TH1: `3-x>=0 <=> x<=3`
`3-x+x^2-x(x+4)=0`
`<=>3-x+x^2-x^2-4x=0`
`<=> -5x=-3`
`<=>x=3/5(TM)`
TH2: `3-x<0 <=> x>3`
`x-3+x^2-x^2-4x=0`
`<=>-3x=3`
`<=>x=-1(L)`
Vậy `S={3/5}`.
a,x+5/x-1+8/x^2-4x+3=x+1/x-3 b,x-4/x-1-x^2+3/1-x^2+5/x+1=0 c,3x/4-5=3-x/2+5x-1/6 d,(x-2)(x+2)-(x-3)(x+4)-2x+3=0 e,(x-1)^2+2(x+1)=5x+5 g,(x-3)(x+4)x=0
1) 2x^4-7x^2-4=0
2)(x62+5x^2)-2(x^2+5x)-24=0
3)x^2-2x-3(x-1)+3=0
4)(x+1/x)^2+2(x+1/x)-8=0
5)x(x+1)(x+2)(2x+3)-18=0
7)(x^2+4x+7)=(x+4)nhân vs căn bậc hai cua x^2 +2
tìm x: part 1 : a,(x^3)^2-(x+1)(x-1)=1 b,(x-2)^2-3(x-2)=0 c,(x+2)(x^2-2x+4)-x(x^2+2)=15 d,(x+1)^2-(x+1)(x-2)=0 e,4x(x-2017)-x+2017=0 f,(x+4)^2-16=0 part 2: a,x^3+27+(x+3)(x-9)=0 b,(2x-1)^2-4x^2+1=0 c,2(x-3)+x^2-3x=0 d,x^2-2x+1=6x-6 e,x^3-9x=0
tìm x , biết :
a, ( x mũ 3 - 4 x mũ 2 ) - ( x -4 ) = 0
b, x mũ 5 - 9x = 0
c, ( x mxu 3 - x mũ 2 ) mũ 2 - 4 x mũ 2 + 8x - 4 = 0
1) (x^3 - x^2)- 4x^2 + 8x - 4 = 0
2) 2x^3 - 50x = 0
3) (x + 1) = ( x + 1)(x - 1)
4) ( 3x+1)^2-4(X-3)^2=0
5)(X+3)(X^2-5X+9)-X^3=2X
6) (4X+3)^2-(4X-3)^2-5X-2=0
7)(X-1)^3-(X-3)(X^2+3X+9)-3X(2-X)=5
Giải PT
a, (9x^2 -4) (x+1) = (3x+2)(x^2+1)
b, (x-1)^2 - 1 +x^2 = (1-x) (x+3)
c, (x^2-1)(x+2) (x-3) = (x-1) (x^2-4) (x+5)
d, x^4 + x^3 +x +1 =0
e,x^3 -7x+6 =0
f, x^4 -4x^3 +12x-9 = 0
g, x^5 - 5x^3 +4x = 0
h, x^4 - 4x^3 + 3x^2 +4x-4 =0
Giải các phương trình sau:
1. x^2-4-(x+5)(2-x)=0
2. 2x^3+4x=x^2 +2x
3. (x^2+3x+2)(x^2+3x+3)-2=0
4.(x+1)(x+2)(x+3)(x+4)-24 =0
1) 2x+6=0
2)(x^2-2x+1)-4=0
3)x-2/x+2+3/x-2=x^2-11/x^2-4
4)x(x^2-1)=0
5)4x+20=0
6)x+3/x+1+x-2/x=2
7)1+2x-5/6=3-x/4
8x+2/x-2-1/x=2/x^2-2x
9)2(x+1)=5x-7
1) 5x^2 = 13x
2) (5x^2 + 3x – 2 )^2 = (4x^2 – 3x – 2 )^2
3) x^3 + 27 + (x + 3)(x – 9) = 0
4) 5x(x – 2000) – x + 2000 = 0
5) 5x(x – 2) – x – 2 = 0
6) 4x(x + 1) = 8( x + 1)
7) x(x – 4) + (x – 4)^2 = 0
8) x^2 – 6x + 8 = 0
9) 9x^2 + 6x – 8 = 0
10) x^3 + x^2 + x + 1 = 0
11) x^3 - x^2 - x + 1 = 0
12) (5 – 2x)(2x + 7) = 4x^2 – 25
13) x(2x - 1) + 1/3 . 2/3x = 0
14) 4(2x + 7) – 9(x + 3)^2 = 0
GIÚP TUI ZỚI MỌI NGƯỜI OIWIII!!!
