\(x^4-2x^3+3x^2-2x+1=0\)

Chia cả hai vé cho \(x^2\)

\(\Leftrightarrow x^2-2x+3-\dfrac{2}{x}+\dfrac{1}{x^2}\)

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

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

Đặt x+1/x = a, ta có:

\(a^2-2a+1=0\)

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

\(\Leftrightarrow a=1\)

\(\Leftrightarrow x+\dfrac{1}{x}=1\)

\(\Leftrightarrow x^2+1=x\)

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

\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)

Do \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+3>0\)

Do đó phương trình vô nghiệm

1) \(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}}=\sqrt{3}+1-\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}+1-\sqrt{3}+1=2\)

2) \(\dfrac{3}{5}\sqrt{25x-50}-\sqrt{x-2}=6\left(đk:x\ge2\right)\)

\(\Leftrightarrow3\sqrt{x-2}-\sqrt{x-2}=6\)

\(\Leftrightarrow2\sqrt{x-2}=6\)

\(\Leftrightarrow\sqrt{x-2}=3\)

\(\Leftrightarrow x-2=9\Leftrightarrow x=11\left(tm\right)\)

Giải

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

\(\frac{2\left(x+1+x-1\right)}{x^2-1}=\frac{2\left(2x\right)}{x^2-1}=\frac{4x}{x^2-1}\)

Tới đây bí rồi

Đợi tí mình giải cho !!!!!!
 

  x^5=x^4+x^3+x^2+x+2? 
<=> 
x^5-1=x^4+x^3+x^2+x+1 
<•>(x-1)(x^4+x^3+x^2+x+1)=x^4+x^3+x^2+... 
<=> 
(x^4+x^3+x^2+x+1)[(x-1)-1]=0 
=> 
(x^4+x^3+x^2+x+1)(x-2)=0 
=> 
x-2=0=>x=2 
[ 
x^4+x^3+x^2+x+1>0 moi x 
nghiêm 
x=2

x thuộc rỗng

Taco:VD:2.3.4.5>24=>x<0

x=-2=>tổng=0

Vayx=-1

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)=0\)