a)

\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)

b)

\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)

c)

\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

d)

\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

e)

\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

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

\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)

f)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)

a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh

a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)

\(x-\dfrac{23}{14}=1\)

\(x=1+\dfrac{23}{14}\)

\(x=\dfrac{37}{14}\)

b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)

\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)

\(x+\dfrac{9}{5}=\dfrac{11}{3}\)

\(x=\dfrac{11}{3}-\dfrac{9}{5}\)

\(x=\dfrac{28}{15}\)

a \(\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)

b \(\dfrac{22}{77}-\dfrac{14}{77}=\dfrac{8}{77}\)

c \(\dfrac{11}{13}\times\dfrac{26}{31}=11\times\dfrac{2}{31}=\dfrac{22}{31}\)

d \(\dfrac{1}{2}\times3\times\dfrac{2}{5}=\dfrac{3}{5}\)

a, 2/5 + 11/15 = 6/15 + 11/15 = 17/15

b,2/7 - 2/11 = 22/77 - 14/77 = 8/77

c, 11/13 x 26/31= 11 x 2/31 =22/31

d, 1/2 : 1/3 x 2/5

=  ( 1/2 x 2/5 ) : 1/3

=     1/5        : 1/3

=         3/5

a) Ta có: \(7\cdot\dfrac{3}{14}-\dfrac{1}{14}\)

\(=\dfrac{3}{2}-\dfrac{1}{14}\)

\(=\dfrac{21}{14}-\dfrac{1}{14}\)

\(=\dfrac{10}{7}\)

b) Ta có: \(\dfrac{3}{2}+\dfrac{7}{4}:\dfrac{5}{2}\)

\(=\dfrac{3}{2}+\dfrac{7}{4}\cdot\dfrac{2}{5}\)

\(=\dfrac{3}{2}+\dfrac{7}{10}\)

\(=\dfrac{15}{10}+\dfrac{7}{10}=\dfrac{22}{10}=\dfrac{11}{5}\)

Lời giải:
\(7\times \frac{3}{14}-\frac{1}{14}=\frac{7\times 3}{14}-\frac{1}{14}=\frac{21}{14}-\frac{1}{14}=\frac{21-1}{14}=\frac{20}{14}=\frac{2\times 10}{2\times 7}=\frac{10}{7}\)

\(\frac{3}{2}+\frac{7}{4}:\frac{5}{2}=\frac{3}{2}+\frac{7}{4}\times \frac{2}{5}=\frac{3}{2}+\frac{7\times 2}{4\times 5}=\frac{3}{2}+\frac{7\times 2}{2\times 2\times 5}\)

\(=\frac{3}{2}+\frac{7}{2\times 5}=\frac{3\times 5}{2\times 5}+\frac{7}{2\times 5}=\frac{3\times 5+7}{2\times 5}=\frac{22}{2\times 5}=\frac{2\times 11}{2\times 5}=\frac{11}{5}\)

a. 7 x \(\dfrac{3}{14}\) - \(\dfrac{1}{14}\)

= \(\dfrac{7}{1}\) x \(\dfrac{3}{14}\) - \(\dfrac{1}{14}\)

= \(\dfrac{98}{14}\) x \(\dfrac{3}{14}\) - \(\dfrac{1}{14}\)

= \(\dfrac{3}{2}\) - \(\dfrac{1}{14}\)

= \(\dfrac{21}{14}\) - \(\dfrac{1}{14}\)

= \(\dfrac{10}{7}\)

b. \(\dfrac{3}{2}\) + \(\dfrac{7}{4}\) : \(\dfrac{5}{2}\)

= \(\dfrac{6}{4}\) + \(\dfrac{7}{4}\) : \(\dfrac{5}{2}\)

= \(\dfrac{13}{4}\) : \(\dfrac{5}{2}\)

= \(\dfrac{13}{4}\) . \(\dfrac{2}{5}\)

= \(\dfrac{13}{10}\)

chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D

\(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\left|1\dfrac{1}{8}-\dfrac{5}{12}\right|\)

\(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\dfrac{17}{24}\)

\(x=3-1\dfrac{1}{2}+\dfrac{5}{4}-2+\dfrac{17}{24}\)

\(x=\dfrac{35}{24}\)

bn làm từng bước nhé :D

\(\dfrac{-4}{x}=\dfrac{x}{-49}\\ \Rightarrow x^2=\left(-4\right)\left(-49\right)\\ \Rightarrow x^2=196\\ \Rightarrow x=\pm14\)

\(\dfrac{3.6}{x-3}=\dfrac{5}{3}\\ \Rightarrow5\left(x-3\right)=3.3.6\\ \Rightarrow5\left(x-3\right)=54\\ \Rightarrow x-3=\dfrac{54}{5}\\ \Rightarrow x=\dfrac{54}{5}+3\\ \Rightarrow x=\dfrac{69}{15}\)

\(\left(2x+1\right):2=12:3\\ \left(2x+1\right):2=4\\2x+1=2\\ 2x=1\\ x=\dfrac{1}{2} \)

\(\left(2x-14\right):3=12:9\\ \left(2x-14\right):3=\dfrac{4}{3}\\ 2x-14=4\\ 2x=16\\ x=8\)

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)

\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)

= \(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)

= \(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)

= \(\dfrac{40}{14}=\dfrac{20}{7}\)

\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)

=\(\dfrac{9}{2}+\dfrac{1}{11}\)

=\(\dfrac{101}{22}\)

\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)

\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)

\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)

\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)

\(x=\dfrac{102}{21}=\dfrac{34}{7}\)