设关于x一次函数y=a1x+b1与y=a2x+b2,我们称函数y=m（a1x+b1）+n（a2x+b2）（其中m+n=1

20.(1)取m=n=1/2,得y=2x.
(2)x=c时y=2c(m+n)=2c.
(3)5=aa1+b1,5=aa2+b2,a1b1=a2b2=1,
∴m(a1^2a^2+b1^2)+n(a2^2a^2+b2^2)+2ma+2na
=m(a1^2a^2+b1^2)+(1-m)(a2^2a^2+b2^2)+2a
=m[(aa1+b1)^2-2aa1b1]+(1-m)[(aa2+b2)^2-2aa2b2]+2a
=m[25-2a]+(1-m)[25-2a]+2a
=25-2a+2a
=25.