Question

求解第八题

Answer 1

[1-√(1-x²)]/x
原卖老弯式=ln[1-√(1-x²)]-lnx
设v=x²中闷,u=1-v,t=1-√u,则dv/dx=2x,du/dv=-1,dt/du=1/2√u
∴dt/dx=dt/du*du/dv*dv/含尺dx=-x/√(1-x²)
∴dy/dx=-x/√(1-x²)-1/x
f(x1)+f(x2) = [(ax1+b)(x2^2+1)+(ax2+b)(x1^2+1)]/(1+x1^2)(1+x2^2) = 0

a(x1x2)(x1+x2)+b(x1^2+x2^2)+2b+a(x1+x2) = 0
b(x1^2+x2^2+2) = 0
x1^2+x2^2 = (4b^2/a^2+2) >0
b=0
x1+x2=0 x1x2=-1
x1=1 x2=-1
f(1) = a/2 f(-1)=-a/2