8: \(=\dfrac{-5}{9}-\dfrac{4}{9}+\dfrac{8}{15}+\dfrac{7}{15}-\dfrac{2}{11}=\dfrac{-2}{11}\)

9: =2/7-2/5+5/7=1-2/5=3/5

10: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{-5}{19}\)

11: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{-5}{7}\)

8 = -2/11

9 = 3/5

10 = -5/19

11 = -5/7

11 = 5/13

a: \(\dfrac{3}{8}+\dfrac{7}{8}=\dfrac{3+7}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)

b: \(\dfrac{7}{9}-\dfrac{4}{9}=\dfrac{7-4}{9}=\dfrac{3}{9}=\dfrac{1}{3}\)

c: \(\dfrac{5}{6}+\dfrac{1}{8}=\dfrac{20}{24}+\dfrac{3}{24}=\dfrac{20+3}{24}=\dfrac{23}{24}\)

d: \(\dfrac{9}{15}-\dfrac{2}{5}=\dfrac{3}{5}-\dfrac{2}{5}=\dfrac{3-2}{5}=\dfrac{1}{5}\)

e: \(\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{9}{5}=\dfrac{2+3+9}{5}=\dfrac{14}{5}\)

g: \(\dfrac{8}{10}-\dfrac{1}{10}-\dfrac{3}{10}=\dfrac{8-1-3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

h: \(\dfrac{23}{7}-\dfrac{4}{7}+\dfrac{2}{7}=\dfrac{23-4+2}{7}=\dfrac{21}{7}=3\)

a) ĐK xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{6}{y}=9\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{7}{x}=16\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{7}{16}\\y=-\dfrac{42}{17}\end{matrix}\right.\)

Vậy S = {(\(\dfrac{7}{16};-\dfrac{42}{17}\))}

b) Đk xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{1}{y}=14\\\dfrac{8}{x}-\dfrac{1}{y}=-8\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{13}{x}=6\\\dfrac{5}{x}+\dfrac{1}{y}=14\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{13}{6}\\y=\dfrac{13}{152}\end{matrix}\right.\)

Vậy S={(\(\dfrac{13}{6};\dfrac{13}{152}\))}

c) ĐK xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{7}{y}=21\\-\dfrac{2}{x}-\dfrac{5}{y}=-11\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{2}{y}=10\\\dfrac{2}{x}+\dfrac{7}{y}=21\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{1}{5}\\x=-\dfrac{1}{7}\end{matrix}\right.\)

Vậy S={(\(-\dfrac{1}{7};\dfrac{1}{5}\))}

d) ĐK xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{9}{x}+\dfrac{2}{y}=22\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{14}{x}=35\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)

Vậy S={(0,4;-4)}

e) ĐKXĐ : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{5}{y}=10\\-\dfrac{3}{x}-\dfrac{7}{y}=8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-\dfrac{2}{y}=18\\\dfrac{3}{x}+\dfrac{5}{y}=10\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{1}{9}\\x=\dfrac{3}{55}\end{matrix}\right.\) 'Vậy....

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

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

= - \(\dfrac{47}{33}\)

Tham khảo:

1) \(\dfrac{5}{6}-\dfrac{6}{7}+\dfrac{7}{8}-\dfrac{8}{9}+\dfrac{10}{9}-\dfrac{5}{6}+\dfrac{6}{7}-\dfrac{7}{8}+\dfrac{8}{9}\)

\(=\left(\dfrac{5}{6}-\dfrac{5}{6}\right)-\left(\dfrac{6}{7}+\dfrac{6}{7}\right)+\left(\dfrac{7}{8}-\dfrac{7}{8}\right)-\left(\dfrac{8}{9}+\dfrac{8}{9}\right)+\dfrac{10}{9}\)

\(=0-0+0-0+\dfrac{10}{9}\)

\(=\dfrac{10}{9}\)

2) \(\dfrac{1}{13}+\dfrac{16}{7}+\dfrac{3}{105}-\dfrac{9}{7}-\dfrac{-12}{13}\)

\(=\left(\dfrac{1}{13}-\left(-\dfrac{12}{13}\right)\right)+\left(\dfrac{16}{7}-\dfrac{9}{7}\right)+\dfrac{3}{105}\)

\(=1+1+\dfrac{3}{105}\)

\(=\dfrac{213}{105}=\dfrac{71}{35}\)

\(\dfrac{3}{7}+\dfrac{4}{1}=\dfrac{3}{7}+\dfrac{28}{7}=\dfrac{31}{7}\)

\(\dfrac{5}{9}\times3=\dfrac{5}{9}\times\dfrac{3}{1}=\dfrac{15}{9}=\dfrac{5}{3}\)

\(\dfrac{7}{8}-\dfrac{2}{9}=\dfrac{63}{72}-\dfrac{16}{72}=\dfrac{47}{72}\)

\(\dfrac{5}{1}-\dfrac{3}{4}=\dfrac{20}{4}-\dfrac{3}{4}=\dfrac{17}{4}\)

\(\dfrac{5}{7}:\dfrac{6}{1}=\dfrac{5}{7}\times\dfrac{1}{6}=\dfrac{5}{42}\)

\(\dfrac{5}{6}\times\dfrac{2}{7}=\dfrac{5\times2}{6\times7}=\dfrac{10}{42}=\dfrac{5}{21}\)

\(\dfrac{3}{1}+\dfrac{3}{4}=\dfrac{12}{4}+\dfrac{3}{4}=\dfrac{15}{4}\)

\(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{3}{5}\times\dfrac{1}{2}=\dfrac{3\times1}{5\times2}=\dfrac{3}{10}\)

\(\dfrac{5}{1}:\dfrac{3}{8}=\dfrac{5}{1}\times\dfrac{8}{3}=\dfrac{40}{3}\)

a) \(\dfrac{3}{7}+4=\dfrac{3}{7}+\dfrac{4}{1}=\dfrac{3}{7}+\dfrac{28}{7}=\dfrac{31}{7}\)

b) \(\dfrac{5}{9}\times3=\dfrac{5\times3}{9}=\dfrac{15}{9}=\dfrac{5}{3}\)

c) \(\dfrac{7}{8}-\dfrac{2}{9}=\dfrac{63}{72}-\dfrac{16}{72}=\dfrac{47}{72}\)

d) \(5-\dfrac{3}{4}=\dfrac{5}{1}-\dfrac{3}{4}=\dfrac{20}{4}-\dfrac{3}{4}=\dfrac{17}{4}\)

e) \(\dfrac{5}{7}:6=\dfrac{5}{7}:\dfrac{6}{1}=\dfrac{5}{7}\times\dfrac{1}{6}=\dfrac{5}{42}\)

f) \(\dfrac{5}{6}\times\dfrac{2}{7}=\dfrac{5\times2}{6\times7}=\dfrac{10}{42}=\dfrac{5}{21}\)

g) \(3+\dfrac{3}{4}=\dfrac{3}{1}+\dfrac{3}{4}=\dfrac{12}{4}+\dfrac{3}{4}=\dfrac{15}{4}\)

h) \(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{3\times4}{5\times8}=\dfrac{12}{40}=\dfrac{3}{10}\)

i) \(5:\dfrac{3}{8}=\dfrac{5}{1}:\dfrac{3}{8}=\dfrac{5}{1}\times\dfrac{8}{3}=\dfrac{40}{3}\)

a) \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\) (1)

\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{z}{3}=\dfrac{y}{7}\) (2)

Từ (1) và (2) suy ra: \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y-z}{9-7-3}=\dfrac{-15}{-1}=15\)

\(\Rightarrow\left\{{}\begin{matrix}x=15.9\\y=15.7\\z=15.3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=135\\y=105\\z=45\end{matrix}\right.\)

Vậy, x = 135, y = 105, z = 45

b, \(\dfrac{x}{-3}=\dfrac{y}{-8}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{64}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x^2}{9}=\dfrac{y^2}{64}=\dfrac{x^2-y^2}{9-64}=-\dfrac{44}{\dfrac{5}{-55}}=-\dfrac{44}{5}:\left(-55\right)=-\dfrac{44}{5}.-\dfrac{1}{55}=\dfrac{44}{275}=0,16\)

+) \(\dfrac{x^2}{9}=0,16\Rightarrow x^2=1,44\Rightarrow x=\pm1,2\)

+) \(\dfrac{y^2}{64}=0,16\Rightarrow y^2=10,24\Rightarrow y=\pm3,2\)

Vậy ...

a) \(\dfrac{8}{9}x=\dfrac{2}{7}-\dfrac{2}{3}=-\dfrac{8}{21}\)

\(x=-\dfrac{8}{21}:\dfrac{8}{9}=-\dfrac{3}{7}\)

b) \(\dfrac{2}{5}x=\dfrac{2}{5}-\dfrac{2}{5}=0\)

\(x=0:\dfrac{2}{5}=0\)

c)\(\dfrac{7}{8}x=\dfrac{2}{9}-\dfrac{1}{3}=-\dfrac{1}{9}\)

\(x=-\dfrac{1}{9}:\dfrac{7}{8}=-\dfrac{8}{63}\)

a) 2/7 - 8/9 . x = 2/3

⇒ 8/9 . x = 2/7 - 2/3 

⇒ 8/9 .x = -8/21

⇒ x = -8/21 : 8/9

⇒ x = -3/7.

Vậy...

b) 2/5 - 2/5x = 2/5

⇒ 2/5x = 0

⇒ x = 0.

Vậy...