Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
a)Xét 1/2-1/3-1/6=3/6-2/6-1/6=0
=> (1+2+3+...+2018).(3/1.2+3/2.3+...+3/2018.2019).(1/2-1/3-1/6)=(1+2+3+...+2018).(3/1.2+3/2.3+...+3/2018.2019).0=0
b) 4A=1.2.3.4+2.3.4.4+..+x(x+1)(x+2)4
=1.2.3.4+2.3.4.5-1.2.3.4+...+x(x+1)(x+2)(x+3)-x(x+1)(x+2)(x-1)
= (x-1)x(x+1)(x+2)
=> A=x(x+1)(x+2)(x-1)/4
\(A=\left(1-\frac{1}{2018}\right)\left(1-\frac{2}{2018}\right)\left(1-\frac{3}{2018}\right)...\left(1-\frac{2020}{2018}\right).\)
\(=\frac{2017}{2018}\cdot\frac{2016}{2018}\cdot\frac{2015}{2018}\cdot...\cdot\left(1-\frac{2018}{2018}\right)\cdot...\cdot\frac{-2}{2018}\)
\(=\frac{2017}{2018}\cdot\frac{2016}{2018}\cdot\frac{2015}{2018}\cdot...\cdot0\cdot...\cdot\frac{-2}{2018}\)
\(=0\)
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 !!
1) Đặt dãy trên là \(A\)
Theo bài ra ta có :
\(A=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)
\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)
2) \(A=\frac{5^{2018}-2017+1}{5^{2018}-2017}=\frac{5^{2018}-2017}{5^{2018}-2017}+\frac{1}{5^{2018}-2017}=1+\frac{1}{5^{2018}-2017}\)( 1 )
\(B=\frac{5^{2018}-2019+1}{5^{2018}-2019}=\frac{5^{2018}-2019}{5^{2018}-2019}+\frac{1}{5^{2018}-2019}=1+\frac{1}{5^{2018}-2019}\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\)\(A=1+\frac{1}{5^{2018}-2017}< 1+\frac{1}{5^{2018}-2019}=B\)
\(\Rightarrow A< B\)
Vậy \(A< B.\)
1) Ta có B =
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) < \(\frac{1}{1.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)= \(\frac{99}{100}\)
=> B < 1 ( chứ không phải \(\frac{1}{2}\) bạn nhé)
Sai thì thôi chứ mk chỉ làm rờ thôi
a)
⇒ \(\frac{11x-1}{4}=\frac{10}{4}\)
⇒ 11x - 1 = 10
11x = 10 + 1 = 11
x = 11 : 11 = 1
b)
\(\left[{}\begin{matrix}3x-6=0\\\frac{x}{9}-\frac{1}{3}=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}3x=0+6\\\frac{x}{9}=0+\frac{1}{3}\end{matrix}\right.\)⇒ \(\left[{}\begin{matrix}3x=6\\\frac{x}{9}=\frac{1}{3}\end{matrix}\right.\)⇒ \(\left[{}\begin{matrix}x=6:3\\\frac{x}{9}=\frac{3}{9}\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy x = 2 hoặc x = 3
c)
\(M=c\left(\frac{5}{7}+\frac{7}{14}-\frac{17}{14}\right)\)
\(M=c\left(\frac{10}{14}+\frac{7}{14}-\frac{17}{14}\right)\)
\(M=\left(\frac{2018}{2019}-\frac{2019}{2020}\right).0\)
M = 0
d)
\(N=\frac{-7}{13}+2-\frac{19}{13}+\frac{2020}{2018}.\frac{2018}{202}\)
\(N=\left(\frac{-7}{13}-\frac{19}{13}\right)+2+10\)
N = \(-2+2+10\)
N = 10