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
30 tháng 7 2018
\(a) \frac{4}{5}.x=\frac{8}{35}\)
\(\implies x= \frac{8}{35}:\frac{4}{5}\)
\(\implies x=\frac{8}{35}.\frac{5}{4}\)
\(\implies x=\frac{2}{7}\). Vậy \(x=\frac{2}{7}\)
\(b) \frac{3}{5}x-\frac{1}{2}=\frac{1}{7}\)
\(\implies \frac{3}{5}x=\frac{1}{7}+\frac{1}{2}\)
\(\implies \frac{3}{5}x=\frac{9}{14}\)
\(\implies x=\frac{9}{14}:\frac{3}{5}\)
\(\implies x=\frac{9}{14}.\frac{5}{3} \)
\(\implies x=\frac{15}{14}\). Vậy \(x=\frac{15}{14}\)
\(c) x-25\% x=0,5\)
\(\implies 75\% x=0,5\)
\(\implies \frac{3}{4}x=\frac{1}{2}\)
\(\implies x=\frac{1}{2}:\frac{3}{4}\)
\(\implies x=\frac{1}{2}.\frac{4}{3}\)
\(\implies x=\frac{2}{3}\). Vậy \(x=\frac{2}{3}\)
\(d) (\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}).120+x:\frac{1}{3}=-4\)
Có: \(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\)
\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)
\(=\frac{1}{24}-\frac{1}{30}\)
\(=\frac{1}{120}\)
Thay vào ta có: \(\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(\implies 1+x:\frac{1}{3}=-4\)
\(\implies x:\frac{1}{3}=-5\)
\(\implies x=-5.\frac{1}{3}\)
\(\implies x=\frac{-5}{3}\). Vậy \(x=\frac{-5}{3}\)
~ Hok tốt a~
30 tháng 7 2018
\(\frac{4}{5}.x=\frac{8}{35}\) \(\frac{3}{5}x-\frac{1}{2}=\frac{1}{7}\)
\(x=\frac{8}{35}:\frac{4}{5}\) \(\frac{3}{5}x=\frac{1}{7}+\frac{1}{2}\)
\(x=\frac{8}{35}.\frac{5}{4}\) \(\frac{3}{5}x=\frac{9}{14}\)
\(x=\frac{2}{7}\) \(x=\frac{9}{14}:\frac{3}{5}\)
Vậy \(x=\frac{2}{7}\) \(x=\frac{9}{14}.\frac{5}{3}\)
\(x-25\%x=0.5\) \(x=\frac{15}{14}\)
\(x-\frac{1}{4}x=\frac{1}{2}\) Vậy \(x=\frac{15}{14}\)
\(x\left(1-\frac{1}{4}\right)=\frac{1}{2}\) \(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)
\(x=\frac{1}{2}:\frac{3}{4}\) \(\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\) \(x=\frac{2}{3}\) \(\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)
Vậy \(x=\frac{2}{3}\) \(\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(1+x:\frac{1}{3}=-4\)
\(x:\frac{1}{3}=\left(-4\right)-1\)
\(x:\frac{1}{3}=-5\)
\(x=\left(-5\right).\frac{1}{3}\)
\(x=-\frac{5}{3}\)
Vậy \(x=-\frac{5}{3}\)
NQ
2
10 tháng 6 2020
a) \(\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)
0,5 x + 0,6 ( x - 2 ) = 3
0,5 x + 0.6 x - 1,2 = 3
1,1 x = 4,2
x = \(\frac{42}{11}\)
Kết luận:
b) \(\frac{1}{3}x-0,5x=0,75\)
\(\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)
\(-\frac{1}{6}x=\frac{3}{4}\)
\(x=-\frac{9}{2}\)
Kết luận:
c) \(\frac{3}{-2}x-0,5x=75\%\)
-1,5x - 0,5x = 0,75
-2x = 0,75
x = -0,375
Kết luận:
d) \(-\frac{2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)
-0,4 x + 0,25 = 0,75 - 0,75 x
-0,4 x + 0,75 x = 0,75 - 0,25
0,35 x = 0,5
x = \(\frac{10}{7}\)
Kết luận:
F
10 tháng 6 2020
\(a,\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)
\(\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)
\(\left(\frac{1}{2}+\frac{3}{5}\right)x-\frac{6}{5}=3\)
\(\frac{11}{10}x-\frac{6}{5}=3\)
\(\frac{11}{10}x=\frac{21}{5}\)
\(x=\frac{42}{11}\)
\(b,\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)
\(\left(\frac{1}{3}-\frac{1}{2}\right)x=\frac{3}{4}\)
\(\frac{-1}{6}x=\frac{3}{4}\)
\(x=\frac{-9}{2}\)
\(c,\frac{3}{-2}x-\frac{1}{2}x=75\%\)
\(\left(\frac{3}{-2}-\frac{1}{2}\right)x=\frac{3}{4}\)
\(-2x=\frac{3}{4}\)
\(x=\frac{-3}{8}\)
\(\frac{-2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)
\(\frac{-2}{5}x+\frac{3}{4}x=\frac{3}{4}-\frac{1}{4}\)
\(\left(\frac{-2}{5}+\frac{3}{4}\right)x=\frac{1}{2}\)
\(\frac{7}{20}x=\frac{1}{2}\)
\(x=\frac{10}{7}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 8 2023
a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)
y = \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)
y = \(\dfrac{4}{3}\)
b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)
y - 0,5 + 0,5 = \(\dfrac{3}{4}\)
y = \(\dfrac{3}{4}\)
c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2
0,8 - 0,4y = 0,2
0,4y = 0,8 - 0,2
0,4y = 0,6
y = 1,5
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 8 2023
d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{14}{9}\)
y = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)
y = \(\dfrac{29}{36}\)
e, y : \(\dfrac{5}{4}\) = \(\dfrac{9}{5}\) + \(\dfrac{1}{2}\)
y : \(\dfrac{5}{4}\) = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{8}\)
f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\)) = \(\dfrac{4}{5}\)
2y = \(\dfrac{4}{5}\)
y = \(\dfrac{2}{5}\)
T
2
14 tháng 4 2017
a) x-30%x=-4/5 b) (x-2).(x+4)=0 c)4/3-(0,5+2/3x)=1/6
1x-3/10x=-4/5 TH1: x-2 =0 =>x=2 4/3-(1/2+2/3x)=1/6
x(1-3/10)=-4/5 TH2: x+4 =0 =>x=-4 1/2+2/3x=4/3-1/6
x.7/10=-4/5 1/2+2/3x=7/6
x=-8/7 2/3x=7/6-1/2
2/3x=2/3
=>x=1
F
15 tháng 6 2020
\(a,\frac{-2}{3}\div\left(\frac{1}{2}-3x\right)=\frac{5}{3}\)
\(\frac{1}{2}-3x=\frac{-2}{3}\div\frac{5}{3}\)
\(\frac{1}{2}-3x=\frac{-2}{5}\)
\(3x=\frac{1}{2}-\frac{-2}{5}\)
\(3x=\frac{9}{10}\)
\(x=\frac{9}{10}\div3=\frac{3}{10}\)
\(b,\frac{1}{7}\left|x\right|-\frac{4}{5}=\frac{1}{5}\)
\(\frac{1}{7}\left|x\right|=\frac{1}{5}+\frac{4}{5}=1\)
\(\left|x\right|=1\div\frac{1}{7}=7\)
\(\Leftrightarrow x=\pm7\)
15 tháng 6 2020
-2/3:(1/2-3x)=5/3
(1/2-3x)=5/3.-3/2
(1/2-3x)=-5/2
3x=-5/2+1/2
3x=-3/2
x=-3/2:3/1
x=-3/2.1/3
x=-1/2
\(\frac{2}{5}.\frac{1}{x}+\frac{1}{x.2}+\frac{2}{5}=0.5\)
\(\frac{2}{5}.\frac{1}{x}+\frac{1}{2}.\frac{1}{x}+\frac{2}{5}=0.5\)
\(\frac{1}{x}\left(\frac{2}{5}+\frac{1}{2}\right)+\frac{2}{5}=0.5\)
\(\frac{1}{x}.\frac{9}{10}+\frac{2}{5}=0.5\)
\(\frac{1}{x}.\frac{9}{10}=\frac{1}{2}-\frac{2}{5}\)
\(\frac{1}{x}.\frac{9}{10}=\frac{1}{10}\)
\(\frac{1}{x}=\frac{1}{10}:\frac{9}{10}\)
\(\frac{1}{x}=\frac{1}{10}.\frac{10}{9}\)
\(\frac{1}{x}=\frac{1}{9}\)
\(\Rightarrow x=\frac{9.1}{1}=9\)