k=4/2=2
y=2x
AO^2=2^2+4^2=20
20-4^2=4
√4+2=4
B(4,0)
設(shè)解析式y(tǒng)=ax+b
代入兩點(diǎn)坐標(biāo)
2a+b=4
4a+b=0
解得a=-2 b=8
解析式為y=-2a+8
2、x²+y²+4x-6y+14
=(x+2)^2+(y-3)^2+1
因?yàn)?x+2)^2+(y-3)^2>=0
所以(x+2)^2+(y-3)^2+1>=1
所以不論x、y取何值,代數(shù)式x²+y²+4x-6y+14的值總是正數(shù)
3、(x^2+5x+5)^2
4、x²-y²+x-3y
=x^2-y^2+x-y-2y
=x^2-(y+1)^2+x-y+1
=(x-y-1)(x+y+1)+x-y+1
=2
5、（x²+2xy-3y²）÷（x-y）
=(x+3y)(x-y)/(x-y)
=x+3y=125
6、a（a-1）-（a²-b）=2
b-a=2
ab-(a²+b²)/2
=(2ab-a^2-b^2)/2
=-(a-b)^2/2=-(2^2)/2=-2