/x+5/phần3 - 1/2 =1/3
Tìm x
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
9 tháng 1 2022
\(\Leftrightarrow2.5:x=2+\dfrac{2}{3}-2-\dfrac{1}{5}=\dfrac{7}{15}\)
hay \(x=\dfrac{5}{2}:\dfrac{7}{15}=\dfrac{5}{2}\cdot\dfrac{15}{7}=\dfrac{75}{14}\)
AN
3
DC
19 tháng 4 2022
= 3/4 + 1/3 = 13/12
5/4 x X = 5/8
X = 5/8 : 5/4
X = 1/2
vậy X = ...
HX
2
NA
10 tháng 8 2016
\(\frac{3}{2}.\frac{4}{5}-x=\frac{2}{3}\)
\(\frac{6}{5}-x=\frac{2}{3}\)
\(x=\frac{6}{5}-\frac{2}{3}\)
\(x=\frac{18}{15}-\frac{10}{15}\)
\(x=\frac{8}{15}\)
2
24 tháng 3 2021
\(\frac{3}{2}\times\frac{4}{5}-x=\frac{2}{3}\)
\(\frac{12}{10}-x=\frac{2}{3}\)
\(\frac{6}{5}-x=\frac{2}{3}\)
\(x=\frac{6}{5}-\frac{2}{3}\)
\(x=\frac{18}{15}-\frac{10}{15}\)
\(x=\frac{8}{15}\)
\(x\times3\frac{1}{3}=\frac{3}{3}:4\frac{1}{4}\)
\(x\times\frac{10}{3}=1:\frac{17}{4}\)
\(x\times\frac{10}{3}=1.\frac{4}{17}\)
\(x\times\frac{10}{3}=\frac{4}{17}\)
\(x=\frac{4}{17}:\frac{10}{3}\)
\(x=\frac{4}{17}\times\frac{3}{10}\)
\(x=\frac{6}{85}\)
SI
1 tháng 7 2021
30% . x - x - \(\dfrac{5}{6}\) = \(\dfrac{1}{3}\)
\(\Rightarrow\) (\(\dfrac{3}{10}\) - 1) . x = \(\dfrac{1}{3}+\dfrac{5}{6}\)
\(\Rightarrow\) \(-\dfrac{7}{10}\) . x = \(\dfrac{7}{6}\)
\(\Rightarrow\) x = \(-\dfrac{3}{5}\)
DD
1 tháng 7 2021
KO ghi đề nhé
\(\dfrac{3}{10}.x-x=\dfrac{1}{3}+\dfrac{5}{6}\)
\(x\left(\dfrac{3}{10}-1\right)=\dfrac{2}{6}+\dfrac{5}{6}\)
\(x.\dfrac{-7}{10}=\dfrac{7}{6}\)
\(x=\dfrac{7}{10}:\dfrac{-7}{6}\)
\(x=\dfrac{7}{10}.\dfrac{6}{-7}\)
\(x=\dfrac{42}{-70}\)
\(x=\dfrac{6}{-10}\)
\(x=\dfrac{3}{-5}\)
Vậy \(x=\dfrac{3}{-5}\)
19 tháng 2 2021
Ta có : \(\dfrac{x}{2}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x=2\left(1-x\right)\)
\(\Leftrightarrow3x=2-2x\)
\(\Leftrightarrow5x=2\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy ...
19 tháng 2 2021
Ta có: \(\dfrac{x}{2}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x=2\left(1-x\right)\)
\(\Leftrightarrow3x=2-2x\)
\(\Leftrightarrow3x+2x=2\)
\(\Leftrightarrow5x=2\)
hay \(x=\dfrac{2}{5}\)
Vậy: \(x=\dfrac{2}{5}\)
VD
20 tháng 3 2022
a,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne\pm1\\x\ne\dfrac{1}{2}\end{matrix}\right.\)
\(A=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}+\dfrac{5}{x^2-1}\right):\dfrac{2x+1}{x^2-1}\\ =\left(\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{5}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{2x-2-x-1+5}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{x+2}{2x+1}\)
\(b,A=3\\ \Leftrightarrow\dfrac{x+2}{2x+1}=3\\ \Leftrightarrow6x+3=x+2\\ \Leftrightarrow5x+1=0\\ \Leftrightarrow x=-\dfrac{1}{5}\left(tm\right)\)
\(c,\dfrac{1}{A}=\dfrac{2x+1}{x+2}=\dfrac{2x+4-3}{x+2}=\dfrac{2\left(x+2\right)-3}{x+2}=2-\dfrac{3}{x+2}\)
Để `1/A` là số nguyên thì `3/(x+2)` nguyên \(\Rightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
Ta có bảng:
| x+2 | -3 | -1 | 1 | 3 |
| x | -5 | -3 | -1(ktm) | 1(ktm) |
Vậy \(x\in\left\{-5;-3\right\}\)
DN
0
TH
1
\(\frac{x+5}{3}-\frac{1}{2}=\frac{1}{3}\)
\(\frac{x+5}{3}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
\(x+5=\frac{5}{6}.3=\frac{5}{2}\)
\(x=\frac{5}{2}-5\)
\(x=\frac{5}{2}-\frac{10}{2}=\frac{-5}{2}\)