24}=\dfrac{3}{2}\\ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{5}{6}+\dfrac{5}{6}=\dfrac{10}{6}\\ \dfrac{3}{5}+\dfrac{3}{2}+2=\dfrac{21}{10}+\dfrac{20}{10}=\dfrac{41}{10}\)
Câu 5:
Gọi số học sinh lớp 6A là x(bạn)
(Điều kiện: \(x\in Z^+\))
Vì số học sinh lớp 6A khi xếp hàng 3;5;9 đều vừa đủ nên \(x\in BC\left(3;5;9\right)\)
=>\(x\in B\left(45\right)\)
=>\(x\in\left\{45;90;...\right\}\)
mà 40<=x<=50
nên x=45(nhận)
Vậy: Lớp 6A có 45 bạn
Câu 4:
a: n đọc là hai nghìn không trăm hai mươi mốt
=>n=2021
M={2;0;1}
b: Đặt *=x
Để \(\overline{2x5}⋮9\) thì \(2+x+5⋮9\)
=>\(x+7⋮9\)
=>x=2
c: \(a\cdot b=\left(-10\right)\cdot5=-50< -49=c\)
Câu 2:
\(\dfrac{232}{24}=9\left(dư16\right)\)
=>Cần thuê ít nhất là 9+1=10 xe để đủ xếp chỗ cho 232 bạn
\(\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{2}{4}\)
= \(\dfrac{24}{60}\) + \(\dfrac{40}{60}\) + \(\dfrac{30}{60}\)
= \(\dfrac{64}{60}\) + \(\dfrac{30}{60}\)
= \(\dfrac{47}{30}\)
\(\dfrac{2}{6}+\dfrac{3}{12}\)
= \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\)
= \(\dfrac{7}{12}\)
\(\dfrac{5}{6}\) + \(\dfrac{1}{3}\)
= \(\dfrac{5}{6}\) + \(\dfrac{2}{6}\)
= \(\dfrac{7}{6}\)
\(\dfrac{1}{3}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{6}\)
= \(\dfrac{4}{12}\) + \(\dfrac{5}{12}\) + \(\dfrac{10}{12}\)
= \(\dfrac{9}{12}\) + \(\dfrac{10}{12}\)
= \(\dfrac{19}{12}\)
\(\dfrac{5}{8}\) + \(\dfrac{4}{7}\)
= \(\dfrac{35}{56}\) + \(\dfrac{32}{56}\)
= \(\dfrac{67}{56}\)
\(\dfrac{7}{3}\) + \(\dfrac{8}{7}\)
= \(\dfrac{49}{21}\) + \(\dfrac{24}{21}\)
= \(\dfrac{73}{21}\)
\(\dfrac{1}{5}+\dfrac{5}{35}\)
= \(\dfrac{7}{35}\) + \(\dfrac{5}{35}\)
= \(\dfrac{12}{35}\)
bài 1
a, 432 x 25 + 46 x 432 - 432
= 432 x ( 25 + 46 - 1)
=432 x 70
=30 240
b, 432 : 6 - 232 : 6 + 102 : 6
=(432 - 232 +102) : 6
=302:6
=50,333333333333..................
bài 2
a, 285 * x + 115 * x = 400
(285 + 115) * x = 400
400 * x = 400
x = 400 : 400
x = 1
b, x - 6 : 2 - (48 - 24 x 2 : 6 - 3) = 0
x - 3 - (48 - 48 : 6 - 3) = 0
x - 3 - ( 48 - 8 - 3) = 0
x - 3 - 37 = 0
x - 3 = 0 + 37
x - 3 =37
x = 37 + 3
x = 40
a) 432x25+46x432-432
=10800+19872-432
=30240
ban oi cau b sai đề hay gì đó
a: \(\dfrac{1}{8}\cdot\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{1}{10}\cdot\dfrac{5}{3}=\dfrac{1}{2\cdot3}=\dfrac{1}{6}\)
b: \(=\dfrac{8-3}{12}\cdot\dfrac{6}{5}=\dfrac{6}{12}=\dfrac{1}{2}\)
c: \(=\dfrac{24}{35}:\dfrac{32}{35}=\dfrac{3}{4}\)
d: =63/21+15/21-7/21=71/21
a \(\dfrac{1}{8}\times\dfrac{4}{5}\times\dfrac{10}{6}=\dfrac{1}{6}\)
b \(\left(\dfrac{8}{12}-\dfrac{3}{12}\right)\times\dfrac{6}{5}=\dfrac{5}{12}\times\dfrac{6}{5}=\dfrac{1}{2}\)
c \(\dfrac{24}{35}:\dfrac{32}{35}=\dfrac{24}{35}\times\dfrac{35}{32}=\dfrac{3}{4}\)
d \(\dfrac{63}{21}+\dfrac{15}{21}-\dfrac{7}{21}=\dfrac{71}{21}\)
a) \(\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{24}{60}+\dfrac{40}{60}+\dfrac{45}{60}=\dfrac{24+40+45}{60}=\dfrac{109}{60}\)
b) \(\dfrac{1}{3}+\dfrac{5}{12}+\dfrac{5}{6}=\dfrac{4}{12}+\dfrac{5}{12}+\dfrac{10}{12}=\dfrac{4+5+10}{12}=\dfrac{19}{12}\)
c) \(\dfrac{1}{6}+\dfrac{5}{24}+\dfrac{2}{3}=\dfrac{4}{24}+\dfrac{5}{24}+\dfrac{16}{24}=\dfrac{4+5+16}{24}=\dfrac{25}{24}\)
Ta thấy: các dãy số đều có khoảng cách là 2
Tính số số hạng có trong các dãy:
( 23 - 1) : 2 + 1 = 12(số)
(24 - 2) : 2 + 1 = 12(số)
Tính tổng các dãy số:
(23 + 1) x 12 : 2 = 144
(24 + 2) x 12 : 2 = 156
*144
*156