pt <=> (x^5-x^4)-(4x^4-4x^3)+(4x^2-4x)-(x-1) = 0

<=> (x-1).(x^4-4x^3+4x-1) = 0

<=> (x-1).[(x^4-x^3)-(3x^3-3x)+(x-1)] = 0

<=> (x-1).(x-1).(x^3-3x^2-3x+1) = 0

<=>(x-1)^2.[(x^3+x^2)-(4x^2+4x)+(x+1)] = 0

<=> (x-1)^2.(x+1).(x^2-4x+1) = 0

<=> x-1=0 hoặc x+1=0 hoặc x^2-4x+1=0

<=> x=1 hoặc x=-1 hoặc x=2+\(\sqrt{3}\)hoặc x = 2-\(\sqrt{3}\)

k mk nha

Bạn nhóm có nhân tử (x-1) là đc

2:

=>x^3-1-2x^3-4x^6+4x^6+4x=6

=>-x^3+4x-7=0

=>x=-2,59

4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50

=>-62x+12=-50

=>x=1

\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...

a: \(\Leftrightarrow\left(2-x\right)\left(x-3\right)+\left(x-1\right)\left(x+3\right)=-4x\)

\(\Leftrightarrow2x-6-x^2+3x+x^2+3x-x-3=-4x\)

=>7x-9=-4x

=>11x=9

hay x=9/11

b: \(\Leftrightarrow\left(5-x\right)\left(x-4\right)+\left(x+2\right)\left(x+4\right)=-3x\)

\(\Leftrightarrow5x-20-x^2+4x+x^2+6x+8=-3x\)

=>15x-12=-3x

=>18x=12

hay x=2/3

tu lam

bạn chia ra thui có kq mmk làm cho