Question

[No ri do]

Tìm x biết: x^5=x^4+x^3+x^2+x+2

Answer 1

[Lightning Farron]

x⁵=x⁴+x³+x²+x+2

<=>x⁵-1=x⁴+x³+x²+x+1

VT=x⁵+x⁴+x³+x²+x-x⁴-x³-x²-x-1

=x(x⁴+x³+x²+x+1)-(x⁴+x³+x²+x+1)

=(x-1)(x⁴+x³+x²+x+1)

pt trở thành 

(x-1)(x⁴+x³+x²+x+1)=x⁴+x³+x²+x+1

<=>[(x-1)-1](x⁴+x³+x²+x+1)=0

<=>(x-2)(x⁴+x³+x²+x+1)=0

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x^4+x^3+x^2+x+1=0\left(2\right)\end{array}\right.\)

\(\Leftrightarrow x=2\). Thấy x=0 không là nghiệm của (2) 

Chia 2 vế cho x² đc:

\(x^2+x+1+\frac{1}{x}+\frac{1}{x^2}=0\)

\(\Leftrightarrow\left(x+\frac{1}{x}\right)+\left(x^2+\frac{1}{x^2}\right)+1=0\)

\(\Leftrightarrow\left(t^2-2\right)+t+1=0\)

\(\Leftrightarrow t^2+t-1=0\)

\(\Leftrightarrow t=\frac{-1\pm\sqrt{5}}{2}\Rightarrow x\in\left\{\varnothing\right\}\)

Vậy pt trên có x=2 thỏa mãn