\(\dfrac{1}{2.\left(x+1\right)}\) = \(\dfrac{1}{6}\) ;

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30 tháng 7 2018

Ta có : 1, 1/ 2.(x+1) = 1/6 .

⇒ 6 = 2.(x+1) .

⇒ x + 1 = 6 : 2 .

⇒ x + 1 = 3 .

⇒ x = 3 - 1 .

⇒ x = 2 .

2, 1 < x/5 ≤ 2

⇒ 5/5 < x/5 ≤ 10/5 .

5 < x ≤ 10 .

⇒ x = 6 ; 7 ; 8 ; 9 ; 10 .

\(\Leftrightarrow5\cdot\dfrac{5-4-2}{20}< \dfrac{x}{20}< =3\cdot\dfrac{10-5-4}{20}\)

=>-1/20<x/20<=3/20

=>-1<x<=3

hay \(x\in\left\{0;1;2;3\right\}\)

2 tháng 5 2017

a) \(\left(2x-3\right)\left(6-2x\right)=0\)

\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)

\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)

Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).

b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)

\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)

\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)

\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)

\(-\dfrac{11}{15}=-x\left(x-1\right)\)

\(\Rightarrow x=1.491631652\)

Vậy \(x=1.491631652\)

c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)

\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)

\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)

Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).

d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)

Vậy \(x=\dfrac{10}{3}\).

e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)

\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)

\(\dfrac{x}{3}=\dfrac{7}{10}\)

\(x=\dfrac{3\cdot7}{10}\)

\(x=\dfrac{21}{10}\)

Vậy \(x=\dfrac{21}{10}\).

f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)

\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)

\(\dfrac{x}{5}=\dfrac{11}{10}\)

\(x=\dfrac{5\cdot11}{10}\)

\(x=\dfrac{55}{10}=\dfrac{11}{2}\)

Vậy \(x=\dfrac{11}{2}\).

g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)

Vậy \(x=2\).

h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)

Vậy \(x=14\).

a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}< x< \dfrac{1}{48}-\dfrac{1}{16}+\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{6}{12}-\dfrac{4}{12}-\dfrac{3}{12}< x< \dfrac{1}{48}-\dfrac{3}{48}+\dfrac{8}{48}\)

\(\Leftrightarrow\dfrac{-1}{12}< x< \dfrac{1}{8}\)

\(\Leftrightarrow-2< 24x< 3\)

=>x=0

b: \(\Leftrightarrow\dfrac{9-10}{12}< \dfrac{x}{12}< 1-\dfrac{8-3}{12}=\dfrac{7}{12}\)

=>-1<x<7

hay \(x\in\left\{0;1;2;3;4;5;6\right\}\)

a: \(\Leftrightarrow70+18< x< 120+126+70\)

=>88<x<316

hay \(x\in\left\{89;90;...;315\right\}\)

b: \(\Leftrightarrow-\dfrac{9}{3}< x< \dfrac{8}{5}+\dfrac{9}{5}=\dfrac{17}{5}\)

=>-3<x<3,4

hay \(x\in\left\{-2;-1;0;1;2;3\right\}\)

21 tháng 4 2017

tìm x a)
\(\dfrac{7}{2}\)-\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\) + \(\dfrac{7}{2}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{12}+\dfrac{7}{12}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-12}{12}=1\)
\(x+\dfrac{7}{10}\)= 1 . \(\dfrac{6}{5}\)
*Rồi tự làm phần tt đi




20 tháng 4 2017

Mình ghi kết quả luôn nha bạn

6 tháng 5 2018

\(\dfrac{1}{7}=\dfrac{8}{-x}\)=> \(-x=56\)

=> \(x=56\)

2) => 18x = 18

=> x = 1

3) \(\dfrac{-4}{3}+x=\dfrac{-11}{6}\)

=> \(x=\dfrac{-11}{6}+\dfrac{4}{3}\)

=> \(x=\dfrac{-1}{2}\)

4) 45%.x =\(\dfrac{3}{5}\)

=> \(x=\dfrac{3}{5}:\dfrac{9}{20}\)

=> \(x=\dfrac{4}{3}\)

a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)

=>2/5x=8/5

=>x=4

b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)

\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)

=>1/3x=-6

=>x=-18

c: =>2|x-1/3|=0,24-4/5=-0,56<0

10 tháng 3 2018

Ta có: 1/3 + −2/5+ 1/6 + −1/5 ≤ x < −3/4+2/7+-1/4+3/5+5/7

⇒10-12+5-6/30≤ x< -105+40-35+84+100/140

⇒-3/30≤ x <84/140

⇒-0,1≤ x < 0,6

⇒x=0

12 tháng 6 2017

K chép lại đề, lm luôn nhé:

*\(\Rightarrow\) \(\left(\dfrac{7}{2}+2x\right)\cdot\dfrac{8}{3}=\dfrac{16}{3}\)

\(\Rightarrow\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)

\(\Rightarrow2x=2-\dfrac{7}{2}=-\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{3}{4}\)

* \(\Rightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{\dfrac{3}{4}-2}{2}=-\dfrac{5}{8}\)

=> K có gt x nào t/m đề

* Đề sai

* \(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)

*\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)

\(\Rightarrow2x-1=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right)=-\dfrac{4}{63}\)

\(\Rightarrow2x=-\dfrac{4}{63}+1=\dfrac{59}{63}\)

\(\Rightarrow x=\dfrac{59}{63}:2=\dfrac{59}{126}\)

* \(\Rightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0\Rightarrow x=0\\2x=-\dfrac{6}{5}\Rightarrow x=-\dfrac{3}{5}\end{matrix}\right.\)

* \(\Rightarrow-5x-1-\dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)

\(\Rightarrow-5x-\dfrac{1}{2}x-\dfrac{3}{2}x=-\dfrac{5}{6}+1-\dfrac{1}{3}\)

\(\Rightarrow-7x=-\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{1}{6}:7=-\dfrac{1}{42}\)

12 tháng 6 2017

a)\(\left(3\dfrac{1}{2}+2x\right).2\dfrac{2}{3}=5\dfrac{1}{3}\)

\(\left(\dfrac{7}{2}+2x\right).\dfrac{8}{3}=\dfrac{16}{3}\)

\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)

\(2x=2-\dfrac{7}{2}=\dfrac{-3}{2}\Rightarrow x=\dfrac{-3}{4}\)

b)\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)

\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2=\dfrac{-1}{4}\)

\(\Rightarrow\left|2x-3\right|=\dfrac{-1}{8}\)

\(\Rightarrow x\in\varnothing\)

c) Đề sai,bạn có viết chữ x đâu,đó là phép tính mà.

d)\(\left(3x-1\right)\left(\dfrac{-1}{2}x+5\right)=0\)

\(\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{-1}{2}x+5=0\Rightarrow x=10\)

e)\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)

\(\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=\dfrac{-21}{4}\)

\(2x-1=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{-4}{63}\)

\(\Rightarrow2x=\dfrac{59}{63}\Rightarrow x=\dfrac{59}{126}\)

g)\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)

\(\left(2x+\dfrac{3}{5}\right)^2=0+\dfrac{9}{25}=\dfrac{9}{25}\)

\(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\)

\(th1:x=0\)

\(th2:x=\dfrac{-3}{5}\)

h)\(-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)

\(-5x+-1-\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)

\(\Leftrightarrow-5x+-1+\dfrac{5}{6}-\dfrac{1}{3}=2x\)

\(-5x+\dfrac{-1}{2}=2x\)

\(\dfrac{-1}{2}=2x+5x\)

\(\dfrac{-1}{2}=7x\Rightarrow x=\dfrac{-1}{14}\)