\(\dfrac{x-1}{9}+\dfrac{1}{3}=\dfrac{1}{y+2}\)

\(\dfrac{x-1}{9}+\dfrac{3}{9}=\dfrac{1}{y+2}\)

\(\dfrac{x-1+3}{9}=\dfrac{1}{y+2}\)

\(\dfrac{x-\left(1-3\right)}{9}=\dfrac{1}{y+2}\)

\(\dfrac{x-\left(-2\right)}{9}=\dfrac{1}{y+2}\)

\(\dfrac{x+2}{9}=\dfrac{1}{y+2}\)

\(\left(x+2\right)\left(y+2\right)=9\)

=> (X+2) ; (y+2) ϵ Ư(9)

TH1: x+2 = 1 => x = -1

y+2=9 => y = 7

TH2: x+2 = 9 => x = 7

=> y +2 = 1 => y =-1

TH3:x+2 = -9 => x = -11

y+2 = -1 => y=-3

TH4: x+2 = -1 => x =-3

y+2 = -9 => x=-11

TH5: x+2 = -3 => x =-5

y+2 = -3 => y=-5

TH6: x+2 =3 =>  x = 1

y+2=3 => y=1

Ta có: \(x+y+z=18\)

\(\dfrac{x+1}{3}=\dfrac{y+2}{5}=\dfrac{z+3}{5}\)

\(\Rightarrow\dfrac{x+1}{3}=\dfrac{y+2}{5}=\dfrac{z+3}{5}and=\dfrac{\left(y+z\right)+\left(2+3\right)}{5}+\dfrac{\left(x+1\right)}{3}\)

\(\Leftrightarrow\dfrac{5+\left(y+z\right)}{5}+\dfrac{1+x}{3}\)

\(and\dfrac{5}{5}=1\)

\(\Rightarrow x=1-\dfrac{1}{3}=\dfrac{2}{3}\) vậy \(x=2\)

Ps: tự làm tiếp nha mình mới làm tới đó

Làm tiếp :

Vì \(x=2\Rightarrow\left(y+z\right)=18-2=16\)

\(\Rightarrow\dfrac{y+2}{4}=\dfrac{z+3}{5}and=\dfrac{2}{4}+\dfrac{3}{5}+\dfrac{y}{4}+\dfrac{z}{5}\)

Vậy \(y=1-\dfrac{2}{4}=\dfrac{2}{4}=2+4=6\)

\(z=16-\left(6+2\right)=8\)

\(\left[{}\begin{matrix}x=2\\y=6\\z=8\end{matrix}\right.\)

Lâu lâu nhai lại dạng này cũng thấy ngon, dù không thích

Xong rồi! Dù sai hay đúng gì thì mình cũng góp công chút nhé! Nhưng ti lệ đúng là 85% thực chất là 95% nhưng (không dám nói hơn, sợ mất mặt)

a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)

\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)

\(\Leftrightarrow x-4=25\)

\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)

b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)

\(\Leftrightarrow x\left(x+1\right)=18.4\)

\(\Leftrightarrow x\left(x+1\right)=72\)

vì \(72=8.9=\left(-8\right).\left(-9\right)\)

\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)

c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)

\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)

\(\Leftrightarrow2x+3-2x-8⋮x+4\)

\(\Leftrightarrow-5⋮x+4\)

\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)

\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)

ai giải giúp mình với 
mình đang cần gấp

Tham khảo

Bạn tham khảo nha!

\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)

\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)

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

\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}=\dfrac{1+2y+1+6y}{18+6x}=\dfrac{2\left(1+4y\right)}{2\left(9+3x\right)}=\dfrac{1+4y}{9+3x}\)

⇒ \(\dfrac{1+4y}{9+3x}=\dfrac{1+4y}{28}\)

⇒\(9+3x=28\)

⇒\(3x=19\)

⇒\(x=\dfrac{19}{3}\)

bạn thay vào là tìm được y

a) \(a\left(b+1\right)=3\left(a;b\inℤ\right)\)

\(\Rightarrow a;\left(b+1\right)\in U\left(3\right)=\left\{-1;1;-3;3\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(-1;-4\right);\left(1;2\right);\left(-3;-2\right);\left(3;0\right)\right\}\)

b) \(2n+7⋮n+1\left(n\inℤ\right)\)

\(\Rightarrow2n+7-2\left(n+1\right)⋮n+1\)

\(\Rightarrow2n+7-2n-2⋮n+1\)

\(\Rightarrow5⋮n+1\)

\(\Rightarrow n+1\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)

\(\Rightarrow n\in\left\{-2;0;-6;4\right\}\)

c) \(xy+x-y=6\left(x;y\inℤ\right)\)

\(\Rightarrow x\left(y+1\right)-y-1+1=6\)

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

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

\(\Rightarrow\left(x-1\right);\left(y+1\right)\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)

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