Ta có :

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-4x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\\left(x-2\right)^2-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\pm\sqrt{3}\end{cases}}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-1;2+\sqrt{3};2-\sqrt{3}\right\}\)

a) \(8x^2-12xy+4y^2-2x-1\)

\(=4y^2-8xy-2y-4xy+8x^2+2x+2y-4x-1\)

\(=\left(4y^2-8xy-2y\right)-\left(4xy-8x^2-2x\right)+\left(2y-4x-1\right)\)

\(=2y\left(2y-4x-1\right)-2x\left(2y-4x-1\right)+\left(2y-4x-1\right)\)

\(=\left(2y-2x+1\right)\left(2y-4x-1\right)\)

b) \(4x^4+16\)

\(=4x^4+16x^2+16-16x^2\)

\(=\left(2x^2+4\right)^2-\left(4x\right)^2\)

\(=\left(2x^2+4x+4\right)\left(2x^2-4x+4\right)\)

\(=4\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)

Áp dụng bđt Cauchy schwarz:

=> 1/x+1/y+4/z+16/t >= [(1+1+2+4)^2] / x+y+z+t=8^2/(x+y+z+t)=64/1=64

=> đpcm.

Áp dụng BĐT Svac - xơ:

\(\frac{1}{x}+\frac{1}{y}+\frac{4}{z}+\frac{16}{t}\ge\frac{\left(1+1+2+4\right)^2}{x+y+z+t}=\frac{64}{1}=64\)

(Dấu "="\(\Leftrightarrow x=y=\frac{1}{22};z=\frac{2}{11};t=\frac{8}{11}\))

\(a+b=1\)\(\Rightarrow\hept{\begin{cases}a-1=-b\\b-1=-a\end{cases}}\)

Ta có: \(\frac{a}{b^3-1}-\frac{b}{a^3-1}=\frac{a}{\left(b-1\right)^3+3b\left(b-1\right)}-\frac{b}{\left(a-1\right)^3+3a\left(a-1\right)}\)

\(=\frac{a}{-a^3-3ab}-\frac{b}{-b^3-3ab}=\frac{a}{-a\left(a^2+3b\right)}-\frac{b}{-b\left(b^2+3a\right)}\)

\(=\frac{-1}{a^2+3b}-\frac{-1}{b^2+3a}=\frac{-1}{a^2+3b}+\frac{1}{b^2+3a}=\frac{-\left(b^2+3a\right)+a^2+3b}{\left(a^2+3b\right)\left(b^2+3a\right)}\)

\(=\frac{-b^2-3a+a^2+3b}{a^2b^2+3a^3+3b^3+9ab}=\frac{-\left(b^2-a^2\right)+\left(3b-3a\right)}{a^2b^2+3\left(a^3+b^3\right)+9ab}\)

\(=\frac{-\left(b-a\right)\left(b+a\right)+3\left(b-a\right)}{a^2b^2+3\left[\left(a+b\right)^3-3ab\left(a+b\right)\right]+9ab}=\frac{-\left(b-a\right)+3\left(b-a\right)}{a^2b^2+3\left[1-3ab\right]+9ab}\)

\(=\frac{2\left(b-a\right)}{a^2b^2+3-9ab+9ab}=\frac{2\left(b-a\right)}{a^2b^2+3}\left(đpcm\right)\)

Tớ học ngu nên chỉ biết cách nhân ra rồi rút gọn chứ không biết cách nào ngắn hơn :)) Hơi dài dòng nên phân tích từng vế 1 nhé :D

2/ \(\left(2x^2+5x-204\right)^2+4\left(x^2-5x-206\right)=4\left(2x^2+5x-204\right)\left(x^2-5x-206\right)\)

*****\(VT=\left(2x^2+5x-204\right)^2+4\left(x^2-5x-206\right)^2\)

\(=4x^4+25x^2+41616+20x^3-816x^2-2040x+4\left(x^4-387x^2+42436-10x^3+2060x\right)\)

\(=4x^2+25x^2+41616+20x^3-816x^2-2040x+4x^2-1548x^2+169744-40x^3+8240x\)

\(=8x^4-1523x^2+6200x+211360\)

*****\(VP=\left(8x^2+20x-816\right)\left(x^2-5x-206\right)\)

\(=8x^4-40x^3-1648x^2-100x^2-4120x-816x^2+4080x+168096\)

\(=8x^4-1748x^2-40x+168096\)

\(\Rightarrow8x^4-1523x^2+6200x+211360=8x^4-1748x^2-40x+168096\)

\(\Leftrightarrow-1523x^2+6200x+211360+1748x^2-40x+168096=0\)

\(\Leftrightarrow255x^2+43264+6240x=0\)

\(\Leftrightarrow\left(15x+208\right)^2=0\)

\(\Leftrightarrow15x+208=0\)

\(\Leftrightarrow x=-\frac{208}{15}\)

+ Ta có: \(x^4-5x^3+6x^2+5x+1=0\)

        \(\Rightarrow x^2-5x+6+\frac{5}{x}+\frac{1}{x^2}=0\)( chia cả hai vế cho \(x^2\))

       \(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-\left(5x-\frac{5}{x}\right)+6=0\)

      \(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-5.\left(x-\frac{1}{x}\right)+6=0\)( *** )

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

- Thay  \(a=x-\frac{1}{x};\)\(a^2+2=x^2+\frac{1}{x^2}\)vào ( *** )

- Ta có: \(a^2+2-5a+6=0\)

     \(\Leftrightarrow a^2-5a+8=0\)

     \(\Leftrightarrow4a^2-20a+32=0\)

     \(\Leftrightarrow\left(4a^2-20a+25\right)+7=0\)

     \(\Leftrightarrow\left(2a-5\right)^2+7=0\)

- Ta lại có: \(\hept{\begin{cases}\left(2a-5\right)^2\ge0\forall a\\7>0\end{cases}}\Rightarrow \left(2a-5\right)^2+7\ge7>0\)mà \(\left(2a-5\right)^2+7=0\)

\(\Rightarrow\left(2a-5\right)^2+7\)( vô nghiệm ) \(\Rightarrow\)\(x^4-5x^3+6x^2+5x+1=0\)( vô nghiệm )

Vậy \(S=\left\{\varnothing\right\}\)

+ Ta có: \(\left(2x^2+5x-204\right)^2+4.\left(x^2-5x-206\right)=4.\left(2x^2+5x-204\right).\left(x^2-5x-206\right)\)( ** )

- Đặt \(a=2x^2+5x-204;\)\(b=x^2-5x-206\)\(\Rightarrow\)\(a.b=\left(2x^2+5x-204\right).\left(x^2-5x-206\right)\)

- Thay \(a=2x^2+5x-204;\)\(b=x^2-5x-206\)\(\Rightarrow\)\(a.b=\left(2x^2+5x-204\right).\left(x^2-5x-206\right)\)

vào ( ** )

- Ta có: \(a^2+4b^2=4ab\)

      \(\Leftrightarrow a^2-4ab+4b^2=0\)

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

      \(\Leftrightarrow a-2b=0\)

      \(\Leftrightarrow a=2b\)( * )

- Thay  \(a=2x^2+5x-204;\)\(b=x^2-5x-206\)vào ( * )

- Ta lại có: \(2x^2+5x-204=2.\left(x^2-5x-206\right)\)

       \(\Leftrightarrow2x^2+5x-204=2x^2-10x-412\)

      \(\Leftrightarrow\left(2x^2-2x^2\right)+\left(5x+10x\right)=-\left(412-204\right)\)

      \(\Leftrightarrow15x=-208\)

      \(\Leftrightarrow x=-\frac{208}{15} \left(TM\right)\)

Vậy \(S=\left\{-\frac{208}{15}\right\}\)