a).\(\dfrac{x}{15}\)=\(\dfrac{4}{3}\)\(\Leftrightarrow\)x.3=15.4\(\Leftrightarrow\)3x=60\(\Leftrightarrow\)x=20

b)x:\(\dfrac{4}{3}\)=\(\dfrac{5}{6}\)\(\Leftrightarrow\)\(\dfrac{3x}{4}=\dfrac{5}{6}\)\(\Leftrightarrow\)3x.6=4.5\(\Leftrightarrow\)18x=20\(\Leftrightarrow\)\(\dfrac{10}{9}\)

c)\(\dfrac{7}{2}\)-x=\(\dfrac{5}{6}\)\(\Leftrightarrow\)\(\dfrac{7}{2}-\dfrac{2x}{2}\)=\(\dfrac{5}{6}\)\(\Leftrightarrow\)\(\dfrac{7-2x}{2}\)=\(\dfrac{5}{6}\)\(\Leftrightarrow\)(7-2x).6=2.5\(\Leftrightarrow\)42-12x=10

\(\Leftrightarrow\)-12x=-32\(\Leftrightarrow\)x=\(\dfrac{8}{3}\)

Hhigh             

sửa đề : \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

áp dụng t/c dãy t/s = nhau

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{58}{3+4+5}=\frac{58}{12}=\frac{29}{6}\)

\(\frac{x}{3}=\frac{29}{6}\Rightarrow x=\frac{29}{2}\)

\(\frac{y}{4}=\frac{29}{6}\Rightarrow y=\frac{58}{3}\)

\(\frac{z}{5}=\frac{29}{6}\Rightarrow z=\frac{145}{6}\)

vậy ...

chết rồi mình nhìn nhầm đề nhé : phải là :  3+4-5 = 58/2=29

từ đây bạn tìm x;y;z nhé!

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{x+y-z}{3+4-5}=\frac{58}{2}=29\)

\(\Rightarrow\hept{\begin{cases}x=28.3=54\\y=28.4=112\\z=28.5=140\end{cases}}\)

Bài 1: 

b) Ta có: \(D=\dfrac{-5}{10}\cdot\dfrac{-4}{10}\cdot\dfrac{-3}{10}\cdot...\cdot\dfrac{3}{10}\cdot\dfrac{4}{10}\cdot\dfrac{5}{10}\)

\(=\dfrac{-5}{10}\cdot\dfrac{-4}{10}\cdot\dfrac{-3}{10}\cdot...\cdot0\cdot...\cdot\dfrac{3}{10}\cdot\dfrac{4}{10}\cdot\dfrac{5}{10}\)

=0

0,2-0,375+5/11/-0,3+9/16-15/22

`x+y-6=0<=>y=-x+6`

Hàm số `y=(4m-5)x+3m //// y=-x+6`

    `<=>{(4m-5=-1),(3m ne 6):}`

   `<=>{(m=1),(m ne 2):}=>m=1`.

2xy - 6x + y = 15

<=> 2x(y - 3) + y - 3 = 12

<=> (2x + 1)(y - 3) = 12 (1)

Từ (1) \(\Rightarrow2x+1\inƯ\left(12\right)\)

Mà 2x + 1 là số lẻ \(\forall x\inℕ\)

=> \(2x+1\in\left\{1;3\right\}\Leftrightarrow x\in\left\{0;1\right\}\)

với x = 0 => y = 15

với x = 1 => y = 7

Vậy (x;y) = (0;15) ; (1;7) 

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