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Ta có:

\(\left|4x\right|-\left|-13,5\right|=\left|2\frac{1}{4}\right|\)

\(\Rightarrow\left|4x\right|=\left|2\frac{1}{4}\right|+\left|-13,5\right|=2\frac{1}{4}+13,5=\frac{63}{4}\)

\(\Rightarrow\left[{}\begin{matrix}4x=\frac{63}{4}\\4x=-\frac{63}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{63}{16}\\x=\frac{-63}{16}\end{matrix}\right.\)

3 tháng 12 2015

1) |4x|-|-13,5|+|9/4|

=> 4x - 13,5 +9/4

=> 4x- 11,25

=> 4x = 11,25

=> x = 11,25:4 = 2,8125

2)  x-24=y => x-y = 24

=> x/7 = y/3 = (x-y)/(7-3) = 24/4 = 6

=> x = 42   ;  y=18

2 tháng 8 2016

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

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

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

\(\Leftrightarrow\)\(\frac{-\left(x-4\right)+x}{x\left(x-4\right)}=\frac{x}{x\left(x-4\right)}\)

\(\Leftrightarrow\)\(4-x+x=x\)

\(\Leftrightarrow x=4\)

lo nói mk làm cách lâu chứ m cx hỏi người khác!!!!!!!!!!! 

 

30 tháng 5 2018

pt đã cho \(\Leftrightarrow4\left|x\right|-13,5=2\dfrac{1}{4}\Leftrightarrow4\left|x\right|=2\dfrac{1}{4}+13,5\Leftrightarrow4\left|x\right|=\dfrac{63}{4}\Leftrightarrow\left|x\right|=\dfrac{63}{16}\Leftrightarrow x=\pm\dfrac{63}{16}\)

kl: x= +- 63/16

30 tháng 5 2018

Giải:

\(\left|4x\right|-\left|-13,5\right|=\left|2\dfrac{1}{4}\right|\)

\(\Leftrightarrow\left|4x\right|-13,5=2\dfrac{1}{4}\)

\(\Leftrightarrow\left|4x\right|-13,5=\dfrac{9}{4}\)

\(\Leftrightarrow\left|4x\right|=\dfrac{9}{4}+13,5\)

\(\Leftrightarrow\left|4x\right|=15,75\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=15,75\\4x=-15,75\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3,9375\\x=-3,9375\end{matrix}\right.\)

Vậy ...

2 tháng 12 2015

dễ thấy |x+2/7| > 0;|x+4/7|>0;|x+3 1/7| >0

=>|x+2/7|+|x+4/7|+|x+3 1/7| > 0;mà VT=VP

nên 4x>0

ta có: \(\left|x+\frac{2}{7}\right|+\left|x+\frac{4}{7}\right|+\left|x+3\frac{1}{7}\right|=4x=>x+\frac{2}{7}+x+\frac{4}{7}+x+\frac{22}{7}=4x=>3x+4=4x=>x=4\)

vậy x=4

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