(1) X^2+X+1>0(2) X^2+X-10
8个回答
1.原式化为:3/4 X^2+(X/2+1)^2恒大于0,所以x为任意值
2.原式化为(X+1/2)^2<5/4,所以-(√5+1)/2
3.原式化为(X-3)(X+2)>0,所以X3
相关问题
1/(x^2+2x+10)+1/(x^2+11x+10)+1/(x^2-13x+10)=0
10(x+1)^2-25(x+1)+10=0 5x^2-2x=0
解方程:1/(x2+2x+10)+1/(x2+11x+10)+1/(x2-13x+10)=0
解方程 (1)[1/2][x-[1/2](x-1)]=[2/3](x-1);(2)[0.1x−0.2/0.02−x+10
2.X(2X-5)=4X-10 2X^2-9X+10=0 (2X-5)(X-2)=0 X1=2.5,X2=2
1+2(1-x)=0 2x-2(10-x)=2 x/3-5=5-x/2 2(x+1)/3=5(x+1)/6-1
X+X^2+……+X^9+X^10=A0+A1(1+X)+A2(1+X)^2……+A9(1+X)^9+A10(1+X)^
(x^2+2x+2)^5=a0+a1(x+1)+a2(x+1)^2+…+a9(x+1)^9+a10(x+1)^10,其中
(1)-2(x-1)=4.(2)[0.1x−0.2/0.02−x+10.5]=3
(x+1)2+(x+1)11=a0+a1(x+2)+a2(x+2)2+…+a10(x+2)10+a11(x+2)11,则