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3 tháng 8 2023
1+2+3+...+2023=\(\dfrac{\left(2023-1\right)+1\cdot\left(1+2023\right)}{2}\)=2047276.
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
Khoảng cách 2 số hạng kề nhau:
3-2=1
Số lượng số hạng của dãy:
(2023-1):1+1=2023
Tổng dãy số trên:
(2023 +1): 2 x 2023= 2047276
Đ.số: 2047276
DC
7
9 tháng 8 2023
`@` `\text {Ans}`
`\downarrow`
`a,`
\(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)
\(\dfrac{6}{5}-x=\dfrac{2}{3}\)
\(x=\dfrac{6}{5}-\dfrac{2}{3}\)
\(x=\dfrac{8}{15}\)
Vậy, `x = \dfrac{8}{15}`
`b,`
\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)
\(x\times3\dfrac{1}{3}=\dfrac{40}{51}\)
\(x=\dfrac{40}{51}\div3\dfrac{1}{3}\)
\(x=\dfrac{4}{17}\)
Vậy, `x=`\(\dfrac{4}{17}\)
`c,`
\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)
\(x=\dfrac{34}{7}\)
Vậy, `x=`\(\dfrac{34}{7}\)
NX
9 tháng 8 2023
a,\(\dfrac{3}{2}.\dfrac{4}{5}-x=\dfrac{2}{3}\)
\(\dfrac{4}{5}-x=\dfrac{2}{3}:\dfrac{3}{2}\)
\(\dfrac{4}{5}-x=\dfrac{4}{9}\)
\(x=\dfrac{4}{5}-\dfrac{4}{9}\)
\(x=\dfrac{16}{45}\)
KV
4 tháng 11 2023
Bài 1
a) 3 2/5 - 1/2
= 17/5 - 1/2
= 34/10 - 5/10
= 29/10
b) 4/5 + 1/5 × 3/4
= 4/5 + 3/20
= 16/20 + 3/20
= 19/20
c) 3 1/2 × 1 1/7
= 7/2 × 8/7
= 4
d) 4 1/6 : 2 1/3
= 25/6 : 7/3
= 25/14
KV
4 tháng 11 2023
Bài 2
a) 3 × 1/2 + 1/4 × 1/3
= 3/2 + 1/12
= 18/12 + 1/12
= 19/12
b) 1 4/5 - 2/3 : 2 1/3
= 9/5 - 2/3 : 7/3
= 9/5 - 2/7
= 63/35 - 10/35
= 53/35
DC
3
9 tháng 8 2023
a) \(x+2\dfrac{3}{4}=5\dfrac{2}{3}\)
\(x+\dfrac{11}{4}=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}-\dfrac{11}{4}\)
\(x=\dfrac{35}{12}\)
b) \(x-1\dfrac{4}{5}=3\dfrac{2}{7}\)
\(x-\dfrac{9}{5}=\dfrac{23}{7}\)
\(x=\dfrac{23}{7}+\dfrac{9}{5}\)
\(x=\dfrac{178}{35}\)
c) \(x\times3\dfrac{1}{2}=4\dfrac{3}{4}\)
\(x\times\dfrac{7}{2}=\dfrac{19}{4}\)
\(x=\dfrac{19}{4}\div\dfrac{7}{2}\)
\(x=\dfrac{19}{14}\)
d) \(x\div2\dfrac{2}{3}=4\dfrac{1}{3}\)
\(x\div\dfrac{8}{3}=\dfrac{13}{3}\)
\(x=\dfrac{13}{3}\times\dfrac{8}{3}\)
\(x=\dfrac{104}{9}\)
9 tháng 8 2023
a) \(...=x+\dfrac{11}{3}=\dfrac{17}{3}\Rightarrow x=\dfrac{17}{3}-\dfrac{11}{3}=\dfrac{6}{3}=2\)
b) \(...\Rightarrow x-\dfrac{9}{5}=\dfrac{23}{7}\Rightarrow x=\dfrac{23}{7}+\dfrac{9}{5}=\dfrac{115}{35}+\dfrac{36}{35}=\dfrac{151}{35}\)
c) \(...\Rightarrow x.\dfrac{7}{2}=\dfrac{19}{4}\Rightarrow x=\dfrac{19}{4}:\dfrac{7}{2}\Rightarrow x=\dfrac{19}{4}.\dfrac{2}{7}=\dfrac{19}{14}\)
d) \(...\Rightarrow x:\dfrac{8}{3}=\dfrac{13}{3}\Rightarrow x=\dfrac{13}{3}.\dfrac{8}{3}=\dfrac{124}{9}\)
DD
0
MH
14 tháng 9 2021
a)\(\dfrac{2}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}+\dfrac{1}{3}\right).\dfrac{4}{5}=1.\dfrac{4}{5}=\dfrac{4}{5}\)
b)\(\dfrac{2}{3}.\dfrac{4}{5}-\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}-\dfrac{1}{3}\right).\dfrac{4}{5}=\dfrac{1}{3}.\dfrac{4}{5}=\dfrac{4}{15}\)
14 tháng 9 2021
a) \(\dfrac{2}{3}\times\dfrac{4}{5}+\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=\dfrac{4}{5}\times1=\dfrac{4}{5}\)
b) \(\dfrac{2}{3}\times\dfrac{4}{5}-\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\left(\dfrac{2}{3}-\dfrac{1}{3}\right)=\dfrac{4}{5}\times\dfrac{1}{3}=\dfrac{4}{15}\)
c) \(\dfrac{1}{2}:\dfrac{3}{4}+\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}+\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\times\left(\dfrac{1}{2}+\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{2}{3}=\dfrac{8}{9}\)
d) \(\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}-\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{1}{3}=\dfrac{4}{9}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 11 2023
Bài 1:
a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\)
= \(\dfrac{17}{5}\) - \(\dfrac{1}{2}\)
= \(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)
= \(\dfrac{29}{10}\)
b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)
= \(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)
= \(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)
= \(\dfrac{19}{20}\)
c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)
= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)
= \(\dfrac{29}{6}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 11 2023
Bài 2:
3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\)
= \(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)
= \(\dfrac{28}{5}\)
b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)
= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)
= \(\dfrac{43}{34}\)
TA
9 tháng 6 2023
a, \(2\dfrac{1}{3}+4\dfrac{1}{5}+4\dfrac{1}{3}\)
\(=\dfrac{7}{3}+\dfrac{21}{5}+\dfrac{13}{3}\)
\(=\dfrac{1}{3}\left(7+13\right)+\dfrac{21}{5}\)
\(=\dfrac{20}{3}+\dfrac{21}{5}=\dfrac{100+63}{15}=\dfrac{163}{15}\)
b, \(5\dfrac{3}{4}-4\dfrac{1}{2}.3\dfrac{7}{8}\)
\(=\dfrac{23}{4}-\dfrac{9}{2}.\dfrac{31}{8}\)
\(=\dfrac{23}{4}-\dfrac{279}{16}=\dfrac{92-279}{16}=-\dfrac{187}{16}\)
c, \(1\dfrac{1}{2}.3\dfrac{2}{3}.4\dfrac{3}{4}\)
\(=\dfrac{3}{2}.\dfrac{11}{3}.\dfrac{19}{4}=\dfrac{209}{8}\)
d, \(6\dfrac{4}{5}:2\dfrac{3}{4}:1\dfrac{1}{2}\)
\(=\dfrac{34}{5}:\dfrac{11}{4}:\dfrac{3}{2}\)
\(=\left(\dfrac{34}{5}.\dfrac{4}{11}\right).\dfrac{2}{3}=\dfrac{136}{55}.\dfrac{2}{3}=\dfrac{272}{165}\)
AH
Akai Haruma
Giáo viên
30 tháng 9 2023
Câu 1:
a. $3\frac{2}{3}+2\frac{1}{2}=(3+2)+(\frac{2}{3}+\frac{1}{2})=5+\frac{7}{6}=6+\frac{1}{6}=6\frac{1}{6}$
b. \(2\frac{1}{2}\times 3\frac{2}{5}=\frac{5}{2}\times \frac{17}{5}=\frac{17}{2}\)
c.
\(3\frac{1}{3}: 4\frac{1}{4}=\frac{10}{3}: \frac{17}{4}=\frac{40}{51}\)
d.
\(3\frac{1}{2}+4\frac{5}{7}-5\frac{5}{14}=(3+4-5)+(\frac{1}{2}+\frac{5}{7}-\frac{5}{14})=2+\frac{6}{7}=2\frac{6}{7}\)
AH
Akai Haruma
Giáo viên
30 tháng 9 2023
Câu 2:
a. $x\times \frac{2}{7}=\frac{6}{11}$
$x=\frac{6}{11}: \frac{2}{7}=\frac{21}{11}$
b. $x: \frac{3}{2}=\frac{1}{4}$
$x=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$
A= 1 - 2 - 22 - 23 + 24 +...+ 22022 (sửa đề)
= -13 + (24 + 25 + 26 + ... + 22022)
2A = -26 + (25 + 26 + 27 + ... + 22023)
2A - A = -26 + (25 + 26 + 27 + ... + 22023) - [-13 + (24 + 25 + 26 + ... + 22022)]
A = -13 +(22023 - 24)
= 22023 - 29
Vậy...
B = 1 + 3 + 32 + 33 + 34 + ... + 32022
3B = 3 + 32 + 33 + 34 + 35 +...+ 32023
3B - B = 3 + 32 + 33 + 34 + 35 +...+ 32023 - (1 + 3 + 32 + 33 + 34 + ... + 32022)
2B = 32023 - 1
=> B = \(\dfrac{3^{2023}-1}{2}\)
Vậy...
#Ayumu