答案:5
f(x1+x2)=ax1^2+bx1+5 +ax2^2+bx2+5 -5+2ax1x2
f(x1+x2)=3+3-5+2ax1x2=1+2ax1x2
ax2+bx+5=3
-->
x1=(b^2+sqrt(b^2-8a))/2a
x2=(b^2-sqrt(b^2-8a))/2a
x1*x2=2/a
f(x1+x2)=1+2ax1x2=5
答案:5
f(x1+x2)=ax1^2+bx1+5 +ax2^2+bx2+5 -5+2ax1x2
f(x1+x2)=3+3-5+2ax1x2=1+2ax1x2
ax2+bx+5=3
-->
x1=(b^2+sqrt(b^2-8a))/2a
x2=(b^2-sqrt(b^2-8a))/2a
x1*x2=2/a
f(x1+x2)=1+2ax1x2=5