a: \(\dfrac{1}{8}\cdot\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{1}{10}\cdot\dfrac{5}{3}=\dfrac{1}{2\cdot3}=\dfrac{1}{6}\)

b: \(=\dfrac{8-3}{12}\cdot\dfrac{6}{5}=\dfrac{6}{12}=\dfrac{1}{2}\)

c: \(=\dfrac{24}{35}:\dfrac{32}{35}=\dfrac{3}{4}\)

d: =63/21+15/21-7/21=71/21

a \(\dfrac{1}{8}\times\dfrac{4}{5}\times\dfrac{10}{6}=\dfrac{1}{6}\)

b \(\left(\dfrac{8}{12}-\dfrac{3}{12}\right)\times\dfrac{6}{5}=\dfrac{5}{12}\times\dfrac{6}{5}=\dfrac{1}{2}\)

c \(\dfrac{24}{35}:\dfrac{32}{35}=\dfrac{24}{35}\times\dfrac{35}{32}=\dfrac{3}{4}\)

d \(\dfrac{63}{21}+\dfrac{15}{21}-\dfrac{7}{21}=\dfrac{71}{21}\)

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

\(\Rightarrow A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|\)

\(\Rightarrow A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2\right|=2\)

dấu "="xảy ra khi \(\left(2x-1\right).\left(3-2x\right)\ge0\)

\(\Rightarrow\frac{1}{2}\le x\le\frac{3}{2}\)

vậy min A=2 khi \(\frac{1}{2}\le x\le\frac{3}{2}\)

\(a,\dfrac{6}{7}+\dfrac{3}{10}=\dfrac{60}{70}+\dfrac{21}{70}=\dfrac{81}{70}\\ b,\dfrac{5}{9}+\dfrac{1}{3}=\dfrac{5}{9}+\dfrac{3}{9}=\dfrac{8}{9}\\ c,\dfrac{5}{8}-\dfrac{2}{5}=\dfrac{25}{40}-\dfrac{16}{40}=\dfrac{9}{40}\\ d,\dfrac{1}{4}-\dfrac{1}{7}=\dfrac{7}{28}-\dfrac{4}{28}=\dfrac{3}{28}\)

e: \(\dfrac{17}{5}:x=\dfrac{34}{5}:\dfrac{4}{3}=\dfrac{34}{5}\cdot\dfrac{3}{4}=\dfrac{102}{20}=\dfrac{51}{10}\)

\(\Leftrightarrow x=\dfrac{17}{5}:\dfrac{51}{10}=\dfrac{17}{5}\cdot\dfrac{10}{51}=\dfrac{10}{5}\cdot\dfrac{17}{51}=2\cdot\dfrac{1}{3}=\dfrac{2}{3}\)

f: \(x:\dfrac{4}{5}=\dfrac{25}{8}:\dfrac{5}{4}=\dfrac{25}{8}\cdot\dfrac{4}{5}=\dfrac{100}{40}=\dfrac{5}{2}\)

\(\Leftrightarrow x=\dfrac{5}{2}\cdot\dfrac{4}{5}=\dfrac{4}{2}=2\)

g: \(\left(0.25x+2012\right)\cdot2013=\left(50+2012\right)\cdot2013\)

\(\Leftrightarrow0.25x+2012=50+2012\)

\(\Leftrightarrow0.25x=50\)

hay x=200

h: \(\left(x-\dfrac{1}{2}\right)\cdot\dfrac{5}{3}=\dfrac{7}{4}-1=\dfrac{3}{4}\)

\(\Leftrightarrow x-\dfrac{1}{2}=\dfrac{3}{4}:\dfrac{5}{3}=\dfrac{9}{20}\)

\(\Leftrightarrow x=\dfrac{9}{20}+\dfrac{1}{2}=\dfrac{9}{20}+\dfrac{10}{20}=\dfrac{19}{20}\)

a: \(\dfrac{4}{5}-\dfrac{5}{6}< =\dfrac{x}{30}< =\dfrac{1}{3}-\dfrac{3}{10}\)

=>\(\dfrac{24-25}{30}< =\dfrac{x}{30}< =\dfrac{10-9}{30}\)

=>\(\dfrac{-1}{30}< =\dfrac{x}{30}< =\dfrac{1}{30}\)

=>-1<=x<=1

mà x nguyên

nên \(x\in\left\{-1;0;1\right\}\)

b: \(\dfrac{a}{7}+\dfrac{1}{14}=\dfrac{-1}{b}\)

=>\(\dfrac{2a+1}{14}=\dfrac{-1}{b}\)

=>\(\left(2a+1\right)\cdot b=-14\)

mà 2a+1 lẻ (do a là số nguyên)

nên \(\left(2a+1\right)\cdot b=1\cdot\left(-14\right)=\left(-1\right)\cdot14=7\cdot\left(-2\right)=\left(-7\right)\cdot2\)

=>\(\left(2a+1;b\right)\in\left\{\left(1;-14\right);\left(-1;14\right);\left(7;-2\right);\left(-7;2\right)\right\}\)

=>\(\left(a;b\right)\in\left\{\left(0;-14\right);\left(-1;14\right);\left(3;-2\right);\left(-4;2\right)\right\}\)

.

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

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

Vậy x = \(-\frac{3}{2}\) hoặc x = 3

b)\(\left(\frac{1}{2}-x\right)^2=\frac{64}{49}\)

\(\Rightarrow\left(\frac{1}{2}-x\right)^2=\left(\frac{8}{7}\right)^2\) hoặc \(\left(\frac{1}{2}-x\right)^2=\left(-\frac{8}{7}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}-x=\frac{8}{7}\\\frac{1}{2}-x=-\frac{8}{7}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}-\frac{8}{7}\\x=\frac{1}{2}+\frac{8}{7}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{9}{14}\\x=\frac{23}{14}\end{cases}}\)

Vậy x = \(-\frac{9}{14}\) hoặc x = \(\frac{23}{14}\)

c) \(\frac{1}{2}.\left(x-4,5\right)=\frac{3}{4}.x=\frac{5}{12}\) ( câu này mik ko hiểu cho lắm)

k mik nha mn!

doi mik sua

la cong 3/4

tớ chỉ làm được phần a)thôi

ta có:\(\left(x-\frac{2}{3}\right):\frac{-3}{7}=\frac{-9}{11}\)

\(x-\frac{2}{3}=\frac{-9}{11}.\frac{-3}{7}\)

\(x-\frac{2}{3}=\frac{27}{77}\)

\(x=\frac{27}{77}+\frac{2}{3}\)

\(x=\frac{235}{231}\)

vậy x=\(\frac{235}{231}\)

a) \(\left(x-\frac{2}{3}\right)\):\(\frac{-3}{7}\)=\(\frac{-9}{1}\)

                        \(x-\frac{2}{3}\)=\(-9\cdot\frac{-3}{7}\)

                        \(x-\frac{2}{3}\)= \(\frac{27}{7}\)

                                     \(x\)= \(\frac{27}{7}\)\(+\frac{2}{3}\)

                                      \(x\)=\(\frac{95}{21}\)

b)\(x:2+x:\frac{5}{2}+x:\frac{1}{3}=\frac{-18}{25}\)

      \(x:\left(2+\frac{5}{2}+\frac{1}{3}\right)=\frac{-18}{25}\)

            \(x:\frac{12+15+2}{6}=\frac{-18}{25}\)

                                \(x:\frac{29}{6}=\frac{-18}{25}\)

                                          \(x=\frac{-18}{25}\cdot\frac{29}{6}\)

                                           \(x=\frac{-87}{25}\)

a)3+x=7

x=7-3

x=4

b)-3x+5=-22

-3x=-22-5

-3x=-27

x=-27:(-3)

x=9

d)(-x+3)(2x+4)=0

-x+3=0 hoặc 2x+4=0

x=3 hoặc x=-2

Phần a;b tính dễ dàng nên bạn tự làm nha

c)(-3).|x-1|=-9

=>|x-1|=(-9):(-3)

=>|x-1|=3

=>x-1 =3 hoặc x-1=-3

=>x =3+1 =>x =-3+1

=>x =4 =>x =-2

Vậy \(x\in\left\{4;-2\right\}\)

d)\(\left(-x+3\right).\left(2x+4\right)=0\)

=>-x+3=0 hoặc 2x+4=0

\(\Rightarrow-x=-3\) \(\Rightarrow2x=-4\)

\(\Rightarrow x=3\) \(\Rightarrow x=-4:2\)

\(\Rightarrow x=-2\)

Vậy \(x\in\left\{3;-2\right\}\)

Chúc bn học tốt

Ko cần đâu bn à mk mong bn đấy

a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

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

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

    \(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)

a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0      hay      \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x     = 1         I\(\Leftrightarrow\)\(\frac{-1}{2}\)x     = -5
\(\Leftrightarrow\)  x     = \(\frac{1}{3}\)  I\(\Leftrightarrow\)            x     = 10

b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\)    I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\)          hay     \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{29}{24}\)        I\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{-13}{24}\)
\(\Leftrightarrow\)      x              = \(\frac{29}{12}\)        I\(\Leftrightarrow\)      x              = \(\frac{-13}{12}\)

c) (2x +\(\frac{3}{5}\))² - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))²       = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\)         = \(\frac{3}{5}\)    hay      2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x                    = 0           I \(\Leftrightarrow\)2x           = \(\frac{-6}{5}\)
\(\Leftrightarrow\)   x                    = 0           I \(\Leftrightarrow\) x           = \(\frac{-3}{5}\)

d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)