∵a(x²-1)-2cx+b(x²+1)=0有两个实数根∴△=4c²+4(a+b)(a-b)=4(a²+c²-b²)≥0即：a²+c²-b²≥0由余弦定理知：cosB= (a²+c²-b²)/2ac≥0∴∠B是非钝角.（无...