( x - 7 )( x² - 9x + 20 )( x - 2 ) = 72

⇔ [ ( x - 7 )( x - 2 ) ]( x² - 9x + 20 ) - 72 = 0

⇔ ( x² - 9x + 14 )( x² - 9x + 20 ) - 72 = 0

Đặt t = x² - 9x + 17

pt ⇔ ( t - 3 )( t + 3 ) - 72 = 0

    ⇔ t² - 9 - 72 = 0

    ⇔ t² - 81 = 0

    ⇔ ( t - 9 )( t + 9 ) = 0

    ⇔ ( x² - 9x + 17 - 9 )( x² - 9x + 17 + 9 ) = 0

    ⇔ ( x² - 9x + 8 )( x² - 9x + 26 ) = 0

    ⇔ ( x² - 8x - x + 8 )( x² - 9x + 26 ) = 0

    ⇔ [ x( x - 8 ) - ( x - 8 ) ]( x² - 9x + 26 ) = 0

    ⇔ ( x - 8 )( x - 1 )( x² - 9x + 26 ) = 0

Vì x² - 9x + 26 = ( x² - 9x + 81/4 ) + 23/4 = ( x - 9/2 )² + 23/4 ≥ 23/4 ∀ x

=> x - 8 = 0 hoặc x - 1 = 0

=> x = 8 hoặc x = 1

\(x^7+x^2+1=x^7-x+x^2+x+1\)

\(=\left(x^7-x\right)+\left(x^2+x+1\right)=x\left(x^6-1\right)+\left(x^2+x+1\right)\)

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

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

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

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

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

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

Cách nhẩm nghiệm: 

\(ax^2+bx+c=0\)

+) Nếu \(a+b+c=0\)thì phương trình có nghiệm là \(1\)

\(\Rightarrow\)Khi phân tích đa thức thành nhân từ thì sẽ chứa hạng tử \(x-1\)

+) Nếu \(a-b+c=0\)thì phương trình có nghiệm là \(-1\)

\(\Rightarrow\)Khi phân tích đa thức thành nhân tử thì sẽ chứa hạng tử \(x+1\)

\(x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}\)\(\Leftrightarrow x^2yz+xz=xy^2z+xy=xyz^2+yz\)

\(\Leftrightarrow\hept{\begin{cases}xyz\left(x-y\right)=x\left(y-z\right)\\xyz\left(y-z\right)=y\left(z-x\right)\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}xyz\left(x-y\right)=x\left(y-z\right)\\xyz\left(y-z\right)=y\left(z-x\right)\\xyz\left(z-x\right)=z\left(y-x\right)\end{cases}}\Leftrightarrow x^3y^3z^3.\left(x-y\right)\left(y-z\right)\left(z-x\right)=-xyz.\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x^3y^3z^3=-xyz\\\left(x-y\right)\left(y-z\right)\left(z-x\right)=0\end{cases}\Leftrightarrow\orbr{\begin{cases}xyz=\pm1\\x=y=z\end{cases}}}\)

\(3\left(x-1\right)^2+\left(x+5\right)\left(2-3x\right)=25\)

\(3x^2-6x+3-3x^2-13x+10=25\)

                                             \(-19x+13=25\)

                                                            \(-19=25-13\)

                                                          \(-19x=12\)

                                                                     \(x=\frac{-12}{19}\)

\(3\left(x-1\right)^2+\left(x+5\right)\left(2-3x\right)=25\)

\(3\left(x^2-2x+1\right)+2x-3x^2+10-15x=25\)

\(3x^2-6x+3+2x-3x^2+10-15x=25\)

\(-18x+13=25\)

\(-18x=12\)

\(x=-\frac{2}{3}\)

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

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

\(=\left(x-1\right)\left(y^2-x-1\right)\)

xy ² − y ² + 1 − x ² = (xy ² − y ² )+ (1 − x ² )

= y ² (x − 1 )− (x ² − 1)

= y ² (x − 1) −( x − 1)(x + 1)

 = (x − 1 ) (y ² − x − 1 )

Ta có

  \(x^2+x+1=x^2+2.\frac{1}{2}x+\frac{1}{4}+1-\frac{1}{4}\)

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

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

  Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

       \(\frac{3}{4}>0\)

  Nên \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)hay \(x^2+x+1>0\)

  mà theo đề bài \(x^2+x+1=0\)

    Vậy \(x\in\varnothing\)

x² - y² - 4 + 4y = x² - ( y² - 4y + 4 ) = x² - ( y - 2 )² = ( x - y + 2 )( x + y - 2 )

x² - y² - 4 + 4y

= x² - ( y ² - 4y + 4 )

= x ² - ( y - 2 )²

= ( x - y + 2 )( x + y - 2 )