a, \(3\frac{12}{13}< x\frac{12}{13}< 5\frac{12}{13}\Rightarrow x=4\)

b, \(x\frac{3}{4}=\frac{21989}{7996}=\frac{11}{4}=2\frac{3}{4}\Rightarrow x=2\)

~ Hok tốt ~

Trả lời : 

a)x=4

b)x=2

\(\downarrow\)

a) \(\frac{561}{143}< x\frac{12}{13}< \frac{1463}{247}\)

\(\Rightarrow3\frac{12}{13}< x\frac{12}{13}< 5\frac{12}{13}\)

\(\Rightarrow x=4\)

b) \(x\frac{3}{4}=\frac{21989}{7996}\)

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

\(\Rightarrow x=2\)

~ Hok tốt ~

Tìm x

a) | x - \(\dfrac{3}{2}\) | = \(\left(\dfrac{12}{13}+7\right)\) + \(\left(-8+\dfrac{1}{13}\right)\)

| x - \(\dfrac{3}{2}\) | = \(\dfrac{12}{13}+7-8+\dfrac{1}{13}\)

| x - \(\dfrac{3}{2}\) | = 1 + 7 - 8

=> | x - \(\dfrac{3}{2}\) | = 0

=> x - \(\dfrac{3}{2}\) = 0

=> x = \(\dfrac{3}{2}\)

b) ( \(\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{1}{6}\) ) . \(\dfrac{1}{30}\) - | 3 . x - 3 | = 2

=> \(\dfrac{27}{4}\) . \(\dfrac{1}{30}\) - | 3x - 3 | = 2

=> \(\dfrac{9}{40}\) - | 3x - 3 | = 2

=> | 3x - 3 | = \(\dfrac{9}{40}\) - 2

=> | 3x - 3 | = \(\dfrac{-71}{40}\)

Th1 :

3x - 3 = \(\dfrac{-71}{40}\)

=> 3x = \(\dfrac{-71}{40}\) + 3

=> 3x = \(\dfrac{49}{40}\)

=> x = \(\dfrac{49}{40}\) : 3

=> x = \(\dfrac{49}{120}\)

TH2 :

3x - 3 = \(\dfrac{71}{40}\)

=> 3x = \(\dfrac{191}{40}\)

=> x = \(\dfrac{191}{120}\)

Vậy x = \(\dfrac{49}{120}\) hoặc \(\dfrac{191}{120}\)

a: =>|x-3/2|=12/13+1/13+7-8=0

=>x-3/2=0

hay x=3/2

b: \(\Leftrightarrow\dfrac{45-12-10}{60}\cdot\dfrac{1}{30}-\left|3x-3\right|=2\)

\(\Leftrightarrow\left|3x-3\right|=\dfrac{23}{60}\cdot\dfrac{1}{30}-2=-\dfrac{1777}{1800}\)(vô lý)

a) Ta có: \(\frac{x}{9}=\frac{-12}{27}\)

=> \(27.x=-12.9\)

=> \(27x=-108\)

=> \(x=108:27\)

=>\(x=4\)

x=2/3

Ta viết phương trình thành dạng \(\left|x-\frac{1}{3}\right|+\left|x-\frac{4}{3}\right|=\frac{9}{2}x\)

+) Xét  khoảng \(x< \frac{1}{3}\)

\(pt\Leftrightarrow\left(\frac{1}{3}-x\right)+\left(\frac{4}{3}-x\right)=\frac{9}{2}x\)

\(\Leftrightarrow\frac{5}{3}-2x=\frac{9}{2}x\Leftrightarrow\frac{13}{2}x=\frac{5}{3}\Leftrightarrow x=\frac{10}{39}\left(tm\right)\)

+) Xét khoảng \(\frac{1}{3}\le x\le\frac{4}{3}\)

\(pt\Leftrightarrow\left(x-\frac{1}{3}\right)+\left(\frac{4}{3}-x\right)=\frac{9}{2}x\)

\(\Leftrightarrow1=\frac{9}{2}x\Leftrightarrow x=\frac{2}{9}\)(L)

Xét khoảng \(x>\frac{4}{3}\)

\(pt\Leftrightarrow\left(x-\frac{1}{3}\right)+\left(x-\frac{4}{3}\right)=\frac{9}{2}x\)

\(\Leftrightarrow2x-\frac{5}{3}=\frac{9}{2}x\Leftrightarrow\frac{5}{3}=\frac{-5}{2}x\)(loại vì x chắc chắn âm)

Vậy tập nghiệm \(S=\left\{\frac{10}{39}\right\}\)

\(\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right):\frac{-5}{6}< x< \frac{4}{21}.\frac{4}{7}\)

\(\Rightarrow\left(\frac{6}{12}+\frac{9}{12}-\frac{4}{12}\right):\frac{-10}{12}< x< \frac{16}{147}\)

\(\Rightarrow\frac{11}{12}.\frac{-12}{10}< x< \frac{16}{147}\)

\(\Rightarrow\frac{-11}{10}< x< \frac{16}{147}\)

\(\Rightarrow\frac{-1617}{1470}< x< \frac{16}{1470}\)

\(x=\left\{-1;0\right\}\)