NH
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
14 tháng 5 2022
X x \(\dfrac{3}{4}\)+ X x\(\dfrac{1}{5}\)+ X x \(\dfrac{1}{20}\)+ X= 1000
BT
1
9 tháng 2 2020
a) \(\frac{8}{9}\cdot x-\frac{2}{3}=\frac{1}{3}\cdot x+1\frac{1}{3}\)
=> \(\frac{8x}{9}-\frac{2}{3}=\frac{x}{3}+\frac{4}{3}\)
=> \(\frac{8x}{9}-\frac{6}{9}=\frac{x+4}{3}\)
=> \(\frac{8x-6}{9}=\frac{x+4}{3}\)
=> \(3\left(8x-6\right)=9\left(x+4\right)\)
=> \(24x-18=9x+36\)
=> \(24x-18-9x=36\)
=> \(24x-9x=54\)
=> \(15x=54\)
=> \(5x=18\)
=> \(x=\frac{18}{5}\)
Vậy x = \(\frac{18}{5}\)
b) \(\left(x-\frac{1}{2}\right)\left(\frac{3}{2}-2x\right)=0\)
=> \(\orbr{\begin{cases}x-\frac{1}{2}=0\\\frac{3}{2}-2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=\frac{3}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}:2=\frac{3}{4}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{2};\frac{3}{4}\right\}\)
KT
0
1 tháng 4 2018
3(x - 1/2) - 5(x + 3/5) = x + 1/5
(3x - 3/2) - (5x + 3) - x = 1/5
3x - 3/2 - 5x - 3 - x = 1/5
(3x - 5x - x) - 3/2 - 3 = 1/5
- 3x = 1/5 + 3 + 3/2 = 47/10
x = 47/10 : (-3) = 47/10 . (-1)/3 = -47/30
26 tháng 3 2022
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{11}{75}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{11}{75}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{11}{75}:\frac{1}{2}=\frac{22}{75}\Leftrightarrow\frac{1}{x+2}=\frac{1}{25}\Leftrightarrow x=23\)
Ta có: \(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right).\left(x+2\right)}+\frac{1}{\left(x+2\right).\left(x+3\right)}-\frac{1}{x}=\frac{1}{2020}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2020}\)
\(\Leftrightarrow-\frac{1}{\left(x+3\right)}=\frac{1}{2020}\)
\(\Rightarrow-\left(x+3\right)=2020\)
\(\Leftrightarrow-x-3=2020\)
\(\Leftrightarrow-x=2023\)
\(\Leftrightarrow x=-2023\)
Vậy \(x=-2023\)
Bài làm:
Ta có: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2020}\)
\(\Leftrightarrow\frac{\left(x+1\right)-x}{x\left(x+1\right)}+\frac{\left(x+2\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}+\frac{\left(x+3\right)-\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2020}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2020}\)
\(\Rightarrow\frac{1}{-x-3}=\frac{1}{2020}\)
\(\Rightarrow-x-3=2020\Rightarrow x=-2023\)