\(\frac{-2}{1,5}=\frac{x-1}{3}\)tim x

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27 tháng 12 2019

\(\frac{-2}{1,5}=\frac{-2.2}{1,5.2}=\frac{-4}{3}=\frac{x-1}{3}\)

\(\Rightarrow x-1=-4\)\(\Rightarrow x=-3\)

Vậy \(x=-3\)

27 tháng 12 2019
-2/1,5=x-1/3 =>(-2)×3=1,5×(x-1) =>-6=1,5x-1,5 1,5x=-6+1,5=-4,5 x=-4,5÷1,5 x=-3 Vậy x=-3
8 tháng 7 2017

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

1 tháng 8 2017

khó lắm

bây h thì bạn giải đc chưa

18 tháng 8 2020

a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)

=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)

=> \(\frac{2}{3}x=-\frac{29}{70}\)

=> \(x=-\frac{29}{70}:\frac{2}{3}\)

=> \(x=-\frac{29}{70}.\frac{3}{2}\)

=> \(x=-\frac{87}{140}\)

b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)

=> \(-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)

=> \(-\frac{21}{13}x=-\frac{3}{3}\)

=> \(-\frac{21}{13}x=1\)

=> \(x=1:\left(-\frac{21}{13}\right)\)

=> \(x=-\frac{13}{21}\)

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

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

=> \(\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)(T/M)

d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)

=> \(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)

=> \(=>\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}\\x=-\frac{1}{2}-\frac{3}{4}\end{matrix}\right.=>\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{matrix}\right.\)(T/M)

HỌC TỐT vui

18 tháng 8 2020

a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)

\(\Leftrightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)

\(\Leftrightarrow\frac{2}{3}x=-\frac{29}{70}\)

\(\Leftrightarrow x=-\frac{29}{70}:\frac{2}{3}\)

\(\Leftrightarrow x=-\frac{87}{140}\)

b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)

\(\Leftrightarrow-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)

\(\Leftrightarrow-\frac{21}{13}x=-1\)

\(\Leftrightarrow x=-1:\left(-\frac{21}{13}\right)\)

\(\Leftrightarrow x=\frac{13}{21}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)

d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)

\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=\frac{5}{4}\end{matrix}\right.\)

a) Ta có: \(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)=\frac{2}{3}\)

\(\frac{2}{3}:\left(x-1\right)=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)

\(x-1=\frac{2}{3}:\frac{1}{6}=\frac{2}{3}\cdot6=4\)

hay x=5

Vậy: x=5

b) \(5,4-3\left[x-120\%\right]=\frac{3}{10}\)

\(\frac{27}{5}-3\cdot\left(x-\frac{6}{5}\right)=\frac{3}{10}\)

\(3\left(x-\frac{6}{5}\right)=\frac{27}{5}-\frac{3}{10}=\frac{51}{10}\)

hay \(x-\frac{6}{5}=\frac{51}{10}\cdot\frac{1}{3}=\frac{17}{10}\)

\(x=\frac{17}{10}+\frac{6}{5}=\frac{29}{10}\)

Vậy: \(x=\frac{29}{10}\)

c) \(10\cdot3^{x+2}-3^x=89\)

\(\Leftrightarrow10\cdot3^2\cdot3^x-3^x=89\)

\(\Leftrightarrow3^x\left(90-1\right)=89\)

\(\Leftrightarrow3^x=1\)

hay x=0

Vậy: x=0

d) \(5\cdot\left(x-0,2\right)=3x+\left(\frac{-2}{3}\right)^3\)

\(5\cdot\left(x-\frac{1}{5}\right)=3x+\frac{-8}{27}\)

\(\Leftrightarrow5x-1-3x-\frac{-8}{27}=0\)

\(\Leftrightarrow2x-\frac{19}{27}=0\)

\(\Leftrightarrow2x=\frac{19}{27}\)

hay \(x=\frac{\frac{19}{27}}{2}=\frac{19}{27}\cdot\frac{1}{2}=\frac{19}{54}\)

Vậy: \(x=\frac{19}{54}\)

e) \(\left(2x+\frac{3}{4}\right)^2-1,5=2\frac{1}{2}\)

\(\Leftrightarrow\left(2x+\frac{3}{4}\right)^2=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{3}{2}=-2\\2x+\frac{3}{2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2-\frac{3}{2}\\2x=2-\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{7}{2}\\2x=\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{2}\cdot\frac{1}{2}\\x=\frac{1}{2}\cdot\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{4}\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{-\frac{7}{4};\frac{1}{4}\right\}\)

3 tháng 3 2018

Câu 2 đây:

\(|x^2+|x-1||=x^2+2\)

\(\Rightarrow\orbr{\begin{cases}x^2+\left|x-1\right|=x^2+2\\x^2+\left|x-1\right|=-x^2-2\left(l\right)\end{cases}}\)

\(\Rightarrow\left|x-1\right|=2\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

3 tháng 3 2018

a)    \(M=\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+0,5}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)

\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{2}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2012}{2013}\)

\(=\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\right):\frac{2012}{2013}\)

\(=\left(\frac{2}{7}-\frac{2}{7}\right):\frac{2012}{2013}\)

\(=0\)

26 tháng 9 2019

a ) \(\left|x-2\right|=2\)

       \(\Rightarrow\orbr{\begin{cases}x-2=2\\x-2=-2\end{cases}}\)

        \(\Rightarrow\orbr{\begin{cases}x=2+2\\x=-2+2\end{cases}}\)

       \(\Rightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\) 

Vậy \(x=4\) hoặc \(x=0\)

b ) \(\left|x-\frac{4}{5}\right|=\frac{3}{4}\)

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

       \(\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}+\frac{4}{5}\\x=-\frac{3}{4}+\frac{4}{5}\end{cases}}\)

       \(\Rightarrow\orbr{\begin{cases}x=\frac{15}{20}+\frac{16}{20}\\-\frac{15}{20}+\frac{16}{20}\end{cases}}\)

         \(\Rightarrow\orbr{\begin{cases}x=\frac{31}{20}\\x=\frac{1}{20}\end{cases}}\)

Vậy \(x=\frac{31}{20}\) hoặc \(x=\frac{1}{20}\)

c ) \(6-\left|\frac{1}{2}-x\right|=\frac{2}{5}\)

                 \(\left|\frac{1}{2}-x\right|=6-\frac{2}{5}\)

              \(\left|\frac{1}{2}-x\right|=\frac{28}{5}\)

 \(\Rightarrow\orbr{\begin{cases}\frac{1}{2}-x=\frac{28}{5}\\\frac{1}{2}-x=-\frac{28}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}-\frac{28}{5}\\x=\frac{1}{2}-\left(-\frac{28}{5}\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{51}{10}\\x=\frac{61}{10}\end{cases}}\)

Vậy \(x=-\frac{51}{10}\) hoặc \(x=\frac{61}{10}\)

d ) \(\frac{1}{5}-\left|\frac{1}{5}.x\right|=\frac{1}{5}\)

                   \(\left|\frac{1}{5}.x\right|=\frac{1}{5}-\frac{1}{5}\)

                    \(\left|\frac{1}{5}.x\right|=0\)

\(\Rightarrow\frac{1}{5}.x=0\)

               \(x=0:\frac{1}{5}\)

                \(x=0\)

Vậy \(x=0\)

e ) \(-2,5+\left|3x+5\right|=-1,5\)

                        \(\left|3x+5\right|=-1,5-\left(-2,5\right)\)

                       \(\left|3x+5\right|=1\)

\(\Rightarrow\orbr{\begin{cases}3x+5=1\\3x+5=-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=1-5\\3x=-1-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=-4\\3x=-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-4:3\\x=-6:3\end{cases}}\)

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

Vậy \(x=-\frac{4}{3}\) hoặc \(x=-2\)

g ) \(\left(\frac{2}{3}x-1\right)+\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\left(\frac{2}{3}x-1\right)=0\\\left(\frac{3}{4}x+\frac{1}{2}\right)=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-1=0\\\frac{3}{4}x+\frac{1}{2}=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x=0+1\\\frac{3}{4}x=0-\frac{1}{2}\end{cases}}\)

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

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

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{2}{3}\end{cases}}\)

Vậy \(x=\frac{3}{2};x=-\frac{2}{3}\)

Bạn bạn hơi dài nhưng đã xong 

Chúc bạn học tốt !!!

28 tháng 2 2021

Trả lời:

a, | x - 2 | = 2

=> x - 2 = 2 hoặc x - 2 = -2

=>    x = 4                x = 0

Vậy x = 4; x = 0

b, \(\left|x-\frac{4}{5}\right|=\frac{3}{4}\)

\(\Rightarrow x-\frac{4}{5}=\frac{3}{4}\)hoặc \(x-\frac{4}{5}=\frac{-3}{4}\)

\(\Rightarrow x=\frac{31}{20}\)hoặc \(x=\frac{1}{20}\)

Vậy \(x=\frac{31}{20};x=\frac{1}{20}\)

c, \(6-\left|\frac{1}{2}-x\right|=\frac{2}{5}\)

\(\Rightarrow\left|\frac{1}{2}-x\right|=\frac{28}{5}\)

\(\Rightarrow\frac{1}{2}-x=\frac{28}{5}\)hoặc \(\frac{1}{2}-x=\frac{-28}{5}\)

\(\Rightarrow x=\frac{-51}{10}\)hoặc \(x=\frac{61}{10}\)

Vậy \(x=\frac{-51}{10};x=\frac{61}{10}\)

d, \(\frac{1}{5}-\left|\frac{1}{5}\cdot x\right|=\frac{1}{5}\)

\(\Rightarrow\left|\frac{1}{5}\cdot x\right|=0\)

\(\Rightarrow\frac{1}{5}\cdot x=0\)

\(\Rightarrow x=0\)

Vậy x = 0
e, \(-2,5+\left|3x+5\right|=-1,5\)

\(\Rightarrow\left|3x+5\right|=1\)

\(\Rightarrow3x+5=1\)hoặc \(3x+5=-1\)

\(\Rightarrow3x=-4\)hoặc \(3x=-6\)

\(\Rightarrow x=\frac{-4}{3}\)hoặc \(x=-2\)

Vậy \(x=\frac{-4}{3};x=-2\)

g, \(\left(\frac{2}{3}x-1\right)+\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)

\(\Rightarrow\frac{2}{3}x-1+\frac{3}{4}x+\frac{1}{2}=0\)

\(\Rightarrow\frac{17}{12}x-\frac{1}{2}=0\)

\(\Rightarrow\frac{17}{12}x=\frac{1}{2}\)

\(\Rightarrow x=\frac{6}{17}\)

Vậy \(x=\frac{6}{17}\)

14 tháng 12 2017

dễ mà bạn

14 tháng 12 2017

a) x=4/7 - 1/3=19/21

b) /x-5/=7 -->x-5=7 hoặc x-5=-7

--> x=12 hoặc x= -2

12 tháng 9 2019

a, \(\left(2x-1\right)=-8\)

\(2x=-8+1\)

\(2x=-7\)

\(x=-7:2\)

\(x=-3,5\)

12 tháng 9 2019

a) (2x - 1) = -8

⇒ 2x = -8 + 1

⇒ 2x = -7

b) (3x - 2)\(^2\) = \(\frac{1}{49}\)

Ta có: \(\frac{1}{49}\) = \(\frac{1}{7}\). \(\frac{1}{7}\) hoặc \(\frac{1}{49}\) = \(\frac{-1}{7}\). \(\frac{-1}{7}\)

TH1: 3x - 2 = \(\frac{1}{7}\) TH2: 3x - 2 = \(\frac{-1}{7}\)

⇒ 3x = \(\frac{1}{7}\)+2 ⇒ 3x = \(\frac{-1}{7}\)+2

⇒ 3x = \(\frac{15}{7}\) ⇒ 3x = \(\frac{13}{7}\)

⇒ x = \(\frac{5}{7}\) ⇒ x = \(\frac{13}{21}\)

Vậy: x = \(\frac{5}{7}\) hoặc x = \(\frac{13}{21}\)

7 tháng 8 2018

\(\frac{x+2}{x+6}=\frac{3}{x+1}\)

\(\Rightarrow\left(x+2\right)\left(x+1\right)=3\left(x+6\right)\)

\(\Rightarrow x^2+x+2x+2=3x+18\)

\(\Rightarrow x^2+x+2x-3x=18-2\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x=\pm4\)

các phần còn lại tương tự :)

7 tháng 8 2018

a)\(\frac{x+2}{x+6}\) =\(\frac{3}{x+1}\)

<=>\(\frac{\left(x+2\right)\left(x+1\right)}{\left(x+6\right)\left(x+1\right)}\) =\(\frac{3\left(x+6\right)}{\left(x+1\right)\left(x+6\right)}\)

=> ( x+2) ( x+1) = 3(x+6)

<=>  x2 +3x +3 = 3x +18

<=> x2 +3x -3x = 18 -3 

<=> x2              = 15

 => x                 = \(\sqrt{15}\)

 Vậy x=\(\sqrt{15}\)

b)

23 tháng 7 2019

a) Quy đồng lên đi.

b) \(\frac{x+2}{0.5}=\frac{2x+1}{2}\Leftrightarrow\frac{x+2}{\left(\frac{1}{2}\right)}=\frac{2x+1}{2}\)

\(\Leftrightarrow2x+4=\frac{2x+1}{2}\Leftrightarrow4x+8=2x+1\)

\(\Leftrightarrow x=-\frac{7}{2}\)

c) \(\Leftrightarrow\left|x+\frac{1}{5}\right|=6\). VỚi x >= -1/5 thì:

\(x+\frac{1}{5}=6\Leftrightarrow x=\frac{29}{5}\left(TM\right)\)

Với x < -1/5 thì \(-x-\frac{1}{5}=6\Leftrightarrow x=-\frac{31}{5}\left(TM\right)\)

d) TƯơng tự ý a, quy đồng lên thôi (mẫu chung là 24 thì phải)

23 tháng 7 2019

c) \(\left|x+\frac{1}{5}\right|-4=2\)

=> \(\left|x+\frac{1}{5}\right|=2+4\)

=> \(\left|x+\frac{1}{5}\right|=6\)

=> \(\left\{{}\begin{matrix}x+\frac{1}{5}=6\\x+\frac{1}{5}=-6\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=6-\frac{1}{5}\\x=\left(-6\right)-\frac{1}{5}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\frac{29}{5}\\x=-\frac{31}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{29}{5};-\frac{31}{5}\right\}\).

Mình chỉ làm câu c) thôi nhé.

Chúc bạn học tốt!