2: Ta có: \(x^4-4x^3-9x^2+8x+4=0\)

\(\Leftrightarrow x^4-x^3-3x^3+3x^2-12x^2+12x-4x+4=0\)

\(\Leftrightarrow x^3\left(x-1\right)-3x^2\left(x-1\right)-12x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2-12x-4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-5x^2-10x-2x-4\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x^2-5x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{5-\sqrt{33}}{2}\\x=\dfrac{5+\sqrt{33}}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-2;\dfrac{5-\sqrt{33}}{2};\dfrac{5+\sqrt{33}}{2}\right\}\)

1: Ta có: \(x^4+5x^3+10x^2+15x+9=0\)

\(\Leftrightarrow x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0\)

\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)+6x\left(x+1\right)+9\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2+6x+9\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[x^3+3x^2+x^2+6x+9\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x+3\right)+\left(x+3\right)^2\right]=0\)

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

mà \(x^2+x+3>0\forall x\)

nên (x+1)(x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

Vậy: S={-1;-3}

1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)

\(\Leftrightarrow\left|x-5\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

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....

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

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

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}\left(x+2\right)^2-3=0\\\left(x+3\right)^2-8=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left(x+2\right)^2=3\\\left(x+3\right)^2=8\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-4\pm\sqrt{12}}{2}\\x=\dfrac{-6\pm\sqrt{32}}{2}\end{matrix}\right.\)