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

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

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: \(S=\left\{-1;-4;1;-6\right\}\)

Ta có: \(2a^2+a=3b^2+b\)

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

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

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

*CM 2a+2b+1 và a-b nguyên tố cùng nhau

=> 2a+2b+1 cũng là 1 SCP

Ta có: 

\(2a^2+a=3b^2+b\)

\(\Leftrightarrow2a^2-2b^2+a-b=b^2\)

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

Ta có: 

Đặt \(d=\left(a-b,2a+2b+1\right)\).

\(\Rightarrow\hept{\begin{cases}a-b⋮d\\2a+2b+1⋮d\end{cases}}\Rightarrow\left(a-b\right)\left(2a+2b+1\right)=b^2⋮d^2\Rightarrow b⋮d\)

\(\Rightarrow\left(a-b\right)+b=a⋮d\)

\(\Rightarrow\left(2a+2b+1\right)-2a-2b=1⋮d\Rightarrow d=1\).

Do đó \(a-b,2a+2b+1\)là hai số chính phương. 

\(A=n^3+n^2-n+2=\left(n+2\right)\left(n^2-n+1\right)\)là số nguyên tố suy ra 

\(\orbr{\begin{cases}n+2=1\\n^2-n+1=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}n=-1\\n=1;n=0\end{cases}}\)

Thử lại đều thỏa mãn.