Question

1.已知一元二次方程X²-(K+1)X+2K=0的根是RT△ABC两锐角正弦值,求K

Answer 1

x1x2=2k

x1+x2=k+1

因为一元二次方程X²-(K+1)X+2K=0的根是RT△ABC两锐角正弦值,

所以

x1²+x2²=1

=(x1+x2)²-2x1x2

=(k+1)²-2×2k

=k²-2k+1

k²-2k=0

k=0或k=2

k=0时,方程变为x²-x=0,可以

k=2时,方程变为x²-3x+4=0,无解

所以

k=0

2.

（sinx-3cosx)/(2cosx+sinx) 分子分母同除以cosx

=（tanx-3)/(2+tanx)

=(4-3)/(2+4)

=1/6