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TH
1
DD
0
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,\(\dfrac{3}{2}.\dfrac{4}{5}-x=\dfrac{2}{3}\)
\(\dfrac{4}{5}-x=\dfrac{2}{3}:\dfrac{3}{2}\)
\(\dfrac{4}{5}-x=\dfrac{4}{9}\)
\(x=\dfrac{4}{5}-\dfrac{4}{9}\)
\(x=\dfrac{16}{45}\)
ND
Nguyễn Đức Trí
VIP
5 tháng 8 2023
\(S=\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...\dfrac{1}{nx\left(n+1\right)}\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(S=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}\)
\(T=\dfrac{3}{1x2}+\dfrac{3}{2x3}+\dfrac{3}{3x4}+\dfrac{3}{4x5}+...\dfrac{3}{nx\left(n+1\right)}\)
\(T=3x\left[\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...\dfrac{1}{nx\left(n+1\right)}\right]\)
\(T=3x\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...\dfrac{1}{n}-\dfrac{1}{n+1}\right]\)
\(T=3x\left(1-\dfrac{1}{n+1}\right)=\dfrac{3xn}{n+1}\)
8 tháng 8 2017
a) = \(\frac{127}{96}\)
b) = \(\frac{255}{256}\)
c) Mik bỏ nha
d) = \(\frac{1023}{512}\)
e) = \(\frac{2343}{625}\)
TA
1
T
8 tháng 8 2023
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
NK
1
4 tháng 6 2019
\(\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)
\(=3\left(\frac{1}{\frac{1\cdot2}{2}}+\frac{1}{\frac{2\cdot3}{2}}+\frac{1}{\frac{3\cdot4}{2}}+...+\frac{1}{\frac{100\cdot101}{2}}\right)\)
\(=3\left(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+...+\frac{2}{100\cdot101}\right)\)
\(=6\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{100\cdot101}\right)\)
\(=6\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=6\left(1-\frac{1}{101}\right)=6-\frac{6}{101}=\frac{606-6}{101}=\frac{600}{101}\)
S
2
NL
3
27 tháng 8 2016
3/1 + 3/1+2 + 3/1+2+3 + 3/1+2+3+4 + ... + 3/1+2+3+4+...+100
= 3 × (1/0+1 + 1/1+2 + 1/1+2+3 + 1/1+2+3+4 + ... + 1/1+2+3+4+...+100)
= 3 × (1/(1+0)×2:2 + 1/(1+2)×2:2 + 1/(1+3)×3:2 + 1/(1+4)×4:2 + ... + 1/(1+100)×100:2)
= 3 × (2/1×2 + 2/2×3 + 2/3×4 + 2/4×5 + ... + 2/100×101)
= 3 × 2 × (1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/100×101)
= 6 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/100 - 1/101)
= 6 × (1 - 1/100)
= 6 × 100/101
= 600/101