`1)x^4 -10x^3 +26x^2 -10x+1=0`
`x=0=>VT=1=>x=0(l)`
Chia 2 vế cho `x^2>0` ta có
`x^2-10x+26-10/x+1/x^2=0`
`=>x^2+1/x^2+26-10(x+1/x)=0`
`=>(x+1/x)^2-10(x+1/x)+24=0`
Đặt `a=x+1/x`
`pt<=>a^2-10a+24=0`
`<=>` $\left[ \begin{array}{l}a=4\\a=6\end{array} \right.$
`a=4<=>x+1/x=4<=>x^2-4x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt3+2\\x=-\sqrt3+2\end{array} \right.$
`a=6<=>x+1/x=6<=>x^2-6x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt8+3\\x=-\sqrt8+3\end{array} \right.$
Vậy `S={\sqrt3+2,-\sqrt3+2,\sqrt8+3,-\sqrt8+3}`

2)Do hệ số chẵn bằng=hệ số lẻ
`=>x=-1`
`pt<=>x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0`
`<=>(x+1)(x^3+4x^2+6x+9)=0`
`<=>(x+1)(x^3+3x^2+x^2+6x+9)=0`
`<=>(x+1)[x^2(x+3)+(x+3)^2]=0`
`<=>(x+1)(x+3)(x^2+x+3)=0`
Do `x^2+x+3=(x+1/2)^2+11/4>0`
`=>` $\left[ \begin{array}{l}x=-3\\x=-1\end{array} \right.$
Vậy `S={-1,-3}`

x4+10x3+26x2+10x+1=0x4+10x3+26x2+10x+1=0

⇔x4+6x3+x2+4x3+24x2+4x+x2+6x+1=0⇔x4+6x3+x2+4x3+24x2+4x+x2+6x+1=0

⇔x2(x2+6x+1)+4x(x2+6x+1)+(x2+6x+1)=0⇔x2(x2+6x+1)+4x(x2+6x+1)+(x2+6x+1)=0

⇔(x2+4x+1)(x2+6x+1)=0⇔(x2+4x+1)(x2+6x+1)=0

⇔(x2+4x+4−3)(x3+6x+9−8)=0⇔(x2+4x+4−3)(x3+6x+9−8)=0

⇔[(x+2)2−3][(x+3)2−8]=0⇔[(x+2)2−3][(x+3)2−8]=0

⇒[(x+2)2−3=0(x+3)2−8=0⇒[(x+2)2−3=0(x+3)2−8=0⇒[(x+2)2=3(x+3)2=8⇒[(x+2)2=3(x+3)2=8⇒⎡⎣⎢⎢⎢x=−4±12−−√2x=−6±32−−√2

Thử phân tích VT thành: \(\left(x^2+6x+1\right)\left(x^2+4x+1\right)=0\) xem sao?

\(x^4+10x^3+26x^2+10x+1=0\)

\(\Leftrightarrow\left(x^4+6x^3+x^2\right)+\left(4x^3+24x^2+4x\right)+\left(x^2+6x+1\right)=0\)

\(\Leftrightarrow x^2\left(x^2+6x+1\right)+4x\left(x^2+6x+1\right)+\left(x^2+6x+1\right)=0\)

\(\Leftrightarrow\left(x^2+6x+1\right)\left(x^2+4x+1\right)=0\)

\(\Leftrightarrow\left(x^2+6x+9-8\right)\left(x^2+4x+4-3\right)=0\)

\(\Leftrightarrow\left[\left(x+3\right)^2-8\right]\left[\left(x+2\right)^2-3\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2-8=0\\\left(x+2\right)^2-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=8\\\left(x+2\right)^2=3\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=\pm\sqrt{8}\\x+2=\pm\sqrt{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{8}-3\\x=\pm\sqrt{3}-2\end{cases}}}\)

Nhận thấy \(x=0\) không phải nghiệm, chia 2 vế cho \(x^2\) ta được:

a/ \(x^2+\frac{1}{x^2}+6\left(x+\frac{1}{x}\right)+11=0\)

Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)

\(\Leftrightarrow t^2-2+6t+11=0\Leftrightarrow\left(t+3\right)^2=0\)

\(\Rightarrow t=-3\Rightarrow x+\frac{1}{x}=-3\Leftrightarrow x^2+3x+1=0\) (casio)

b/ \(x^2+\frac{1}{x^2}-10\left(x+\frac{1}{x}\right)+26=0\)

Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)

\(\Leftrightarrow t^2-2-10t+26=0\)

\(\Leftrightarrow t^2-10t+24=0\Rightarrow\left[{}\begin{matrix}t=6\\t=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{x}=4\\x+\frac{1}{x}=6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-4x=1=0\\x^2-6x+1=0\end{matrix}\right.\) (casio)

          \(x^4-10x^3+26x^2-10x+1=0\)

\(\Leftrightarrow\)\(\left(x^4-4x^3+x^2\right)-\left(6x^3-24x+6x\right)+\left(x^2-4x+1\right)=0\)

\(\Leftrightarrow\)\(x^2\left(x^2-4x+1\right)-6x\left(x^2-4x+1\right)+\left(x^2-4x+1\right)=0\)

\(\Leftrightarrow\)\(\left(x^2-6x+1\right)\left(x^2-4x+1\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2-6x+1=0\\x^2-4x+1=0\end{cases}}\)

Nếu   \(x^2-6x+1=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3-\sqrt{8}\\x=\sqrt{8}+3\end{cases}}\)

Nếu  \(x^2-4x+1=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2-\sqrt{3}\\x=\sqrt{3}+2\end{cases}}\)

Vậy....

Lời giải:
a.

$x^4+10x^3+26x^2+10x+1$

$=(x^4+10x^3+25x^2)+x^2+10x+1$

$=(x^2+5x)^2+2(x^2+5x)+1-x^2$

$=(x^2+5x+1)^2-x^2=(x^2+5x+1-x)(x^2+5x+1+x)$

$=(x^2+4x+1)(x^2+6x+1)$

b.

$x^4+x^3-4x^2+x+1$

$=(x^4-x^2)+(x^3-x^2)+(x-x^2)+(1-x^2)$

$=x^2(x-1)(x+1)+x^2(x-1)-x(x-1)-(x-1)(x+1)$

$=(x-1)[x^2(x+1)+x^2-x-(x+1)]$

$=(x-1)(x^3+2x^2-2x-1)$

$=(x-1)[(x^3-1)+(2x^2-2x)]=(x-1)[(x-1)(x^2+x+1)+2x(x-1)]$

$=(x-1)(x-1)(x^2+x+1+2x)=(x-1)^2(x^2+3x+1)$

\(x^4+10x^3+25x^2+x^2+1=0\)

\(\Leftrightarrow\left(x^2+5x\right)^2+x^2+1=0\)

Do \(\left(x^2+5x\right)^2+x^2+1>0\) \(\forall x\)

\(\Rightarrow\) Phương trình vô nghiệm

$x^4-10x^3+26x^2-10x+\mathrm{1=}$0

⇔x²(x²-10x +26 -\(\dfrac{10}{x}+\dfrac{1}{x^2}\))=0

⇔x²-10x+26-\(\dfrac{10}{x}+\dfrac{1}{x^2}=0\)

⇔\(\left(-10x-\dfrac{10}{x}\right)+\left(x^2+\dfrac{1}{x^2}\right)+26=0\)

⇔\(-10\left(x+\dfrac{1}{x}\right)+\left(x^2+\dfrac{1}{x^2}\right)+26=0\)

đặt \(t=\left(x+\dfrac{1}{x}\right)\) thì \(\left(x^2+\dfrac{1}{x^2}\right)=t-2\)

ta có

-10t +t²-2+26=0

=>t²-10t+24=0

=>t²-4t-6t+24=0

=>(t²-4t)-(6t-24)=0

=>t(t-4)-6(t-4)=0

=>(t-4)(t-6)=0

=>t=4 và t=6

* với t=4 thì

\(x+\dfrac{1}{x}=4\Rightarrow x^2-4x+1=0\)(vô nghiệm)

* với t=6 thì

\(x+\dfrac{1}{x}=6\Rightarrow x^2-6x+1=0\) (vô n^o)

vậy S=∅

$x^4-10x^3+26x^2-10x+\mathrm{1\; =0\; à}$

$\mathrm{mk\; là\; theo}$

$x^4-10x^3+26x^2-10x+\mathrm{1=0\; nha}$