\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)

\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)

\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)

\(\Rightarrow-1\le x< 6\)

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

Bài b tương tự

bạn ơi bạn giải câu b được ko. mk ko biết làm câu b

a)

\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)                

Vậy \(x = \frac{{ - 3}}{2}\).

b)

\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)

Vậy \(x = \frac{{ - 5}}{6}\).

c)

\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)      

Vậy \(x = \frac{4}{5}\)

d)

\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)

Vậy \(x = \frac{{ - 2}}{5}\).

Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.

a) (x + 1/2) . (2/3 − 2x) = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)

\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)

\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)

\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)

\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)

\(\Rightarrow x=-\frac{2}{11}\)

c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)

\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)

\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)

\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)

\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)

\(\Rightarrow x=1\)

d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)

\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)

\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)

\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)

\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)

\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)

\(\Rightarrow x\approx28,7\) (số hơi lẻ)

e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)

\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)

a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)

\(=2\)

b) \(\frac{5}{9}\times\frac{1}{2}\times\frac{5}{9}\times\frac{6}{4}=\frac{25}{81}\times\frac{3}{4}=\frac{25}{108}\)

c) \(\frac{7}{8}\div\frac{1}{2}+\frac{9}{8}\div\frac{1}{2}=\left(\frac{7}{8}+\frac{9}{8}\right)\div\frac{1}{2}\)

\(=2\div\frac{1}{2}=4\)

d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)

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

a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}\)

\(=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)

\(=1+1\)

\(=2\)

b) \(\frac{5}{9}.\frac{1}{2}.\frac{5}{9}.\frac{6}{4}\)

\(=\left(\frac{5}{9}\right)^2\left(\frac{1}{2}.\frac{6}{4}\right)\)

\(=\frac{25}{81}.\frac{3}{4}\)

\(=\frac{25}{108}\)

c) \(\frac{7}{8}:\frac{1}{2}+\frac{9}{8}:\frac{1}{2}\)

\(=\frac{7}{8}.2+\frac{9}{8}.2\)

\(=2\left(\frac{7}{8}+\frac{9}{8}\right)\)

\(=2.\frac{16}{8}\)

\(=2.2\)

\(=4\)

d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}\)

\(=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)

\(=1+\frac{1}{2}\)

\(=\frac{2}{2}+\frac{1}{2}\)

\(=\frac{3}{2}\)

a/ \(\frac{6}{7}x=\frac{18}{23}\)

\(x=\frac{18}{23}:\frac{6}{7}=\frac{21}{23}\)

b/ \(2\frac{1}{2}x=\frac{5}{6}\)

\(=>\frac{5}{2}x=\frac{5}{6}\)

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

c/\(x:2\frac{3}{4}=9\frac{5}{8}\)

\(x:\frac{11}{4}=\frac{77}{8}\)

\(x=\frac{77}{8}\cdot\frac{11}{4}=\frac{847}{32}\)

d/\(7\frac{1}{7}\cdot\frac{1}{7}\cdot x=22\frac{1}{8}\)

\(\frac{50}{49}x=\frac{177}{8}\)

\(x=\frac{177}{8}:\frac{50}{49}=\frac{8673}{400}\)

\(a,\frac{6}{7}.x=\frac{18}{23}\) \(\Rightarrow x=\frac{18}{23}:\frac{6}{7}=\frac{18}{23}.\frac{7}{6}=\frac{21}{23}\)

\(b,2\frac{1}{2}.x=\frac{5}{6}\Rightarrow\frac{5}{2}.x=\frac{5}{6}\Rightarrow x=\frac{5}{6}:\frac{5}{2}=\frac{5}{6}.\frac{2}{5}=\frac{1}{3}\)

\(c,x:2\frac{3}{4}=9\frac{5}{8}\Rightarrow x:\frac{11}{4}=\frac{77}{8}\Rightarrow x=\frac{77}{8}.\frac{11}{4}=\frac{847}{32}\)

\(d,7\frac{1}{7}.\frac{1}{7}.x=22\frac{1}{8}\Rightarrow\frac{50}{49}.x=\frac{177}{8}\Rightarrow x=\frac{177}{8}:\frac{50}{49}=\frac{177}{8}.\frac{49}{50}=\frac{8673}{400}\)

a Đ

b S

c S

d Đ

a ) S 

b ) Đ

c ) S

d ) Đ

k cho mk nhé 

\(\frac{3x-7}{5}=\frac{2x-1}{3}\)

\(\Leftrightarrow9x-21=10x-5\)

\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)

\(\frac{4x-7}{12}-x=\frac{3x}{8}\)

\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow-56-64x=36x\)

\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)

\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)

Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0

Vậy x = 2019

\(\frac{5x-8}{3}=\frac{1-3x}{2}\)

\(\Leftrightarrow10x-16=3-9x\)

\(\Leftrightarrow19x=19\Leftrightarrow x=1\)

a) \(x+\left(-7\right)=-20\)

\(\Rightarrow x=-20+7\)

\(\Rightarrow x=-13\)

Vậy \(x=-13\)

b) \(8-x=-12\)

\(\Rightarrow x=8-\left(-12\right)\)

\(\Rightarrow x=20\)

Vậy \(x=20\)

c) \(|x|-7=-6\)

\(\Rightarrow|x|=-6+7\)

\(\Rightarrow|x|=1\)

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

Vậy \(x\in\left\{1;-1\right\}\)

d) \(5^2.2^2-7.|x|=65\)

\(\Rightarrow\left(5.2\right)^2-7.|x|=65\)

\(\Rightarrow10^2-7.|x|=65\)

\(\Rightarrow100-7.|x|=65\)

\(\Rightarrow7.|x|=35\)

\(\Rightarrow|x|=5\)

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

Vậy \(x\in\left\{5;-5\right\}\)

e) \(37-3.|x|=2^3-4\)

\(\Rightarrow37-3.|x|=8-4\)

\(\Rightarrow37-3.|x|=4\)

\(\Rightarrow3.|x|=33\)

\(\Rightarrow|x|=11\)

\(\Rightarrow\orbr{\begin{cases}x=11\\x=-11\end{cases}}\)

Vậy \(x\in\left\{11;-11\right\}\)

f) \(|x|+|-5|=|-37|\)

\(\Rightarrow|x|+5=37\)

\(\Rightarrow|x|=32\)

\(\Rightarrow\orbr{\begin{cases}x=32\\x=-32\end{cases}}\)

Vậy \(x\in\left\{32;-32\right\}\)

g)\(5.|x+9|=40\)

\(\Rightarrow|x+9|=8\)

\(\Rightarrow\orbr{\begin{cases}x+9=8\\x+9=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-17\end{cases}}\)

Vậy \(x\in\left\{-1;-17\right\}\)

h) \(-\frac{5}{6}+\frac{8}{3}+\frac{-29}{6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)

\(\Rightarrow\frac{-5}{6}+\frac{16}{6}+\frac{-29}{6}\le x\le\frac{-1}{2}+\frac{4}{2}+\frac{5}{2}\)

\(\Rightarrow-3\le x\le4\)

Vậy \(-3\le x\le4\)

câu a

x+(-7)=-20

x=-20-(-7)

x=-13

a,(11/15+4/15)+(5/7+2/7)

=1+1

=2

b,5/9x(1/2+6/4)

=5/9x2

=10/9

c,1/2:(7/8+9/8)

=1/2:2

=1

d,(17/10-7/10)+1/2

=1+1/2

=3/2