\(x^3+x^2-x-1=0\)
\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\\x=-1\end{matrix}\right.\)
Vậy \(S=\left\{1;-1\right\}\)
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
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
Tìm x biết:
1> (x+1)^2-(x+2)^2=3
2> (x-1)(x+1)-(x-3)^2=0
3> (x+1)^3-x^2(x+2)-(x-1)^2=0
4> (x+1)(x^2-x+1)-x^2(x+2)+2(x+3)^2=0
1) Tim x
a) 3x(x-1)+x-1=0
b)2(x+3)-x^2-3x=0
làm tương tự như bài này nè!
x(x-2)+x-2=0
=>x*((x-2)-x-3)=0
=>(x-2)(x+10=0
hoac x-3=0 =>x=3
hoac 5x-1=0 =>x=1 phan 5
vậy x = 1 phần 5 ;x=3
hoac 5x-1=0 => x=1 phan 5
tìm x bt a) x^3=x^5 b) 4x.(x+1)=(x+1) c) x.(x-1)-2(1-x)=0 d) 2x.(x-2)-(2-x)^2 e) (x-3)^2+3-x=0 f) 5x.(x-2)-(2-x)=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
Tìm x, biết:
a) 2-x = 2 ( x - 2 ) 3 ; b) 8 x 3 - 72x = 0;
c) ( x - 1 , 5 ) 6 + 2 ( 1 , 5 - x ) 2 = 0; d) 2 x 3 +3 x 2 +3 + 2x = 0;
e) x 3 - 4x- 14x(x - 2) = 0; g) x 2 (x + 1)- x(x + 1) + x(x - 1) = 0.
Tìm x
a)x^2-x-1/4=0
b)x(x-2)x-2=0
c)5x(x-3)-x+3=0
d)x^3-1/4x=0
e)(2x-1)^2-(x+3)=0
giúp e nha
1) (x+2)(x-3) = 0 2) (2x + 3)(-x + 7) = 0
3) (x-1)(x+5)(-3x+8) = 0 4) x(x2-1) = 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
