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9 tháng 10 2016

Đoạn đó ( dưới ) :

\(\Rightarrow\left[\begin{array}{nghiempt}\frac{1}{4}x=2\\\frac{1}{4}x=-2\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=8\\x=-8\end{array}\right.\)

9 tháng 10 2016

:)

Giải :

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

\(\Rightarrow\frac{1}{4}\left|x\right|=2\)

\(\Rightarrow\left[\begin{array}{nghiempt}\frac{1}{4}.x=2\\\frac{1}{4}.x=-2\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=8\\x=-8\end{array}\right.}\)

26 tháng 10 2016

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

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

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

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)

        \(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)

        \(\frac{2}{5}-x=-3\)

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

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

                   \(x=\frac{3}{5}-\frac{15}{5}\)

                   \(x=-\frac{12}{5}\)

Vay \(x=-\frac{12}{5}\) 

    

  

26 tháng 10 2016

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

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)

        \(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)

        \(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)

        \(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)

        \(-3+\frac{3}{x}=\frac{-25}{12}\)

                     \(\frac{3}{x}=\frac{-25}{12}+3\)

                      \(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)

                      \(\frac{3}{x}=\frac{5}{6}\)

                      \(\frac{18}{6x}=\frac{5x}{6x}\)

Đèn dây , bạn tự làm tiếp nhé , de rồi chứ

11 tháng 9 2018

a) 

( 4x - 9 ) ( 2,5 + (-7/3) . x ) = 0

\(\Rightarrow\orbr{\begin{cases}4x-9=0\\2,5+\frac{-7}{3}x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)

P/s: đợi xíu làm câu b

11 tháng 9 2018

b) \(\frac{1}{x\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{-1}{x+3}=\frac{1}{2015}\)

\(\Leftrightarrow x+3=-2015\)

\(\Leftrightarrow x=-2018\)

Vậy,.........

13 tháng 10 2016

1)\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)

\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow\left(x+1\right)=0\).Do \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\ne0\)

\(\Leftrightarrow x=-1\)

13 tháng 10 2016

anh ơi, câu 1 anh viết 0 . DO là ntn ạ ?

 

14 tháng 10 2016

e)

=> (x-2) . (x+7) = ( x-1 ) . ( x +4)

=> x2 +7x - 2x -14 = x2 - x + 4x - 4

x2 + 5x - 14 = x2 + 3x - 4

=> 5x - 14  = 3x - 4

=> 5x  - 3x = 14-4

=> 2x         = 10 => x = 10 : 2 => x = 5

c)

=>( x-1) . 7 = ( x + 5 ) . 6

=> 7x - 7 = 6x + 30

=> 7x - 6x=  30 + 7

=> x         = 37

13 tháng 10 2016

a,x=\(\frac{5}{2}\)

b,x=\(\frac{13}{176}\)

c,x=37

d, x=\(\frac{12}{5}\)

e, x=5

16 tháng 8 2019

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

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

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

16 tháng 8 2019

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

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

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

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

5 tháng 9 2016

3.

a) \(\left|x\right|=2,1\Leftrightarrow\left[\begin{array}{nghiempt}x=2,1\\x=-2,1\end{array}\right.\)

b)\(\left|x\right|=\frac{17}{9}\Leftrightarrow x=-\frac{17}{9}\) (vì x/<0)

c) \(\left|x\right|=1\frac{2}{5}\Leftrightarrow\left|x\right|=\frac{7}{5}\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{7}{5}\\x=-\frac{7}{5}\end{array}\right.\)

d) \(\left|x\right|=0,35\Leftrightarrow x=0,35\) (Vì x>0)

2.

a) \(\left|x-1,7\right|=2,3\)

\(\Leftrightarrow\begin{cases}x\ge1,7\\x-1,7=2,3\end{cases}\) hoặc \(\begin{cases}x< 1,7\\1,7-x=2,3\end{cases}\)

\(\Leftrightarrow\begin{cases}x\ge1,7\\x=4\left(tm\right)\end{cases}\) hoặc \(\begin{cases}x< 1,7\\x=-0,6\left(tm\right)\end{cases}\)

Vậy x={4;-0,6}

b) đề thiếu

 

5 tháng 9 2016

a) |- 2,5| = 2,5 b) |- 2,5| = - 2,5 c) |- 2,5| = - |- 2,5| 

>.<

\(\left(-\frac{28}{19}\right)\times\left(-\frac{38}{14}\right)=\frac{14\times2\times19\times2}{19\times14}=4\)

>.<

\(\left(-\frac{21}{16}\right)\times\left(-\frac{24}{7}\right)=\frac{7\times3\times8\times3}{8\times2\times7}=\frac{9}{2}\)

>.<

\(\left(-\frac{12}{17}\right)\times\left(-\frac{34}{9}\right)=\frac{3\times4\times17\times2}{17\times3\times3}=\frac{8}{3}\)

>.<

\(\left|x\right|=2,1\)

\(x=\pm2,1\)

>.<

\(\left|x\right|=\frac{17}{9}\)

\(x=\pm\frac{17}{9}\)

x < 0

\(x=-\frac{17}{9}\)

>.<

\(\left|x\right|=1^2_5\)

\(x=\pm\frac{7}{5}\)

>.<

\(\left|x\right|=0,35\)

\(x=\pm0,35\)

x > 0

x = 0,35

>.<

\(\left|x-1,7\right|=2,3\)

\(x-1,7=\pm2,3\)

Th1:

x - 1,7 = 2,3

x = 2,3 + 1,7

x = 4

Th2:

x - 1,7 = - 2,3

x = - 2,3 + 1,7

x = - 0,6

>.<

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

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

\(x+\frac{3}{4}=\pm\frac{1}{3}\)

Th1:

x + 3/4 = 1/3

x = 1/3 - 3/4

x = \(-\frac{5}{12}\)

Th2:

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

\(x=-\frac{1}{3}-\frac{3}{4}\)

\(x=-\frac{13}{12}\)

25 tháng 11 2017

1)   \(\frac{x+4}{2005}\)\(+\)\(\frac{x+3}{2006}\)\(\frac{x+2}{2007}\)\(+\)\(\frac{x+1}{2008}\)

\(\Leftrightarrow\)   \(\frac{x+4}{2005}\)\(+\)\(+\)\(\frac{x+3}{2006}\)\(+\)1 = \(\frac{x+2}{2007}\)\(+\)\(+\)\(\frac{x+1}{2008}\)\(+\)1

\(\Leftrightarrow\)\(\frac{x+2009}{2005}\)\(\frac{x +2009}{2006}\)\(\frac{x+2009}{2007}\)+\(\frac{x+2009}{2008}\)

\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006) = (x + 2009)(1/2007 + 1/2008)

\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006 - 1/2007 - 1/2008) = 0

Ta thấy:  1/2005 + 1/2006 - 1/2007 - 1/2008 \(\ne\)0

\(\Leftrightarrow\)x + 2009 = 0

\(\Leftrightarrow\)x = -2009

16 tháng 2 2021

\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=\left(x-1\right)\left(x-2\right)x=0\)

tìm đc x=0;1;2

26 tháng 5 2016

\(PT\Leftrightarrow\frac{x+4+2000}{2000}+\frac{x+3+2001}{2001}=\frac{x+2+2002}{2002}+\frac{x+1+2003}{2003}\)

<=> \(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

<=> \(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

<=> x + 2004 = 0

<=> x = -2004.

26 tháng 5 2016

\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)

\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(x+2004=0\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\right)\)

\(\Rightarrow x=-2004\)