a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)

=>x=-3; y=28; z=-60

b: 5/12=x/-72

=>x=-72*5/12=-6*5=-30

c: =>x+3=-5

=>x=-8

a) \(\left(x+1\right)\left(y+4\right)=7\).

-Vì \(x,y\in Z\) nên ta có thể viết:

\(\left(x+1\right)\left(y+4\right)=1.7\) hay \(\left(x+1\right)\left(y+4\right)=7.1\) hay \(\left(x+1\right)\left(y+4\right)=\left(-1\right).\left(-7\right)\) hay \(\left(x+1\right)\left(y+4\right)=\left(-7\right).\left(-1\right)\)

+Xét trường hợp \(\left(x+1\right)\left(y+4\right)=1.7\):

\(\Rightarrow x+1=1\) và \(y+4=7\) 

\(\Rightarrow x=0\left(tmđk\right)\) và \(y=3\left(tmđk\right)\).

+Xét trường hợp \(\left(x+1\right)\left(y+4\right)=7.1\):

\(\Rightarrow x+1=7\) và \(y+4=1\) 

\(\Rightarrow x=6\left(tmđk\right)\) và \(y=-3\left(tmđk\right)\).

+Xét trường hợp \(\left(x+1\right)\left(y+4\right)=\left(-1\right).\left(-7\right)\):

\(\Rightarrow x+1=-1\) và \(y+4=-7\)

\(\Rightarrow x=-2\left(tmđk\right)\) và \(y=-11\left(tmđk\right)\).

+Xét trường hợp \(\left(x+1\right)\left(y+4\right)=\left(-7\right).\left(-1\right)\):

\(\Rightarrow x+1=-7\) và \(y+4=-1\)

\(\Rightarrow x=-8\left(tmđk\right)\) và \(y=-5\left(tmđk\right)\).

b) \(xy+2x-3y=-1\)

\(\Rightarrow xy+2x-3y+1=0\)

\(\Rightarrow y\left(x-3\right)=-2x-1\)

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

-Vì \(y\in Z\) \(\Rightarrow5⋮\left(x-3\right)\).

\(\Rightarrow\left(x-3\right)\inƯ\left(5\right)\)

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

\(\Rightarrow x\in\left\{4;2;8;-2\right\}\) (đều thỏa mãn điều kiện).

+Với \(x=4\) thì \(y=\dfrac{5}{4-3}=5\) (tmđk).

+Với \(x=2\) thì \(y=\dfrac{5}{2-3}=-5\) (tmđk).

+Với \(x=8\) thì \(y=\dfrac{5}{8-3}=1\) (tmđk)

+Với \(x=-2\) thì \(y=\dfrac{5}{-2-3}=-1\) (tmđk).

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

x+1;y-2 thuộc Ư(3){1;-1;3;-3}

ta có bảng sau :

vậy cặp x;y thuộc {(2;3);(0;1);(4;5);(-2;-1)}
 

a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)

b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)

c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)

\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)

c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)

d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)

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Lời giải:
a.

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x-y}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)

\(\Rightarrow \left\{\begin{matrix} x=60\\ y=45\\ z=40\end{matrix}\right.\)

b)

Từ đkđb suy ra \(\frac{10x}{1}=\frac{5y}{\frac{1}{3}}=\frac{z}{\frac{1}{6}}=\frac{10x-5y+z}{1-\frac{1}{3}+\frac{1}{6}}=\frac{25}{\frac{5}{6}}=30\)

\(\Rightarrow \left\{\begin{matrix} x=3\\ y=2\\ z=5\end{matrix}\right.\)

lỗi