已知x,y满足x²-4xy+4y²-5(x-2y)=3,求2x-4y值
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已知x,y满足x²-4xy+4y²-5(x-2y)=3,求2x-4y值
已知x,y满足x²-4xy+4y²-5(x-2y)=3,求2x-4y值
已知x,y满足x²-4xy+4y²-5(x-2y)=3,求2x-4y值
x²-4xy+4y²-5(x-2y)=3
(x-2y)²-5(x-2y)=3
令x-2y=t,方程化为:
t²-5t=3
t²-5t-3=0
t=(5±√37)/2
所以:2x-4y=2(x-2y)=2t=5±√37
x²-4xy+4y²-5(x-2y)=3,
(x-2y)²-5(x-2y)-3=0,
设x-2y=t,
原方程为t²-5t-3=0,
t=(5±√37)/2
所以x-2y=(5±√37)/2
所以2x-4y=2(x-2y)=5±√37
提示:运用换元法,知识点:一元二次方程求根公式
x²-4xy+4y²-5(x-2y)=(x-2y)²-5(x-2y)=3
(x-2y)²-5(x-2y)-3=0
b²-4ac=25+12=37, x-2y=(5±根号37)/2
2x-4y=5±根号37
x²-4xy+4y²-5(x-2y)=3
x²-4xy+4y²-5x+10y-3=0
(x-2y)²-5(x-2y)-3=0
x-2y=(5±√37)/2
2x-4y=5±√37
设x-2y=t
则原方程写为t^2-5t-3=0,解出t的值,然后求2t即为2x-4y的值
e 很简单,x²-4xy+4y²=(x-2y)² 式子=(x-2y)² -5(x-2y)=3
设X-2Y =Z ,那么有Z² -5Z=3 即使Z² -5Z+2.5²=3+2.5² (Z-2.5)² =正负9.25
x²-4xy+4y²-5(x-2y)-3=0
=>(x-2y)²-5(x-2y)-3=0
=>(x-2y)²-5(x-2y)+25/4-25/4-3=0
=>(x-2y-5/2)²=37/4
=>x-2y-5/2=(±√37)/2
=>x-2y=(5±√37)/2
=>2x-4y=5±√37
x²-4xy+4y²=(x-2y)²
原式可变为(x-2y)²-5(x-2y)-3=0
设2x-4y=a 则x-2y=a/2
原式变为 a²/4-5/2*a-3=0 方程两边同乘12.
得a²-10a-12=0 解这个方程可得a 即2x-4y
x-2y)²-5(x-2y)=3
令x-2y=t,方程化为:
t²-5t=3
t²-5t-3=0
(t-5/2)²-25/4-3=0
(t-5/2)²=37/4
t=(5±√37)/2
所以:2x-4y=2(x-2y)=2t=5±√37