1.2+2.3+3.4+...+x(x+1)=29700
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\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)=2023x\)
\(\Rightarrow2022x+\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)=2023x\)\(\Rightarrow2022x-2023x=-\left(1-\dfrac{1}{2023}\right)\)
\(\Rightarrow-x=-\dfrac{2022}{2023}\Leftrightarrow x=\dfrac{2022}{2023}\)
(x + 1/1.2) + (x + 1/2.3) + (x + 1/3.4) + ... + (x + 1/2022.2023) = 2023x
x + x + x + ... + x + 1/1.2 + 1/2.3 + ... + 1/2022.2023 = 2023x
2022x + 1 - 1/2 + 1/2 - 1/3 + ... + 1/2022 - 2023 = 2023x
2023x - 2022x = 1 - 1/2023
x = 2022/2023
`x/(1.2)+x/(2.3)+x/(3.4)+.....+x/(2017.2018)=1`
`-> x/1 - x/2 +x/2-x/3+x/3-x/4+........+x/2017-x/2018=1`
`-> x-x/2018=1`
`-> 2017/2018 .x=1`
`-> x=2018/2017`
Do mỗi số hạng ở vế trái nằm trong dấu giá trị tuyệt đối mà vế phải 100 là số dương nên x cũng phải dương.
Do x dương và trong mỗi dấu giá trị tuyệt đối đều dương nên ta lập được kết quả sau:
x+1/1.2+x+1/2.3+1/3.4+....+x+1/99.100=100x
Dãy trên có 99 số x nên:
99x+(1-1/2+1/2-1/3+1/3-1/4+....+1/99-1/100)=100x
1-1/100=x
x=99/100
Vậy x=99/100
Chúc em học tốt^^
Ta có: 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2/3
=> 1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2/3
=>1-1/x+1=2/3
=>1/x+1=1/3
=>3=x+1
=>x=2
Ta có\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)
=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{3}\)
=>\(1-\frac{1}{x+1}=\frac{2}{3}\)
=>\(\frac{1}{x+1}=1-\frac{2}{3}\)
=>\(\frac{1}{x+1}=\frac{1}{3}\)
=>\(x+1=3\)
=>\(x=2\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=201\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=201\)
\(1-\frac{1}{x+1}=201\)
\(\frac{1}{x+1}=1-201\)
\(\frac{1}{x+1}=-200\)
\(\Rightarrow x+1=-\frac{1}{200}\)
\(x=-\frac{1}{200}-1\)
\(x=-\frac{201}{200}\)
Vậy \(x=-\frac{201}{200}\)
Gọi A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{19}{20}\)
\(\Rightarrow\) A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)
\(\Rightarrow\) A = 1 - \(\dfrac{1}{x+1}\)
\(\Rightarrow\) 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{19}{20}\)
\(\Rightarrow1-\dfrac{19}{20}=\dfrac{1}{x+1}\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{20}\)
\(\Rightarrow\) x + 1 = 20\(\Rightarrow\) x=19
1/1.2 +1/2.3 +...+ 1/x(x+1) = 2015/2016
<=> 1-1/2 + 1/2 - 1/3 + ... + 1/x - 1/x+1 = 2015/2016
<=> 1 - 1/x+1 = 2015/2016
<=> 1/x+1 = 1/2016
<=> x + 1 = 2016
<=> x = 2015
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)
\(\Leftrightarrow x+1=2016\Rightarrow x=2015\)
1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +......+ 1/x - 1/x+1 = 99/100
1- 1/x+1= 99/100
1/x+1= 1- 99/100
1/x+1=1/100
=> x+1 = 100
x= 100-1
x=99
Đặt A = 1.2+2.3+3.4+...+x(x+1)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + x.(x + 1).(x + 2)
=> 3A = x.(x + 1).(x + 2)
=> A= x.(x + 1).(x + 2) : 3
Ta có: x.(x + 1).(x + 2) : 3 = 29700
=> x.(x + 1).(x + 2) = 89100
=> x.(x + 1).(x + 2) =
Sai đề
Đề sai nhé : mình sủa lại đề nha
1.2 + 2.3 + .... + x(x + 1) = 29760
\(\Leftrightarrow\frac{x\left(x+1\right)\left(x+2\right)}{3}=29760\)
\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=29760\)
\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=30.31.32\)
\(\Rightarrow x=30\)