Chia đa thức cho đơn thức

Ta có $ \left[ {{\left( x+1 \right)}^{3}}-4{{\left( x+1 \right)}^{2}}+3\left( x+1 \right) \right]:\left( x+1 \right)=0 $

$ \begin{array}{l} \Leftrightarrow {{\left( x+1 \right)}^{2}}-4\left( x+1 \right)+3=0 \\ \Leftrightarrow {{x}^{2}}+2x+1-4x-4+3=0 \\ \Leftrightarrow {{x}^{2}}-2x=0 \\ \Leftrightarrow x\left( x-2 \right)=0 \\ \Leftrightarrow \left[ \begin{array}{l} x=0 \\ x=2 \end{array} \right. \end{array} $

Ta có $ \left[ 3{{\left( a-b \right)}^{5}}+{{\left( a-b \right)}^{4}}-2{{\left( b-a \right)}^{2}} \right]:{{\left( b-a \right)}^{2}} $

$ \begin{array}{l} =\left[ 3{{\left( a-b \right)}^{5}}+{{\left( a-b \right)}^{4}}-2{{\left( a-b \right)}^{2}} \right]:{{\left( a-b \right)}^{2}} \\ =3{{\left( a-b \right)}^{3}}+{{\left( a-b \right)}^{2}}-2 \end{array} $

Ta có

$ \left( -12{{x}^{3}}{{y}^{5}}+18{{x}^{3}}{{y}^{4}}+6{{x}^{7}}{{y}^{2}} \right):3x{{y}^{2}} $

$ \begin{array}{l} =\left( -12{{x}^{3}}{{y}^{5}}:3x{{y}^{2}} \right)+\left( 18{{x}^{3}}{{y}^{4}}:3x{{y}^{2}} \right)+\left( 6{{x}^{7}}{{y}^{2}}:3x{{y}^{2}} \right) \\ =-4{{x}^{2}}{{y}^{3}}+6{{x}^{2}}{{y}^{2}}+2{{x}^{6}} \end{array} $

Ta có

$ \begin{array}{l} \left( {{x}^{3}}+8{{y}^{3}} \right):\left( x+2y \right) \\ =\left[ {{x}^{3}}+{{\left( 2y \right)}^{3}} \right]:\left( x+2y \right) \\ =\left( x+2y \right)\left( {{x}^{2}}-2xy+4{{y}^{2}} \right):\left( x+2y \right) \\ ={{x}^{2}}-2xy+4{{y}^{2}} \end{array} $

Ta có $ (8{{x}^{5}}{{y}^{7}}+5{{x}^{8}}{{y}^{4}}-{{x}^{4}}{{y}^{3}}):7{{x}^{n}}y $ là phép chia hết khi $ 0 < n\le 4 $

Vậy có 4 giá trị của n thỏa mãn : $ n\in \left\{ 1;2;3;\left. 4 \right\} \right. $ .

Ta có

$ \begin{array}{l} \left[ 12{{\left( 2x-y \right)}^{3}}+4{{x}^{2}}-4xy+{{y}^{2}} \right]:\left( 6x-3y \right) \\ =\left[ 12{{\left( 2x-y \right)}^{3}}+{{\left( 2x-y \right)}^{2}} \right]:\left[ 3\left( 2x-y \right) \right] \\ =12{{\left( 2x-y \right)}^{3}}:\left[ 3\left( 2x-y \right) \right]+{{\left( 2x-y \right)}^{2}}:\left[ 3\left( 2x-y \right) \right] \\ =4{{\left( 2x-y \right)}^{2}}+\dfrac{1}{3}\left( 2x-y \right) \end{array} $

Ta có $ \left( {{x}^{5}}~+7{{x}^{2}}-4{{x}^{3}}+{{x}^{4}} \right):2{{x}^{2}} $

$ \begin{array}{l} =\left( {{x}^{5}}:2{{x}^{2}}~ \right)+\left( 7{{x}^{2}}:2{{x}^{2}} \right)-\left( 4{{x}^{3}}:2{{x}^{2}} \right)+\left( {{x}^{4}}:2{{x}^{2}} \right) \\ =\dfrac{1}{2}{{x}^{3}}+\dfrac{7}{2}-2x+\dfrac{1}{2}{{x}^{2}} \end{array} $

Ta có

$ \begin{array}{l} \left( 9{{x}^{3}}{{y}^{2}}+6{{x}^{5}}{{y}^{3}}-12{{x}^{4}}y \right):3{{x}^{2}}y \\ =\left( 9{{x}^{3}}{{y}^{2}}:3{{x}^{2}}y \right)+\left( 6{{x}^{5}}{{y}^{3}}:3{{x}^{2}}y \right)-\left( 12{{x}^{4}}y:3{{x}^{2}}y \right) \\ =3xy+2{{x}^{3}}{{y}^{2}}-4{{x}^{2}} \end{array} $

Với $ x\ne 0 $ , ta có

$ \begin{array}{l} \left( {{x}^{3}}+4{{x}^{2}}+4x \right):x=0 \\ \Leftrightarrow {{x}^{2}}+4x+4=0 \\ \Leftrightarrow {{\left( x+2 \right)}^{2}}=0 \\ \Leftrightarrow x=-2 \end{array} $

Ta có

$ \left[ 3{{\left( x-y \right)}^{4}}+2{{\left( x-y \right)}^{3}}-5{{\left( x-y \right)}^{2}} \right]:{{\left( y-x \right)}^{2}} $

$ =\left[ 3{{\left( x-y \right)}^{4}}+2{{\left( x-y \right)}^{3}}-5{{\left( x-y \right)}^{2}} \right]:\left[ {{\left( x-y \right)}^{2}} \right] $

$ =\left[ 3{{\left( x-y \right)}^{4}}+2{{\left( x-y \right)}^{3}}-5{{\left( x-y \right)}^{2}} \right]:{{\left( x-y \right)}^{2}} $

$ =3{{\left( x-y \right)}^{4}}:{{\left( x-y \right)}^{2}}+2{{\left( x-y \right)}^{3}}:{{\left( x-y \right)}^{2}}+\left[ -5{{\left( x-y \right)}^{2}}:{{\left( x-y \right)}^{2}} \right] $

$ =3{{\left( x-y \right)}^{2}}+2\left( x-y \right)-5 $

Ta có $ (3{{x}^{2}}{{y}^{2}}-4x{{y}^{3}}+2x{{y}^{4}}):(-\dfrac{1}{2}x{{y}^{2}})=-6x+8y-4{{y}^{2}} $ .

Ta có $ 20{{x}^{4}}y-25{{x}^{2}}{{y}^{2}}-3{{x}^{2}}y=5{{x}^{2}}y\left( 4{{x}^{2}}-5y-\dfrac{3}{5} \right) $

Nên $ \left( 20{{x}^{4}}y-25{{x}^{2}}{{y}^{2}}-3{{x}^{2}}y \right):5{{x}^{2}}y=4{{x}^{2}}-5y-\dfrac{3}{5} $

Với $ x\ne 0 $ , ta có

$ \begin{array}{l} \left( 2{{x}^{7}}{{y}^{5}}-5{{x}^{2}}{{y}^{3}}+4xy \right):\dfrac{1}{2}xy \\ =2{{x}^{7}}{{y}^{5}}:\dfrac{1}{2}xy-5{{x}^{2}}{{y}^{3}}:\dfrac{1}{2}xy+4xy:\dfrac{1}{2}xy \\ =4{{x}^{6}}{{y}^{4}}-10x{{y}^{2}}+8 \end{array} $

Ta có $ (3{{x}^{5}}{{y}^{2}}+4{{x}^{3}}{{y}^{2}}-8{{x}^{2}}{{y}^{2}}):2{{x}^{2}}{{y}^{2}} $

$ =(3{{x}^{5}}{{y}^{2}}:2{{x}^{2}}{{y}^{2}})+(4{{x}^{3}}{{y}^{2}}:2{{x}^{2}}{{y}^{2}})-(8{{x}^{2}}{{y}^{2}}:2{{x}^{2}}{{y}^{2}}) $

$ =\dfrac{3}{2}{{x}^{3}}+2x-4 $ .