giúp mik vs các cậu ơi mik sắp phải nộp r
help me
Tính B=1.2.3+2.3.4+...+(n-1)n(n+1)
cảm ơn mn
K
Khách
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H
26 tháng 11 2021
9. Has the work been done by him?
10. The boxes were opened and cigarettes were taken out by us
11. She was given a new one
12. He is proved wrong
13. We were promised higher wages
14. This is the third time we have been written to about this by them
15. We were asked to be there at 8 o'clock
16. She is being shown how to do it
NT
6 tháng 9 2021
c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\)
Ở nơi x=9/4-1/2 là x-9/4-1/2 nha
KK
6 tháng 9 2021
a. -1,5 + 2x = 2,5
<=> 2x = 2,5 + 1,5
<=> 2x = 4
<=> x = 2
b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)
<=> 9x + 45 - 3 = 8
<=> 9x = 8 + 3 - 45
<=> 9x = -34
<=> x = \(\dfrac{-34}{9}\)
VM
6 tháng 1 2020
Đặt \(A=1^3+2^3+3^3+...+99^3+100^3\)
\(\Rightarrow A=\left(1-1\right).1.\left(1+1\right)+1+\left(2-1\right).2.\left(2+1\right)+2+...+\left(99-1\right).99.\left(99+1\right)+99+\left(100-1\right).100.\left(100+1\right)+100\)
\(\Rightarrow A=1+2+1.2.3+3+2.3.4+...+100+99.100.101\)
\(\Rightarrow A=\left(1+2+3+...+100\right)+\left(1.2.3+2.3.4+...+99.100.101\right)\)
\(\Rightarrow A=5050+101989800\)
\(\Rightarrow A=101994850.\)
Vậy \(A=101994850.\)
Chúc bạn học tốt!
LH
6 tháng 1 2020
Mik đội ơn pạn nhìu lém!!! Chả ai thèm dzúp mik j cả, chỉ mỗi pạn thui!!! hic hic *xúc cmn động*
3 tháng 4 2019
a, \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot3}+...+\frac{1}{x\cdot\left(x+1\right)}-\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)
\(=1-\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=1-\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2019}{2019}-\frac{2018}{2019}=\frac{1}{2019}\)
Đến đây bn tự tính nhé !!
7 tháng 5 2020
b)
Phân số là số nguyên <=> \(-3⋮2n-1\)
<=> \(2n-1\inƯ\left(-3\right)\)
Ta có bảng:
| 2n-1 | 1 | -1 | 3 | -3 |
| n | 1 | 0 | 2 | -1 |
| Thử lại | Chọn | Chọn | Chọn | Chọn |
KL: \(n\in\left\{1;0;2;-1\right\}\)
c) \(\frac{n+4}{n+1}=\frac{\left(n+1\right)+3}{n+1}\) = \(1+\frac{3}{n+1}\)
Để phân số là số nguyên <=> \(1+\frac{3}{n+1}\) là số nguyên
<=> \(\frac{3}{n+1}\) là số nguyên
Rồi bạn lm tương tự như ở câu b)
T
1
19 tháng 2 2023
a: =>12x+5(x-1)=100-10(3x-1)
=>12x+5x-5=100-30x+30
=>17x-5=-30x+130
=>47x=135
=>x=135/47
b: \(\Leftrightarrow6\left(2x-1\right)+2\left(5-x\right)=24-9\left(x+1\right)\)
=>12x-6+10-2x=24-9x-9
=>10x+4=-9x+15
=>19x=11
=>x=11/19
B=1.2.3+2.3.4+.........+(n−1)n(n+1)B=1.2.3+2.3.4+.........+(n−1)n(n+1)
⇔4B=1.2.3.4+2.3.4.4+........+(n−1)n(n+1).4⇔4B=1.2.3.4+2.3.4.4+........+(n−1)n(n+1).4
⇔4B=(4−0).1.2.3+(5−1).2.3.4+.........+[(n+2)−(n−2)](n−1)n(n+1)⇔4B=(4−0).1.2.3+(5−1).2.3.4+.........+[(n+2)−(n−2)](n−1)n(n+1)
⇔4B=1.2.3.4−0.1.2.3+2.3.4.5−1.2.3.4+.......+(n−1)n(n+1)(n+2)(n+3)−(n−2)(n−1)n(n+1)⇔4B=1.2.3.4−0.1.2.3+2.3.4.5−1.2.3.4+.......+(n−1)n(n+1)(n+2)(n+3)−(n−2)(n−1)n(n+1)
⇔4B=(n−1)n(n+1)(n+2)⇔4B=(n−1)n(n+1)(n+2)
⇔B=(n−1)n(n+1)(n+2)4
Băng Băng 2k6 giúp mik lm ik, mik bận