\(\left(x^2+x-2\right)\left(x^2+x-3\right)=12\)

Đặt \(x^2+x-2=t\)

\(\Leftrightarrow t\left(t-1\right)=12\)

\(\Leftrightarrow t^2-t-12=0\Leftrightarrow\left(t-4\right)\left(t-3\right)=0\Leftrightarrow t=4;t=3\)

hay \(x^2+x-6=0\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\Leftrightarrow x=2;x=-3\)

\(x^2+x-5\ne0\)tự chứng minh 

Vậy tập nghiệm của phương trình là S = { -3 ; 2 } 

Đặt \(a=x^2+x-2\)\(\Rightarrow\)\(a-1=x^2+x-3\)

Ta có: \(a.\left(a-1\right)=12\)

    \(\Leftrightarrow a^2-a-12=0\)

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

    \(\Leftrightarrow\left(a-4\right).\left(a+3\right)=0\)

    \(\Leftrightarrow\orbr{\begin{cases}a-4=0\\a+3=0\end{cases}}\)

    \(\Leftrightarrow\orbr{\begin{cases}a=4\\a=-3\end{cases}}\)

+ \(a=4\)\(\Rightarrow\)\(x^2+x-2=4\)

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

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

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

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

+ \(a=-3\)\(\Rightarrow\)\(x^2+x-2=-3\)

                   \(\Leftrightarrow\)\(x^2+x+1=0\)

                   \(\Leftrightarrow\)\(\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=0\)

                   \(\Leftrightarrow\)\(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\)( * )

 Vì \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)\(\Rightarrow\)Đa thức ( * ) ko có giá trị

Vậy ............

\(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)

Đặt \(x^2+x+1=t\)

\(\Leftrightarrow t\left(t+1\right)=12\Leftrightarrow t^2+t-12=0\)

\(\Leftrightarrow\left(t-3\right)\left(t+4\right)=0\)

\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x+5\ne0\right)=0\)( tự chứng minh )

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

\(\Leftrightarrow x=1;x=-2\)

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

Đặt \(x^2+5x=t\)

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

hay \(x^2+5x-6=0\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\Leftrightarrow x=1;x=-6\)

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

Vậy tập nghiệm phương trình là S = { \(\pm\)1 ; -6 ; -4 }

Ta có: \(\left(x^2+5x\right)^2-2.\left(x^2+5x\right)-24=0\)

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

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

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

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

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

Suy ra: \(x=-1\)hoặc \(x=-4\)hoặc \(x=1\)hoặc \(x=-6\)

Vậy .........

(x² - 5x)² + 10(x² - 5x) + 24 = 0

<=> (x² - 5x)² + 10(x² - 5x) + 25 - 1 = 0

<=> (x² - 5x + 5)² - 1 = 0

<=> (x² - 5x + 4)(x² - 5x + 6) = 0

<=> (x - 1)(x - 4)(x - 2)(x - 3) = 0

<=> x - 1 = 0 hoặc x - 4 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0

<=> x = 1 hoặc x = 4 hoặc x = 2 hoặc x = 3

Vậy \(x\in\left\{1;2;3;4\right\}\)là nghiệm phương trình

Đặt x² - 5x = t

pt <=> t² + 10t + 24 = 0

<=> t² + 4t + 6t + 24 = 0

<=> t( t + 4 ) + 6( t + 4 ) = 0

<=> ( t + 4 )( t + 6 ) = 0

<=> ( x² - 5x + 4 )( x² - 5x + 6 ) = 0

<=> ( x² - x - 4x + 4 )( x² - 2x - 3x + 6 ) = 0

<=> [ x( x - 1 ) - 4( x - 1 ) ][ x( x - 2 ) - 3( x - 2 ) ] = 0

<=> ( x - 1 )( x - 2 )( x - 3 )( x - 4 ) = 0

<=> x = 1 hoặc x = 2 hoặc x = 3 hoặc x = 4

Vậy phương trình có tập nghiệm S = { 1 ; 2 ; 3 ; 4 }

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-2x-1=0\\x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=2\\\left(x+\frac{1}{2}\right)^2=\frac{5}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=\pm\sqrt{2}\\x+\frac{1}{2}=\pm\frac{\sqrt{5}}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\pm\sqrt{2}\\x=-\frac{1\pm\sqrt{5}}{2}\end{cases}}\)

2) Ta có: \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=0\)

\(\Leftrightarrow\left[\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)\right]+\left[2x\left(x^2+4x+8\right)+2x^2\right]=0\)

\(\Leftrightarrow\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)=0\)

\(\Leftrightarrow\left(x^2+6x+8\right)\left(x^2+5x+8\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)=0\)

Vì \(x^2+5x+8=\left(x^2+5x+\frac{25}{4}\right)+\frac{7}{4}=\left(x+\frac{5}{2}\right)^2+\frac{7}{4}>0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}\)

Vậy x = -2 hoặc x = -4

\(A=\frac{4x}{x^2+2x}+\frac{3}{2-x}+\frac{12}{x^2-4}\)(ĐK: \(x\ne0,x\ne\pm2\))

\(A=\frac{4}{x+2}+\frac{3}{2-x}+\frac{12}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4.\left(x-2\right)-3\left(x+2\right)+12}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4x-8-3x-6+12}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{x-2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x+2}\)

\(A=\frac{4x}{x^2+2x}+\frac{3}{2-x}+\frac{12}{x^2-4x}\)

\(=\frac{4x}{x\left(x+2\right)}-\frac{3}{x-2}+\frac{12}{x\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}-\frac{3x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{12}{x\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x^2-8x-3x^2-6x+12}{x\left(x-2\right)\left(x+2\right)}=\frac{x^2-14x+12}{x\left(x-2\right)\left(x+2\right)}\)

Ta có: \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right).\left(\frac{x+2003}{x}\right)\)    \(\left(ĐK:x\ne\pm1;x\ne0\right)\)

    \(\Leftrightarrow A=\left(\frac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x^2-4x-1\right)}{\left(x-1\right).\left(x+1\right)}\right).\left(\frac{x+2003}{x}\right)\)

    \(\Leftrightarrow A=\left(\frac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right).\left(x+1\right)}\right).\left(\frac{x+2003}{x}\right)\)

    \(\Leftrightarrow A=\left(\frac{x^2-1}{x^2-1}\right).\left(\frac{x+2003}{x}\right)\)

    \(\Leftrightarrow A=\frac{x+2003}{x}\)

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right)\left(\frac{x+2003}{x}\right)\)

\(=\left(\frac{\left(x+1\right)^2-\left(x-1\right)^2+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\left(\frac{x-2003}{x}\right)\)

\(=\left(\frac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\left(\frac{x-2003}{x}\right)\)

\(=\left(\frac{x^2-1}{\left(x-1\right)\left(x+1\right)}\right)\left(\frac{x-2003}{x}\right)=\frac{x-2003}{x}\)

\(3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)

Ta có: \(3x^2-6xy+3y^2-12z^2\)

    \(=3.\left(x^2-2xy+y^2-4z^2\right)\)

    \(=3.\left[\left(x-y\right)^2-4z^2\right]\)

    \(=3.\left(x-y-2z\right).\left(x-y+2z\right)\)

Theo bài ra ta có : 

\(x^3-4x^2+x=x\left(x^2-4x+1\right)\)

Gớm Tú ơi, làm gì mà Dis nhiều thế :)) Nghiếp khiếp vậy mày:))))