f(X)=(4^X)一(5x2^X)十4>0 求X范围
x2-5x+1=0则x2+x2/1
解方程:1/x2+x +1/x2+3x+2 +1/x2+5x+6 +1/x2+7x+12 +1/x2+9x+20=5/x
解方程5-x/x2-3x+2=5-x/x2-7x+h
(x-2)(x2-6x-9)-x(x-5)(x-3)其中x= -1/3
1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)=1/(x+4)
已知x2-5x=14,求2x2-x-2x+1-[x2+2x+1]+1的值.
(x2+5x+4)2+2(x2+5x+4)2+1=(x2+.
解分式方程:2/x2+5x+6 + 3/x2+x-6=4/x2-4
高一数学A={x|x2-ax+a2-19+0},B={x|x2-5x+6=0},C={x|x2+2x-8=0}且满足A∩
已知5x2-3x-5=0,则5x2-2x-15x
已知X2-5X=3,求(X-1)(2x-1)-x(x+3)+15/x2-3的值