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1.
a.
x(y-1)-(y-1)=0
(x-1)(y-1)=0
x=1
y=1
b.x,y khác không
(x+y)/xy=1/2
2x+2y=xy
x(2-y)-2(2-y)=4
(x-2)(y-2)=4
x-2={-4,-2,1,4}=>x={-2,0,3,6}
y-2={-1,-2,4,1}=>y={3,0,6,3}
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt x = 4k
y = 7k
=> 4k.7k = 112
=> 28.k^2 = 112
=> k^2 = 112 : 28 = 4
=> k = 2
=> x = 4.2 = 8
y = 7.2 = 14
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Hướng dẫn thôi nhé:
Lời giải:
a)\(xy+x+y+1=0\)
\(\Rightarrow x\left(y+1\right)+1\left(y+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(y+1\right)=0\)
b)\(xy-x-y=0\)
\(\Rightarrow xy-x-y+1=1\)
\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=1\)
\(\Rightarrow\left(x-1\right)\left(y-1\right)=1\)
c)\(xy-x-y-1=0\)
\(\Rightarrow xy-x-y+1=2\)
\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=2\)
\(\Rightarrow\left(x-1\right)\left(y-1\right)=2\)
d) \(xy-x-y+1=0\)
\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(y-1\right)=0\)
e)\(xy+2x+y+11=0\)
\(\Rightarrow xy+2x+y+2=-9\)
\(\Rightarrow x\left(y+2\right)+1\left(y+2\right)=-9\)
\(\Rightarrow\left(x+1\right)\left(y+2\right)=-9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=> \(\frac{x-y}{1}=\frac{x+y}{7}=\frac{xy}{24}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{xy}{24}=\frac{x-y}{1}=\frac{x+y}{7}=\frac{\left(x-y\right)+\left(x+y\right)}{1+7}=\frac{\left(x-y\right)-\left(x+y\right)}{1-7}\)=> \(\frac{xy}{24}=\frac{x}{4}=\frac{y}{3}\)
\(\frac{xy}{24}=\frac{x}{4}\)=>\(\frac{x}{4}.\frac{y}{6}=\frac{x}{4}\)=> \(\frac{y}{6}=\frac{x}{4}:\frac{x}{4}=1\) ( do x khác 0) => y = 6
\(\frac{xy}{24}=\frac{y}{3}\Rightarrow\frac{x}{8}.\frac{y}{3}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{3}:\frac{y}{3}=1\) ( do y khác 0) => x = 8
Vậy...
xy + x + y = 11
=> x(y + 1) + y + 1 = 11 + 1
=> x(y + 1) + 1(y + 1) = 12
=> (x + 1)(y + 1) = 12
lập bảng đi
sao x(y+1) thi hiu nhung ma sao lai y+1 nua vay ban