(1)13-4x2≥0
-√13/2≤x≤√13/2
(2) x2-6x+4≥0
x≥3+√5或x≤3-√5
(3)(x-1)(2-x)≥0
1≤x≤2
(4) x(1-x)>x(2x-3)+1
(3x-1)(x-1)
13-4x2≥0 x2-6x+4≥0 (x-1)(2-x)≥0 x(1-x)>x(2x-3)+1 -6x2-x+1≤0
13-4x2≥0 x2-6x+4≥0 (x-1)(2-x)≥0 x(1-x)>x(2x-3)+1 -6x2-x+1≤0
數(shù)學(xué)人氣:680 ℃時間:2020-09-26 21:04:25
優(yōu)質(zhì)解答
我來回答
類似推薦
- 解下列不等式 X2+2X3-13X -4X2+√2X+3
- x4-2x3-6x2+6x+9=0求x的值,x的4次方-2x的3...
- ①√6x2-x-2√6=0②﹙x+2﹚2-2x-3③x2+2﹙√3+1﹚x+2√3=0④﹙2x-1﹚﹙x-3﹚=4
- 解不等式: (1)x2+2x-3≤0; (2)x-x2+6<0.
- 1.x2+2x-3≤0 2.x-x2+6<0 3.4x2+4x+1≥0 4.x2-6x-9≤05.4+x-x2<0 6.x2+2x+3<0求不等式解集
- 已知實數(shù)x,y滿足2x+3y≤14,2x+y≤9,x≥0,y≥0,S=3x+ay,若S取得最大值時的最優(yōu)解有無窮多個,則實數(shù)a=?
- 請問這種成分還屬301不銹鋼嗎?(C-0.1003;Si-0.2467;Mn-2.2387;p-0.358;S-0.169;Cr-14.6342;Ni-6.0215)
- X=2*3*5*7*11*13*17*19*23*29*.N(N為質(zhì)數(shù)),求證:X+1為質(zhì)數(shù)
- 若√2007n是個非零整數(shù),則最小整數(shù)n是?
- Either I or he ( )soccer with Tom 四個選項 play are plays is
- .the music festival was great!Many famous people (attended) it.
- 如果(M)表示m的全部因數(shù)的和,如(4)=1+2+4=7,則(18)-(21)=()