K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

6 tháng 12 2018

\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{x}.\left(1+2+3+...+x\right)=115\)

\(\Rightarrow1.\left(\frac{1.2}{2}\right)+\frac{1}{2}.\left(\frac{2.3}{2}\right)+\frac{1}{3}.\left(\frac{3.4}{2}\right)+....+\frac{1}{x}.\left[\frac{x\left(x+1\right)}{2}\right]=115\)

\(\Rightarrow\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{x+1}{2}=115\Rightarrow2+3+...+\left(x+1\right)=230\)

\(\frac{\Rightarrow\left[\frac{\left(x+1-2\right)}{1}+1\right].\left(x+1+2\right)}{2}=\frac{x.\left(x+3\right)}{2}=230\Rightarrow x.\left(x+3\right)=460\)

vì x và x+3 là 2 số tự nhiên cách nhau 3 đơn vị => \(x.\left(x+3\right)=460=20.23\Rightarrow x=20\)

Vậy x=20

13 tháng 1 2019

Mik ko hieu giai ki hon dc ko

18 tháng 6 2016

a) \(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\Leftrightarrow\orbr{\begin{cases}x\ge\frac{3}{2}\mid:2x-3=\frac{1}{4}\Rightarrow2x=\frac{13}{4}\Rightarrow x=\frac{13}{8}\left(TM\right)\\x< \frac{3}{2}\mid:3-2x=\frac{1}{4}\Rightarrow2x=\frac{11}{4}\Rightarrow x=\frac{11}{8}\left(TM\right)\end{cases}.}\)

b) \(\Leftrightarrow\left|x-1\right|=\frac{3}{4}\Leftrightarrow\orbr{\begin{cases}x\ge1\mid:x-1=\frac{3}{4}\Rightarrow x=\frac{7}{4}\left(TM\right)\\x< 1\mid:1-x=\frac{3}{4}=>x=\frac{1}{4}\left(TM\right)\end{cases}}\)

c) \(\frac{3}{5\left(x-\frac{5}{6}\right)}-\frac{1}{2\left(\frac{3}{2}-1\right)}=-\frac{1}{4}\Leftrightarrow\frac{3}{\frac{5\left(6x-5\right)}{6}}-\frac{1}{2\cdot\frac{1}{2}}=-\frac{1}{4}\Leftrightarrow\frac{18}{5\left(6x-5\right)}=-\frac{1}{4}+1\)

\(\Leftrightarrow\frac{18}{5\left(6x-5\right)}=\frac{3}{4}\Leftrightarrow6x-5=\frac{24}{5}\Leftrightarrow6x=\frac{49}{5}\Leftrightarrow x=\frac{49}{30}\)

d) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(\Leftrightarrow\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(\Leftrightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\Leftrightarrow2\cdot\frac{x+1-2}{2\left(x+1\right)}=\frac{2015}{2016}\Leftrightarrow\frac{x-1}{x+1}=\frac{2015}{2016}\)

\(\Leftrightarrow2016x-2016=2015x+2015\Leftrightarrow x=2015+2016=4031\)

Vậy x = 4031.

6 tháng 8 2017

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x.\left(x+1\right)}=\frac{2005}{2010}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{401}{402}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{401}{402}\)

\(\Leftrightarrow\frac{1}{x+1}=1-\frac{401}{402}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{402}\)

\(\Leftrightarrow x+1=402\Rightarrow x=401\)

3 tháng 8 2017

câu trên mk làm rồi

\(\dfrac{2x-1}{x-3}=\dfrac{2x+3}{x-1}\)

\(\Rightarrow\left(2x-1\right)\left(x-1\right)=\left(x-3\right)\left(2x+3\right)\)

\(\Rightarrow2x^2-x-2x+1=2x^2-6x+3x-9\)

\(\Rightarrow-x-2x+6x-3x=-1-9\)

\(\Rightarrow0=-10\) (vô lí)

Vậy ko tồn tại giá trị của x.