Đặt A = x³ + y³ + xy 

= (x + y)(x² - xy + y²) + xy

= x² - xy + y² + xy (Vì x + y = 1)

= x² + y² 

Lại có x +y = 1

=> x = 1 - y

Khi đó A = x² + y²

= (1 - y)² + y²

= 1 - 2y + y² + y²

= 2y² - 2y  +1 = \(2\left(y^2-y+\frac{1}{2}\right)=2\left(y^2-2.\frac{1}{2}y+\frac{1}{4}+\frac{1}{4}\right)=2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

Dấu "=" xảy ra <=> \(y-\frac{1}{2}=0\Leftrightarrow y=\frac{1}{2}\Leftrightarrow x=\frac{1}{2}\)

Vậy Min A = \(\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

ta có

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

                            \(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

   \(\Rightarrow x^2+x+1>0\)

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

 \(\Leftrightarrow6-2x=0\)

\(\Leftrightarrow x=3\)

   Vậy ....................

Ta có : \(\left(x^2+x+1\right)\left(6-2x\right)=0\)

Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\ne0\forall x\)

\(\Rightarrow6-2x=0\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

Vậy \(x=3\)

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

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

Đặt \(x^2-x=t\)

\(\Leftrightarrow t\left(t+1\right)-6=0\)

\(\Leftrightarrow t^2+t-6=0\)

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

\(\Leftrightarrow\left(x^2-x-2\right)\left(x^2-x+3\ne0\right)=0\)

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

Vậy tập nghiệm của phương trình là S = { -1 ; 2 }

x( x - 1 )( x² - x + 1 ) - 6 = 0

<=> ( x² - x )( x² - x + 1 ) - 6 = 0

Đặt t = x² - x

pt <=> t( t + 1 ) - 6 = 0

<=> t² + t - 6 = 0

<=> t² - 2t + 3t - 6 = 0

<=> t( t - 2 ) + 3( t - 2 ) = 0

<=> ( t - 2 )( t + 3 ) = 0

<=> ( x² - x - 2 )( x² - x + 3 ) = 0

<=> ( x² - 2x + x - 2 )( x² - x + 3 ) = 0

<=> [ x( x - 2 ) + ( x - 2 ) ]( x² - x + 3 ) = 0

<=> ( x - 2 )( x + 1 )( x² - x + 3 ) = 0

Vì x² - x + 3 > 0 ( bạn tự cm vì lười quá @@ )

=> ( x - 2 )( x + 1 ) = 0

<=> x = 2 hoặc x = -1

Vậy ...

 \(a^3-a^2b+ab^2-6b^3=0\)

\(\Leftrightarrow\left(a^3-a^2b\right)+\left(a^2b-ab^2\right)+\left(3ab^2-6b^3\right)=0\)

\(\Leftrightarrow a^2\left(a-2b\right)+ab\left(a-2b\right)+3b^2\left(a-2b\right)=0\)          

\(\Leftrightarrow\left(a-2b\right)\left(a^2+ab+3b^2\right)=0\left(1\right)\)

Vì \(a>b>0\Rightarrow a^2+ab+3b^2>0\)nên từ (1) ta có \(a-2b=0\Leftrightarrow a=2b\)

Giá trị biểu thức \(P=\frac{a^4-4b^4}{b^4-4a^4}=\frac{16b^4-4b^4}{b^4-64b^4}=\frac{12b^4}{-63b^4}=-\frac{4}{21}\)

Từ \(x=by+cz\)\(\Rightarrow by=x-cz\)

\(\Rightarrow b=\frac{x-cz}{y}\)

\(\Rightarrow1+b=\frac{x+y-cz}{y}=\frac{ax+by+2cz-cz}{y}\)

                 \(=\frac{ax+by+cz}{y}=\frac{\frac{x+y+z}{2}}{y}=\frac{x+y+z}{2y}\)

Chứng minh tương tự , ta được : 

\(1+a=\frac{x+y+z}{2x}\)

\(1+c=\frac{x+y+z}{2z}\)

\(\Rightarrow Q=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\)

\(=\frac{1}{\frac{x+y+z}{2x}}+\frac{1}{\frac{x+y+z}{2y}}+\frac{1}{\frac{x+y+z}{2z}}\)

\(=\frac{2x}{x+y+z}+\frac{2y}{x+y+z}+\frac{2z}{x+y+z}\)

\(=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

Vậy \(Q=2\)

Ta có: \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=1\frac{1993}{1995}\)   ( ĐK: \(x\ne0,\)\(x\ne-1\))

    \(\Leftrightarrow2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)=1\frac{1993}{1995}\)

    \(\Leftrightarrow2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{3988}{1995}\)

    \(\Leftrightarrow1-\frac{1}{x+1}=\frac{1994}{1995}\)

    \(\Leftrightarrow\frac{1}{x+1}=\frac{1}{1995}\)

    \(\Leftrightarrow x+1=1995\)

    \(\Leftrightarrow x=1994\)\(\left(TM\right)\)

Vậy..........

Ta có: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)=24\)

\(\Leftrightarrow\left[\left(x^2+5x\right)^2+4\left(x^2+5x\right)\right]-\left[6\left(x^2+5x\right)+24\right]=0\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+4\right)-6\left(x^2+5x+4\right)=0\)

\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)

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

\(\Rightarrow x\in\left\{-6;-4;-1;1\right\}\)

\(\left(x^2+5x\right)^2-2\left(x^2+5x\right)=24\)

\(\Leftrightarrow\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)

Đặt: \(x^2+5x=t\)

\(\Rightarrow t^2-2t-24=0\)

\(\Leftrightarrow t^2+4t-6t-24=0\)

\(\Leftrightarrow t\left(t+4\right)-6\left(t+4\right)=0\)

\(\Leftrightarrow\left(t+4\right)\left(t-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+4=0\\t-6=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}t=-4\\t=6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2+5x=-4\\x^2+5x=6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+5x+4=0\\x^2+5x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1;x=-4\\x=1;x=-6\end{cases}}}\)

Vậy: \(S=\left\{-1;-4;1;-6\right\}\)