\(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-1\end{matrix}\right.\)
\(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-1\end{matrix}\right.\)
tìm x biết :
4x(x+1) = 8(x+1)
x(2x+1) +\(\dfrac{1}{3}-\dfrac{2}{3}x=0\)
x(x-4) +(x-4)2 =0
BT2: Tìm x 2, 3x(x-4)+2x-8=0 3, 4x(x-3)+x^2-9=0 4, x(x-1)-x^2+3x=0 5, x(2x-1)-2x^2+5x=16
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:
\(a,x^3+2x^2+x+2=0\)
\(b,x^3+4x^2+\dfrac{1}{4}x+1=0\)
bài 7 tìm x
1,x(x+3)-5(x+3)=0 2,5x(x-1)=x-1
3,(x+1)=(x+1)\(^2\) 4,x(2x-3)-2(3-2x)=0
5,\(\left(x-2\right)^2-4=0\) 6,\(36x^2=49\)
7,\(2x\left(x-6\right)-x+6=0\) 8,\(3x\left(2x-1\right)-24x+12=0\)
9,\(x^2-6x+8=0\) 10,\(x^2+2x-15=0\)
Bài 4: Tìm x:
1) x2 - 9x = 0 2) x(x - 4) – x2 = 7 3) 3x + 2(x – 5) = 5
4) 25x2 - 1 = 0 5) 3x(x - 2) - 5(x - 2) = 0 6) 3x(x - 7) + 4(x – 7) = 0
7) 4x2 – 9 = 0 8) 10x(x - 4) + 2x - 8 = 0 9) x(2x - 5) - 2x2 = 0
10) 2x2 – 4x = 0 11) 2x(3 - 4x) + 3(4x - 3) = 0 12) 2x (x – 5) – 2x2 = 3
mọi người giúp mình vs chiều 1g mình thi rồi! cảm ơn!
tìm x
1)x(x-1)-2(1-x)=0
2)(x+1)^2=x+1
3)(x-3)^3+(3-x)=0
4)x^3=x^5
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
Tìm x, biết.
a) x+ 5x2 = 0 b)(x+3)2+(4+x)(4-x)=10
c) 5x( x – 1) = x - 1 d) x2 -2x -3 = 0