\(2x=3y\text{⇒}\dfrac{x}{3}=\dfrac{y}{2}\text{⇒}\dfrac{x}{15}=\dfrac{y}{10}\)

\(2y=5z\text{⇒}\dfrac{y}{5}=\dfrac{z}{2}\text{⇒}\dfrac{y}{10}=\dfrac{z}{4}\)

⇒\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{4}\)

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

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{4}=\dfrac{\left|x+y+z\right|}{\left|15+10+4\right|}=\dfrac{29}{29}=1\)

⇒x=15;y=10;z=4

\(2x=3y=5z\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)(Cùng chia cho 30 )

Áp dụng t/c  dãy tsbn \(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

\(\Rightarrow\hept{\begin{cases}x=5.15=75\\y=5.10=50\\z=5.6=30\end{cases}}\)

Ta có 2x=3y=5z

\(\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

\(\frac{x}{15}=5\Rightarrow x=75\)

\(\frac{y}{10}=5\Rightarrow y=50\)

\(\frac{z}{6}=5\Rightarrow z=30\)

Vậy x = 75 ; y = 50 ; z = 30

cảm ơn hai bạn

a)  ( 2 x + 1 ) ( 3 y − 2 ) = − 55

Suy ra  ( 2 x + 1 )   v à   ( 3 y − 2 ) ∈ Ư ( - 55 )   =   1 ;   − 1 ;   5 ;   − 5 ;   11 ;   − 11 ;   55 ;   − 55

Khi đó ta có bảng sau:

b)  ( x − 3 ) ( 2 y + 1 ) = 7

Suy ra  ( x − 3 ) và  ( 2 y + 1 ) ∈ Ư ( 7 )   =   1 ;   − 1 ;  7 ;   − 7

Khi đó ta có bảng sau

c)  y ( y 4 + 12 ) = − 5

Suy ra  ( y 4 + 12 ) ∈ Ư ( - 5 ) =   1 ;   − 1 ;  5 ;   − 5

Vì  y 4 ≥ 0 ⇒ y 4 + 12 ≥ 12 ⇒ không có giá trị của y thỏa mãn ycbt.

a)

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{3x-2y}{3.5-2.2}=\dfrac{-55}{11}=-5\)

=> \(\left\{{}\begin{matrix}x=-5.5=-25\\y=-5.2=-10\end{matrix}\right.\)

b)

\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{2x+5y}{2.3+5.2}=\dfrac{48}{16}=3\)

=> \(\left\{{}\begin{matrix}x=3.3=9\\y=3.2=6\end{matrix}\right.\)

c)

Có: \(\dfrac{x}{y}=-\dfrac{5}{2}\Leftrightarrow-\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x+y}{-5+2}=\dfrac{30}{-3}=-10\)

=> \(\left\{{}\begin{matrix}x=-10.-5=50\\y=-10.2=-20\end{matrix}\right.\)

d)

Có: \(\dfrac{x}{y}=\dfrac{4}{3}\Leftrightarrow\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{2x+3y}{2.4+3.3}=\dfrac{34}{17}=2\)

=> \(\left\{{}\begin{matrix}x=2.4=8\\y=2.3=6\end{matrix}\right.\)

1)

a/ \(12\left(x-5\right)+7\left(3-x\right)=15\)

\(\Rightarrow12x-12.5+7.3-7x=15\)

\(\Rightarrow12x-60+21-7x=15\)

\(\Rightarrow12x-7x=15+60-21\)

\(\Rightarrow5x=54\)

\(\Rightarrow x=\frac{54}{5}=10,8\)

b/ \(30\left(x+2\right)-6\left(x-5\right)-24x=100\)

\(\Rightarrow30x+30.2-6x-6.5-24x=100\)

\(\Rightarrow30x+60-6x+30-24x=100\)

\(\Rightarrow30x-6x-24x=100-60-30\)

\(\Rightarrow0x=10\)

\(\Rightarrow\) k có giá trị \(x\) nào thỏa mãn đề bài

\(12\left(x-5\right)+7\left(3-x\right)=15\)

\(12x-60+21-7x=15\)

\(12x-\left(60-21+7x\right)=15\)

\(12x-\left(39+7x=15\right)\)

\(12x-39-7x=15\)

\(5x=15+39\)

\(5x=54\)

\(x=10,8\)

vì \( (x-3)(2y+1)=7\) nên x-3 và 2y+1 là 2 số nguyên cùng dấu nhau

\(\Rightarrow\left\{\begin{matrix}x-3>0\\2y+1>0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x>3\\2y>1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x\in\left\{4,5,6,7,...\right\}\\y\in\left\{1,2,3,4,...\right\}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x-3< 0\\2y+1< 0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x< 3\\2y< 1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x\in\left\{2,1,0,...\right\}\\y\in\left\{0,-1,...\right\}\end{matrix}\right.\)

a) (x-3) .(2y+1)=7

=> (x-3) và (2y+1) thuộc Ư(7)={-1;1;-7;7}

Xét các trường hợp rồi tìm ra x là đc

(2x+1).(3y-2)=-55

=> (2x+1) và (3y-2) thuộc Ư(-55)={.....}

tự làm , xét ước và các trường hợp là đc

c)-12.(x-5)+7.(3-x)=5

<=>-12x+60+21-7x=5

<=>-19x=5-60-21

<=>-19x=-86

<=> x=86/19=...

a: (x-1)(y+1)=5

nên \(\left(x-1;y+1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(2;4\right);\left(6;0\right);\left(0;-6\right);\left(-4;-2\right)\right\}\)

b: (x-3)(2y+1)=7

\(\Leftrightarrow\left(x-3;2y+1\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(4;3\right);\left(10;0\right);\left(2;-4\right);\left(-4;-1\right)\right\}\)