K
Khách

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

25 tháng 2 2017

a) \(\left|2x-1\right|=5\)

\(\Rightarrow\left[\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy \(x\left[\begin{matrix}=3\\=-2\end{matrix}\right.\)

b) \(\left(5^x-1\right)3-2=70\)

\(\Rightarrow5^x.3-3=72\)

\(\Rightarrow5^x.3=75\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)

Vậy \(x=2.\)

c) \(\left(x-1\frac{1}{2}\right)^2+\frac{3}{4}=\frac{1}{4}\)

\(\Rightarrow\left(x-1\frac{1}{2}\right)^2=\frac{-1}{2}\)

............. Làm tiếp nhé!

d) \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\) \(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\) \(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{23}{45}\) \(\Rightarrow\frac{22}{45}x=\frac{23}{45}\) \(\Rightarrow x=\frac{23}{45}:\frac{22}{45}\) \(\Rightarrow x=\frac{23}{22}\) Vậy \(x=\frac{23}{22}.\)
25 tháng 2 2017

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)

\(\Rightarrow x=\frac{23}{11}\)

14 tháng 5 2017

a, \(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=0+\frac{1}{16}\)

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2=\left(\frac{-1}{4}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\\\frac{1}{x}-\frac{2}{3}=\frac{-1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{11}{12}\\\frac{1}{x}=\frac{5}{12}\end{cases}\Rightarrow\orbr{\begin{cases}11x=12\\5x=12\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}}\)

b, \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

Đặt S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\)

2S = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{8.9.10}\)

2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)

2S = \(\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)

S = \(\frac{22}{45}:2=\frac{11}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{11}{45}\Rightarrow x=\frac{23}{11}\)

14 tháng 5 2017

a/ (1/x -2/3)2=1/16=(1/4)2

Có 2 trường hợp:

+/ 1/x -2/3= - 1/4

<=> 1/x =2/3 -1/4 = 5/12

=> x1=12/5

+/ 1/x - 2/3 =1/4

<=> 1/x = 2/3 +1/4= 11/12

=> x2=12/11

b/ Ta có: 

2/(1.2.3)=1/(1.2) - 1/2.3 ;  2/(2.3.4)=1/2.3 -1/3.4 ; ...; 2/(8.9.10)=1/8.9 -1/9.10

=> (1/1.2.3 + 1/2.3.4 +...+1/8.9.10)=23/45

<=> (1/1.2 -1/2.3 +1/2.3 -1/3.4 +...+1/8.9-1/9.10).x/2=23/45

<=> (1/1.2 -1/9.10).x/2 =23/45

<=> x.11/45=23/45

=> x=23/11

12 tháng 1 2017

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)

\(\Rightarrow x=\frac{23}{11}\)

5 tháng 4 2017

đặt \(A=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\)

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)

\(2A=\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)

\(A=\frac{22}{45}:2=\frac{11}{45}\)

thay A vào ta được

\(\frac{11}{45}.x=\frac{23}{45}\)

        \(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)

13 tháng 8 2019

bạn vào câu hỏi tương tự tham khảo nha

13 tháng 8 2019

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)

\(\Rightarrow x=\frac{23}{11}\)

22 tháng 5 2017

Ta có:

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}\div\frac{11}{45}=\frac{23}{11}\)

Vậy \(x=\frac{23}{11}\)

23 tháng 5 2017

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{8.9}+\frac{1}{9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\left(\frac{1}{2}+\frac{1}{6}+....+\frac{1}{72}+\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\left(\frac{1}{2}-\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\frac{22}{45}.\frac{1}{2}x=\frac{23}{45}\)

\(\frac{11}{45}.x=\frac{23}{45}\)

 \(x=\frac{23}{45}\div\frac{11}{45}\)

\(x=\frac{23}{11}\)

=> \(x=\frac{23}{11}\)

24 tháng 1 2016

x = 2 nha bạn !!!

23 tháng 2 2020

\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)+x=\frac{23}{45}\)

Cho \(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\)

Ta có công thức sau: \(\frac{1}{n\cdot\left(n+1\right)}+\frac{1}{\left(n+1\right)\cdot\left(n+2\right)}=\frac{2}{n\cdot\left(n+1\right)\left(n+1\right)}\)

\(\Rightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{9\cdot10}=\frac{22}{45}\)

\(\Rightarrow A=\frac{22}{45}:2=\frac{11}{45}\)

Thay vào phép tính trên ta được:

\(\frac{11}{45}\cdot x=\frac{23}{45}\\ x=\frac{23}{45}:\frac{11}{45}\\ x=\frac{23}{11}\)

Vậy \(x=\frac{23}{11}\)

7 tháng 6 2020

\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{23}{45}\)

=> \(\left[\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\right)\right]x=\frac{23}{45}\)

=>\(\left[\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)\right]x=\frac{23}{45}\)

=> \(\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9\cdot10}\right)\right]x=\frac{23}{45}\)

=> \(\left[\frac{1}{2}\cdot\frac{22}{45}\right]x=\frac{23}{45}\)

=> \(\frac{11}{45}x=\frac{23}{45}\)

=> \(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{45}\cdot\frac{45}{11}=\frac{23}{11}\)

Vậy x = 23/11

Ez :))

2 tháng 8 2015

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{46}{45}\)

\(=\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x\)

\(=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\right).x\)

\(=\left(\frac{1}{2}-\frac{1}{90}\right).x=\left(\frac{45}{90}-\frac{1}{90}\right)x=\frac{44}{90}.x=\frac{22x}{45}=\frac{46}{45}\)

=> 22x=46

=> x=\(46:22=\frac{23}{11}\)