\(=\dfrac{9}{7}\cdot\dfrac{7}{9}=1\)

\(\dfrac{9}{8}x\dfrac{8}{7}x\dfrac{7}{3}x\dfrac{1}{3}=\dfrac{9x8x7x1}{8x7x3x3}=1\)

\(=\dfrac{9x8}{8x7}x\)\(\dfrac{7}{3.3}\) 
= \(\dfrac{9}{7}.\)\(\dfrac{7}{9}=1\)

 60x [7/12+4/15]

60x153/180

=9180/180

b 1/2x2/3x3/4x4/5x5/6x6/7x7/8x8/9=40320/4032

a) ĐKXĐ: \(x\ne1\)

Ta có: \(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)

\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)

\(\Leftrightarrow21x-9=2x-2\)

\(\Leftrightarrow21x-2x=-2+9\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\dfrac{7}{19}\)

Vậy: \(S=\left\{\dfrac{7}{19}\right\}\)

`2 : x = x : 8/49`

`<=> x^2 = 16/49`

`<=> x = +-4/7`

a) \(\Leftrightarrow\dfrac{3}{2}:x=\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{3}{2}:\dfrac{1}{2}\\ \Leftrightarrow x=3\)

b) \(\Leftrightarrow x=\dfrac{7}{9}-\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)

c) \(\Leftrightarrow x=\dfrac{8}{7}:\dfrac{6}{7}\\ \Leftrightarrow x=\dfrac{4}{3}\)

d) \(\Leftrightarrow x=\dfrac{9}{5}-\dfrac{3}{7}\\ \Leftrightarrow x=\dfrac{48}{35}\)

a) x = 3

b) x = \(\dfrac{1}{9}\)

c) x = \(\dfrac{4}{3}\)

d)\(\dfrac{48}{35}\)

a) ĐKXD: x ≠ 2

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

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

\(\Leftrightarrow\dfrac{1-3+x}{x-2}=-3\)

\(\Leftrightarrow\dfrac{-2+x}{x-2}=-3\)

\(\Leftrightarrow-2+x=-3\left(x-2\right)\)

\(\Leftrightarrow-2+x=-3x+6\)

\(\Leftrightarrow x+3x=6+2\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\) (loại vì không thỏa mãn điều kiện)

Vậy S = ∅

b) ĐKXĐ: x ≠ 7

 \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

\(\Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{1}{x-7}=8\)

\(\Leftrightarrow\dfrac{7-x}{x-7}=8\)

\(\Leftrightarrow-1=8\left(vô-lý\right)\)

Vậy S = ∅ 

P/s: Ko chắc ạ! 

c) ĐKXĐ: x ≠ 1

\(\dfrac{1}{x-1}+\dfrac{2x}{x^2+x+1}=\dfrac{3x^2}{x^3-1}\)

Quy đồng và khử mẫu ta được:

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

\(\Leftrightarrow x^2+x+1+2x^2-2x-3x^2=0\)

\(\Leftrightarrow-x+1=0\)

\(\Leftrightarrow x=1\) (loại vì ko t/m đk)

Vậy S = ∅

\(a,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \Rightarrow\dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{26}{21}\\ \Rightarrow\dfrac{5}{2}-x=\dfrac{46}{105}\\ \Rightarrow x=\dfrac{433}{210}\\ b,-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow2x-\dfrac{4}{7}-x=\dfrac{3}{5}\\ \Rightarrow2x-x=\dfrac{41}{35}\\ \Rightarrow x=\dfrac{41}{35}\\ c,\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{40}+\dfrac{5}{8}-x=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{5}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{3}{5}.\)

a: x=38/21

a.\(x-\dfrac{2}{3}=\dfrac{8}{7}\)
\(x=\dfrac{8}{7}+\dfrac{2}{3}\)
x=\(\dfrac{38}{21}\)
b.\(\left(x+\dfrac{1}{3}\right)=\dfrac{4}{25} \)
x=\(\dfrac{4}{25}-\dfrac{1}{3}\)
x=\(-\dfrac{13}{75}\)
c.\(-\dfrac{2}{3}:x+\dfrac{5}{8}=-\dfrac{7}{12}\)
\(-\dfrac{2}{3}:x=-\dfrac{29}{24}\)
x=\(\dfrac{16}{29}\)

ai giúp mik ik T_T

a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)

= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))

= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))

= \(\dfrac{1}{13}\) \(\times\)  \(\dfrac{317}{73}\)

= \(\dfrac{317}{949}\)

b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)

=   \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)

= \(\dfrac{199}{45}\)

c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)

=   \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)

= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)

= \(\dfrac{20696}{315}\)

d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)

= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)

= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))

= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)

= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)

e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))

= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)

=  -1 - \(\dfrac{2}{11}\)  - \(\dfrac{8}{33}\)

= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) -  \(\dfrac{8}{33}\)

= - \(\dfrac{47}{33}\)