a) Thay x=6-y

ta có: \(\frac{6-y+3}{y+5}=\frac{3-y}{y+5}=\frac{3}{5}\)

\(\Rightarrow\left(3-y\right).5=\left(y+5\right).3\)

15 - 5y= 3y + 15

-5y-3y=15-15

-8y=0

y=0

Vậy x=6

b) Thay x= y-4

ta có: \(\frac{y-4-7}{y-6}=\frac{y-11}{y-6}=\frac{7}{6}\)

\(\Rightarrow\left(y-11\right).6=\left(y-6\right).7\)

6y-66 = 7y -42

-y=24

y=-24

Vậy x=-4+-24=-28

\(\frac{x+3}{y+5}=\frac{3}{5}\\ \Leftrightarrow5\left(x+3\right)=3\left(y+5\right)\\ \Leftrightarrow5x+15=3y+15\\ \Leftrightarrow5x=3y\\ M\text{à};x+y=6\Rightarrow x=6-y\\ \Rightarrow5\left(6-y\right)=3y\\ \Leftrightarrow30-5y=3y\\ \Rightarrow30=8y\\ \Rightarrow y=\frac{30}{8}\\ \)

\(\frac{x-7}{y-6}=\frac{7}{6}\\ \Leftrightarrow6\left(x-7\right)=7\left(y-6\right)\\ \Leftrightarrow6x-42=7y-42\\ \Leftrightarrow6x=7y\\ M\text{à}x-y=-4;\Rightarrow x=-4+y\\ \Rightarrow6\left(-4+y\right)=7y\\ \Rightarrow-24+6y=7y\\ \Rightarrow y=-24\\ \)

Từ y bạn từ tìm x nhé!!!

bài này lớp 7 à,mình mới học lớp 6 đã có bài này rùi .???

\(\frac{x-5}{y-4}=\frac{5}{4}\)

1. \(\frac{x-5}{y-4}\) = \(\frac{5}{4}\)

=> ( x - 5 )4 = ( y - 4 )5

     4x - 20   = 5y - 20

     4x          = 5y - 20 + 20

     4x          = 5y (1)

Theo bài ra , ta có x - y = 6 nên x = y + 6 (2)

Thay (2) vào (1) , có 4x = 5y <=> 4( y + 6 ) = 5y <=> 4y + 24 = 5y

=> 24 = 5y - 4y => 5y - 4y = 24 => y = 24

Thay y = 24 vào (2) ta đc : x = 24 + 6 = 30

Vậy \(\frac{x}{y}\) = \(\frac{30}{24}\) = \(\frac{5}{4}\)

b/ Có \(\dfrac{x-7}{y-6}=\dfrac{7}{6}\)

nên \(6.\left(x-7\right)=7.\left(y-6\right)\)

\(\rightarrow\) \(6.x-6.7=7.y-7.6\)

\(\Rightarrow\) \(6x=7y\). Mà \(x-y=-4\) nên \(6x-6y=-24\)

\(\rightarrow\) \(7y-6x=-24\)

\(\rightarrow1y=-24\)

Và \(x-y=-4\) \(\Rightarrow\) \(x=\left(-4\right)+y\) \(=\left(-4\right)+\left(-24\right)\)\(=-28\)

Vậy \(x=-28\) \(;\) \(y=-24\)

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 \(y=8\).

Cách 2:

+) Ta có: \(\frac{x-5}{y-4}=\frac{5}{4}\)

\(\Rightarrow4\left(x-5\right)=5\left(y-4\right)\)

\(\Rightarrow4x-20=5y-20\)

\(\Rightarrow4x=5y\)

\(\Rightarrow\frac{x}{5}=\frac{y}{4}\) (1)

+) \(x-y=6\Rightarrow x=6+y\) (2)

Thay (2) vào (1) ta có: \(\frac{y+6}{5}=\frac{y}{4}\)

\(\Rightarrow4\left(y+6\right)=5y\)

\(\Rightarrow4y+24=5y\)

\(\Rightarrow y=24\)

\(\Rightarrow x=30\)

Vậy \(y=24;x=30\)

\(\frac{x-5}{y-4}=\frac{5}{4}\\ \Rightarrow4\left(x-5\right)=5\left(y-4\right)\\ \Rightarrow4x-20=5y-20\\ \Rightarrow4x=5y\\ \Rightarrow\frac{x}{5}=\frac{y}{4}\\ \)

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

\(\frac{x}{5}=\frac{y}{4}=\frac{x-y}{5-4}=\frac{6}{1}=6\\ \Rightarrow x=30;y=24\)

Chúc bạn học tốt !!!

 Mình không biết.

(\(\frac45:\frac65\) + \(\frac15\) : \(\frac{1}{x}\)) x 30 - 26 = 54

(\(\frac45\times\frac56\) + \(\frac15\times x\)) x 30 = 54 + 26

(\(\frac23+\frac15x\)) x 30 = 80

\(\frac23+\frac15x\) = 80 : 30

\(\frac23+\frac15x=\frac83\)

\(\frac15x=\frac83-\frac23\)

\(\frac15x\) = 2

\(x=2:\frac15\)

\(x=2\times5\)

\(x=10\)

Vậy \(x=10\)

6,3 x y + 3,7 x y = 100

y x (6,3 + 3,7) = 100

y x 10 = 100

y = 100 : 10

y = 10

Vậy y = 10

a) Ta có: \(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{y}{3}\)

\(\Rightarrow\dfrac{4}{x}=\dfrac{5-2y}{6}\)

\(\Rightarrow\left(5-2y\right)x=24\)

Vì \(x,y\in Z\Rightarrow\left[{}\begin{matrix}5-2y\in Z\\x\in Z\end{matrix}\right.\)

\(\Rightarrow5-2y\inƯ\left(24\right);x\inƯ\left(24\right)\)

Tự lập bảng xét các giá trị của \(x,y\) nhé.

b) Lại có: \(\dfrac{5}{x}-\dfrac{y}{3}=\dfrac{1}{6}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{6}+\dfrac{y}{3}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1+2y}{6}\)

\(\Rightarrow\left(1+2y\right)x=30\)

Lí luận rồi lập bảng như câu \(a\)).

c) \(\dfrac{x}{6}-\dfrac{2}{y}=\dfrac{1}{30}\)

\(\Rightarrow\dfrac{2}{y}=\dfrac{x}{6}-\dfrac{1}{30}\)

\(\Rightarrow\dfrac{2}{y}=\dfrac{5x-1}{30}\)

\(\Rightarrow\left(5x-1\right)y=60\)

\(......Tương\) \(tự\) \(như\) \(câu\) \(a\))\(b\)).

thank you