\(\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2x+6}{x-1}+\dfrac{3y+14}{y+3}=18\end{matrix}\right.\left(x\ne1;y\ne-3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2x-2+8}{x-1}+\dfrac{3y+9+5}{y+3}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2\left(x-1\right)}{x-1}+\dfrac{8}{x-1}+\dfrac{3\left(y+3\right)}{y+3}+\dfrac{5}{y+3}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\2+\dfrac{8}{x-1}+3+\dfrac{5}{y+3}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{8}{x-1}+\dfrac{5}{y+3}=13\end{matrix}\right.\) (I) 

Đặt: \(\left\{{}\begin{matrix}u=\dfrac{1}{x-1}\\v=\dfrac{1}{y+3}\end{matrix}\right.\)

Hệ (I) trở thành: 

\(\Leftrightarrow\left\{{}\begin{matrix}12u+7v=19\\8u+5v=13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}24u+14v=38\\24u+15v=39\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}12u+7=19\\v=1\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}12u=12\\v=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u=1\\v=1\end{matrix}\right.\) 

Trả ẩn phụ: 

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}=1\\\dfrac{1}{y+3}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y+3=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\left(tm\right)\)

Vậy hệ pt có 1 cặp nghiệm duy nhất là: (2;-2) 

⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=192x−2+8x−1+3y+9+5y+3=18⇔{12�−1+7�+3=192�−2+8�−1+3�+9+5�+3=18

⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=192+8x−1+3+5y+3=18⇔{12�−1+7�+3=192+8�−1+3+5�+3=18

⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩u=1x−1v=1y+3{�=1�−1�=1�+3

Hệ (I) trở thành: 

⇔{12u+7v=198u+5v=13⇔{12�+7�=198�+5�=13

⇔{24u+14v=3824u+15v=39⇔{24�+14�=3824�+15�=39

⇔{12u+7=19v=1⇔{12�+7=19�=1 

⇔{12u=12v=1⇔{12�=12�=1

⇔{u=1v=1⇔{�=1�=1 

Trả ẩn phụ: 

Bài 2:

\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)

\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)

\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)

\(a,\dfrac{1}{7}+\dfrac{1}{3}=\dfrac{3}{21}+\dfrac{7}{21}=\dfrac{10}{21}\\ b,\dfrac{-2}{3}+\dfrac{-5}{7}=\dfrac{-14+\left(-15\right)}{21}=\dfrac{-29}{21}\\ c,\dfrac{1}{2}-\dfrac{-1}{2}-\dfrac{1}{3}=\dfrac{3+3-2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

\(a.\dfrac{4}{28}+\dfrac{12}{36}=\dfrac{1}{7}+\dfrac{1}{3}=\dfrac{3}{21}+\dfrac{7}{21}=\dfrac{10}{21}\\ b.\dfrac{-12}{18}+\dfrac{-15}{21}=\dfrac{-2}{3}+\dfrac{-5}{7}=\dfrac{-14}{21}+\dfrac{-15}{21}=\dfrac{-29}{21}\\ c.\dfrac{14}{28}+\dfrac{16}{32}-\dfrac{17}{51}=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{17}{51}=1-\dfrac{17}{51}=\dfrac{2}{3}\)

\(\dfrac{4}{28}+\dfrac{12}{36}=\dfrac{144}{1008}+\dfrac{336}{1008}=\\ \dfrac{480}{1008}=\dfrac{10}{21}\)

Câu b và c em làm tương tự.

a) 9/18 - (-7/12) + 13/32

= 13/12 + 13/32

= 143/96

b) (5/-8) + 14/25 - 6/10

= (-13/200) - 6/10

= -133/200

Chúc bạn học tốt!! ^^

a: \(\dfrac{9}{18}-\dfrac{-7}{12}+\dfrac{13}{32}\)

\(=\dfrac{1}{2}+\dfrac{7}{12}+\dfrac{13}{32}\)

\(=\dfrac{48}{96}+\dfrac{56}{96}+\dfrac{39}{96}\)

\(=\dfrac{143}{96}\)

b: \(\dfrac{-5}{8}+\dfrac{14}{25}-\dfrac{6}{10}\)

\(=\dfrac{-125}{200}+\dfrac{112}{200}-\dfrac{120}{200}\)

\(=\dfrac{-133}{200}\)

a: =>x-5/14=6/14-1/14=5/14

=>x=10/14=5/7

b; =>2/9:x=3/6+4/6=7/6

=>x=2/9:7/6=2/9*6/7=4/21

c: =>32:x=8

=>x=4

a

\(x-\dfrac{5}{14}=\dfrac{3}{7}-\dfrac{1}{14}\\ x-\dfrac{5}{14}-\dfrac{3}{7}+\dfrac{1}{14}=0\\ x+\dfrac{1}{14}-\dfrac{5}{14}-\dfrac{6}{14}=0\\ x+\dfrac{1-5-6}{14}=0\\ x-\dfrac{5}{7}=0\\ x=0+\dfrac{5}{7}\\ x=\dfrac{5}{7}\)

b

\(\dfrac{2}{9}:x=\dfrac{1}{2}+\dfrac{2}{3}\\ \dfrac{2}{9}:x=\dfrac{3}{6}+\dfrac{4}{6}\\ \dfrac{2}{9}:x=\dfrac{3+4}{6}=\dfrac{7}{6}\\ x=\dfrac{2}{9}:\dfrac{7}{6}\\ x=\dfrac{2}{9}\times\dfrac{6}{7}=\dfrac{2.3.2}{3.3.7}=\dfrac{4}{21}\)

c

\(\dfrac{6}{32:x}=\dfrac{12}{16}\\ 32:x=6:\dfrac{12}{16}\\ 32:x=6\times\dfrac{16}{12}\\ 32:x=\dfrac{3\times2\times4\times4}{3\times4}\\ 32:x=8\\ x=\dfrac{32}{8}\\ x=4\)

`@` `\text {Answer}`

`\downarrow`

`a,`

`x - 5/14 = 3/7 - 1/14`

`x - 5/14 = 5/14`

`=> x = 5/14 + 5/14`

`=> x = 5/7`

Vậy, `x = 5/7`

`b,`

`2/9 \div x = 1/2 + 2/3`

`2/9 \div x = 7/6`

`x = 2/9 \div 7/6`

`x = 4/21`

Vậy, `x = 4/21`

`c,`

\(\dfrac{6}{32\div x}=\dfrac{12}{16}\)

`6/(32 \div x) = 3/4`

`32 \div x = 6 \div 3/4`

`32 \div x = 8`

` x = 32 \div 8`

`x = 4`

Vậy, `x = 4`

\(\dfrac{14}{7}>\dfrac{9}{7}\)

\(\dfrac{6}{13}< 1\)

\(\dfrac{7}{9}=\dfrac{14}{18}\)

\(\dfrac{4}{9}=\dfrac{20}{45}< \dfrac{26}{45}\)

\(\dfrac{11}{24}>\dfrac{9}{24}=\dfrac{3}{8}\)

\(\dfrac{42}{91}=\dfrac{42:7}{91:7}=\dfrac{6}{13}< \dfrac{7}{13}\)

\(\dfrac{34}{64}< \dfrac{52}{64}=\dfrac{13}{16}\)

\(\dfrac{17}{30}>\dfrac{16}{30}=\dfrac{8}{15}\)

a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)

\(\Rightarrow-4< x< \dfrac{-3}{10}\)

\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)

\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)

b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)

\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)

\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)

\(\Rightarrow x=\varnothing\)

c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)

\(\Rightarrow x\in\left\{1;2\right\}\)

+) Với \(x=1\)

\(\Rightarrow y\in\left\{1;2\right\}\)

+) Với \(x=2\)

\(\Rightarrow y=2\)

Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).

Tham khảo!

\(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)

\(\dfrac{-7}{6}=\dfrac{x}{18}\Leftrightarrow6.x=\left(-7\right).18\Rightarrow x=\dfrac{\left(-7\right).18}{6}=-21\)

\(\dfrac{-7}{6}=\dfrac{-98}{y}\Leftrightarrow\left(-7\right).y=6.\left(-98\right)\Rightarrow y=\dfrac{6.\left(-98\right)}{-7}=84\)

\(\dfrac{-7}{6}=\dfrac{-14}{z}\Leftrightarrow\left(-7\right).z=6.\left(-14\right)\Rightarrow z=\dfrac{6.\left(-14\right)}{-7}=12\)

\(\dfrac{-7}{6}=\dfrac{t}{102}\Leftrightarrow6.t=\left(-7\right).102\Rightarrow t=\dfrac{\left(-7\right).102}{6}=-119\)

\(\dfrac{-7}{6}=\dfrac{u}{-78}\Leftrightarrow6.u=\left(-7\right).\left(-78\right)\Rightarrow u=\dfrac{\left(-7\right).\left(-78\right)}{6}=91\)

\(\text{Vậy }x=-21;y=84;y=84;z=12;t=-119;u=91\)

a, \(\dfrac{9}{18}-\dfrac{-7}{12}+\dfrac{13}{32}\)

\(=\dfrac{1}{2}+\dfrac{7}{12}+\dfrac{13}{32}\)

\(=\dfrac{13}{12}+\dfrac{13}{32}=\dfrac{143}{96}\)

b, \(\dfrac{5}{-8}+\dfrac{14}{39}-\dfrac{6}{10}\)

\(\dfrac{-5}{8}+\dfrac{14}{39}-\dfrac{3}{5}\)

\(=\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{14}{39}\)

\(=\dfrac{-49}{40}+\dfrac{14}{39}=\dfrac{-1351}{1560}\)