解方程组
{ x + y = 2 x y − z 2 = 1 (1) \left\{ \begin{aligned} x+y=2 \tag{1} \\ xy-z^2=1 \end{aligned} \right. {x+y=2xy−z2=1(1)
解法一：韦达定理
x
y
−
z
2
=
1
x
y
=
z
2
+
1
xy-z^2=1 \\ xy=z^2+1
xy−z2=1xy=z2+1
根据韦达定理构造方程
t
2
−
2
t
+
z
2
+
1
=
0
(2)
t^2-2t+z^2+1=0 \tag{2}
t2−2t+z2+1=0(2)
x
x
x和
y
y
y是方程的两个解
Δ
=
(
−
2
)
2
−
4
×
(
z
2
+
1
)
=
4
−
4
(
z
2
+
1
)
=
−
4
z
2
⩾
0
z
2
⩽
0
z
2
=
0
z
=
0
\Delta=(-2)^2-4\times(z^2+1) =4-4(z^2+1) =-4z^2\geqslant0 \\ z^2\leqslant0 \\ z^2 = 0 \\ z = 0
Δ=(−2)2−4×(z2+1)=4−4(z2+1)=−4z2⩾0z2⩽0z2=0z=0
代入方程(2)
t
2
−
2
t
+
1
=
0
(
t
−
1
)
2
=
0
t
=
1
t^2-2t+1=0 \\ (t-1)^2=0 \\ t=1
t2−2t+1=0(t−1)2=0t=1
求得结果是
{
x
=
1
y
=
1
z
=
0
\left\{ \begin{aligned} x=1 \\ y=1 \\ z=0 \end{aligned} \right.
⎩⎪⎨⎪⎧x=1y=1z=0
解法二：
x
+
y
=
2
x+y=2
x+y=2
左右平方
x
2
+
2
x
y
+
y
2
=
4
x^2+2xy+y^2=4
x2+2xy+y2=4
$$
xy-z^2=1 \
xy-1=z^2 \
4xy-4=4z^2 \
4xy-(x2+2xy+y2)=4z^2 \
-(x2-2xy+y2)=4z^2 \
-(x-y)2=4z2 \
4z2+(x-y)2=0 \
\because z^2 \geqslant 0 \And (x-y)^2 \geqslant 0 \
z=0 \
x-y=0 \
x=y \
\because x+y=2 \
\therefore x=y=1
$$
解法三：
x
+
y
=
2
x
=
2
−
y
x+y=2 \\ x=2-y \\
x+y=2x=2−y
代入
x
y
−
z
2
=
1
(
2
−
y
)
y
−
z
2
=
1
2
y
−
y
2
−
z
2
=
1
y
2
−
2
y
+
1
+
z
2
=
0
(
y
−
1
)
2
+
z
2
=
0
(
y
−
1
)
2
=
0
z
2
=
0
y
=
1
z
=
0
x
+
y
=
2
x
=
1
xy-z^2=1 \\ (2-y)y-z^2=1 \\ 2y-y^2-z^2=1 \\ y^2-2y+1+z^2=0 \\ (y-1)^2+z^2=0 \\ (y-1)^2=0 \\ z^2=0 \\ y=1 \\ z=0 \\ x+y=2 \\ x=1
xy−z2=1(2−y)y−z2=12y−y2−z2=1y2−2y+1+z2=0(y−1)2+z2=0(y−1)2=0z2=0y=1z=0x+y=2x=1