Chia đơn thức cho đơn thức

Ta có

  \(\begin{array}{l}\left( {7{x^4} - 21{x^3}} \right):7{x^2} + \left( {10x + 5{x^2}} \right):5x\\ = 7{x^4}:\left( {7{x^2}} \right) - 21{x^3}:\left( {7{x^2}} \right) + 10x:\left( {5x} \right) + 5{x^2}:\left( {5x} \right)\\ = {x^2} - 3x + 2 + x\\ = {x^2} - 2x + 2\end{array}\)

Để phép chia \({x^{2n}}:{x^4} = {x^{2n - 4}}\) thực hiện được thì \(n \in \mathbb{N};\,2n - 4 \ge 0 \Leftrightarrow \,n \ge 2;\,n \in \mathbb{N}\) .

Ta có: \(B:C = \left( {4{x^4}{y^4}} \right):\left( {{x^{n - 1}}{y^4}} \right)\)

Đơn thức B chia hết cho đơn thức C khi: \(4 \ge n - 1\)\( \Rightarrow n \le 5\)

Hay \(0<n\le 5.\)

\(D = \left( {15x{y^2} + 18x{y^3} + 16{y^2}} \right):6{y^2} - 7{x^4}{y^3}:{x^4}y\)

\( \Leftrightarrow D = 15x{y^2}:\left( {6{y^2}} \right) + 18x{y^3}:\left( {6{y^2}} \right) + 16{y^2}:\left( {6{y^2}} \right) \)\(- 7{x^4}{y^3}:{x^4}y\)

\( \Leftrightarrow D = \dfrac{5}{2}x + 3xy + \dfrac{8}{3} - 7{y^2}.\)

Tại \(x = \dfrac{2}{3}\) và \(y = 1\) ta có:

 \(D = \dfrac{5}{2}.\dfrac{2}{3} + 3.\dfrac{2}{3}.1 + \dfrac{8}{3} - {7.1^2} = \dfrac{5}{3} + 2 + \dfrac{8}{3} - 7 = \dfrac{{13}}{3} - 5 =  - \dfrac{2}{3}.\)

\(D = \left( {9{x^2}{y^2} - 6{x^2}{y^3}} \right):{\left( { - 3xy} \right)^2} + \left( {6{x^2}y + 2{x^4}} \right):\left( {2{x^2}} \right)\)

\( \Leftrightarrow D = 9{x^2}{y^2}:{\left( { - 3xy} \right)^2} - 6{x^2}{y^3}:{\left( { - 3xy} \right)^2} \)\(+ 6{x^2}y:\left( {2{x^2}} \right) + 2{x^4}:\left( {2{x^2}} \right)\)

\( \Leftrightarrow D = 9{x^2}{y^2}:\left( {9{x^2}{y^2}} \right) - 6{x^2}{y^3}:\left( {9{x^2}{y^2}} \right) \)\(+ 6{x^2}y:\left( {2{x^2}} \right) + 2{x^4}:\left( {2{x^2}} \right)\)

\( \Leftrightarrow D = 1 - \dfrac{2}{3}y + 3y + {x^2}\)

\( \Leftrightarrow D = {x^2} + \dfrac{7}{3}y + 1\)

Đa thức \(D = {x^2} + \dfrac{7}{3}y + 1\) có bậc \(2.\)

\(\begin{array}{l}\,\,\,\,\,{\left( { - xy} \right)^6}:{\left( {2xy} \right)^4} = \left( {{x^6}{y^6}} \right):\left( {{2^4}{x^4}{y^4}} \right) = \dfrac{1}{{{2^4}}}{x^{6 - 4}}{y^{6 - 4}}\\ = \dfrac{1}{{{2^4}}}{x^2}{y^2} = {\left( {\dfrac{1}{{{2^2}}}xy} \right)^2} = {\left( {\dfrac{1}{4}xy} \right)^2}.\end{array}\)

\(\begin{array}{l}\,\,\,\,\left( { - 12{x^4}y + 4{x^3} - 8{x^2}{y^2}} \right):\left( { - 4{x^2}} \right)\\ = \left( { - 12{x^4}y} \right):\left( { - 4{x^2}} \right) + \left( {4{x^3}} \right):\left( { - 4{x^2}} \right) - \left( {8{x^2}{y^2}} \right):\left( { - 4{x^2}} \right)\\ = 3{x^2}y - x + 2{y^2}.\end{array}\)

\(\begin{array}{l}\,\,\,\,\left( {6x{y^2} + 4{x^2}y - 2{x^3}} \right):2x\\ = 6x{y^2}:2x + 4{x^2}y:2x - 2{x^3}:2x\\ = 3{y^2} + 2x - {x^2}.\end{array}\)

Ta có

 \(\begin{array}{l}\left( {4{x^2}y{z^4} + 2{x^2}{y^2}{z^2} - 3xyz} \right):xy\\ = \left( {4{x^2}y{z^4}} \right):\left( {xy} \right) + \left( {2{x^2}{y^2}{z^2}} \right):\left( {xy} \right) - \left( {3xyz} \right):\left( {xy} \right)\\ = 4x{z^4} + 2xy{z^2} - 3z.\end{array}\)

Ta có: \(20{x^5}{y^3}:4{x^2}{y^2} = \left( {20:4} \right).\left( {{x^5}:{x^2}} \right).\left( {{y^3}:{y^2}} \right)\) \( = 5{x^3}y\) nên A sai.

+ \(12{x^3}{y^4}:\dfrac{2}{5}x{y^4} = \left( {12:\dfrac{2}{5}} \right)\left( {{x^3}:x} \right)\left( {{y^4}:{y^4}} \right)\) \( = 30{x^2}\) nên B sai.

+ \(2{\left( {x + y} \right)^3}:5\left( {x + y} \right) = \dfrac{2}{5}\left[ {{{\left( {x + y} \right)}^3}:\left( {x + y} \right)} \right] = \dfrac{2}{5}{\left( {x + y} \right)^2}\) nên C sai.

+ \({x^2}y{z^3}:\left( { - {x^2}{z^3}} \right) =  - \left( {{x^2}:{x^2}} \right)y\left( {{z^3}:{z^3}} \right) =  - y\) nên D đúng.

Ta có: \(\left( { - 15{x^2}{y^6}} \right):\left( { - 5x{y^2}} \right) \)\(= \left( { - 15} \right):\left( { - 5} \right)\left( {{x^2}:x} \right)\left( {{y^6}:{y^2}} \right) = 3x{y^4}\)  nên A đúng.

+ \(\left( {8{x^5} - 4{x^3}} \right):\left( { - 2{x^3}} \right) \)\(= \left[ {8{x^5}:\left( { - 2{x^3}} \right)} \right] - \left[ {4{x^3}:\left( { - 2{x^3}} \right)} \right]\) \( =  - 4{x^2} + 2\) nên B đúng.

+ \(\left( {3{x^2}{y^2} - 4x{y^2} + 2x{y^4}} \right):\left( { - \dfrac{1}{2}x{y^2}} \right)\)\( = \left[ {3{x^2}{y^2}:\left( { - \dfrac{1}{2}x{y^2}} \right)} \right] - \left[ {4x{y^2}:\left( { - \dfrac{1}{2}x{y^2}} \right)} \right] + \left[ {2x{y^4}:\left( { - \dfrac{1}{2}x{y^2}} \right)} \right]\)

\( =  - 6x + 8 - 4{y^2}\) nên C sai.

+ \({\left( {3x - y} \right)^{19}}:{\left( {3x - y} \right)^{10}} = {\left( {3x - y} \right)^{19 - 10}} = {\left( {3x - y} \right)^9}\) nên D đúng.

Ta có

\(\left( {\dfrac{1}{2}{a^2}{x^4} + \dfrac{4}{3}a{x^3} - \dfrac{2}{3}a{x^2}} \right):\left( { - \dfrac{2}{3}a{x^2}} \right)\)

\( = \left( {\dfrac{1}{2}{a^2}{x^4}} \right):\left( { - \dfrac{2}{3}a{x^2}} \right) + \left( {\dfrac{4}{3}a{x^3}} \right):\left( { - \dfrac{2}{3}a{x^2}} \right) \)\(- \left( {\dfrac{2}{3}a{x^2}} \right):\left( { - \dfrac{2}{3}a{x^2}} \right)\)

\( =  - \dfrac{3}{4}a{x^2} - 2x + 1.\)

Và

\(\left( {\dfrac{1}{2}{a^2}{x^4} + \dfrac{4}{3}a{x^3} - \dfrac{2}{3}a{x^2}} \right):\left( {a{x^2}} \right)\)

\( = \left( {\dfrac{1}{2}{a^2}{x^4}} \right):\left( {a{x^2}} \right) + \left( {\dfrac{4}{3}a{x^3}} \right):\left( {a{x^2}} \right) \)\(- \left( {\dfrac{2}{3}a{x^2}} \right):\left( {a{x^2}} \right)\)

\( = \dfrac{1}{2}a{x^2} + \dfrac{4}{3}x - \dfrac{2}{3}\)

Ta có \(A = {\left( {4{x^2}{y^2}} \right)^2}{\left( {x{y^3}} \right)^3} = {4^2}{\left( {{x^2}} \right)^2}{\left( {{y^2}} \right)^2}{x^3}{\left( {{y^3}} \right)^3}\)\( = 16{x^4}{y^4}{x^3}{y^9} = 16{x^7}{y^{13}}\) ; \(B = {\left( {{x^2}{y^3}} \right)^2} = {\left( {{x^2}} \right)^2}{\left( {{y^3}} \right)^2} = {x^4}{y^6}\)

Khi đó \(A:B = 16{x^7}{y^{13}}:{x^4}{y^6} = 16{x^3}{y^7}\) .

Ta có \(27{x^3} - 64{y^3} = {\left( {3x} \right)^3} - {\left( {4y} \right)^3} = \left( {3x - 4y} \right)\left( {{{\left( {3x} \right)}^2} + 3x.4y + {{\left( {4y} \right)}^2}} \right) = \left( {3x - 4y} \right)\left( {9{x^2} + 12xy + 16{y^2}} \right)\)

Vậy đa thức cần điền là \(9{x^2} + 12xy + 16{y^2}\).

Ta có  \(15{x^3}{y^4}:5{x^2}{y^2}\)\( = \left( {15:5} \right).\left( {{x^3}:{x^2}} \right).\left( {{y^4}:{y^2}} \right) = 3x{y^2}\) .

Ta có   \({\left( { - 3x} \right)^5}:{\left( { - 3x} \right)^2} \)\(= {\left( { - 3x} \right)^3} \)\(= {\left( { - 3} \right)^3}.{x^3} =  - 27{x^3}\).

Ta có  \(\left( {2{x^3} - {x^2} + 10x} \right):x\)\( = \left( {2{x^3}:x} \right) - \left( {{x^2}:x} \right) + \left( {10x:x} \right) = 2{x^2} - x + 10\)

Ta có  \(\left( {3{x^5}{y^2} + 4{x^3}{y^2} - 8{x^2}{y^2}} \right):2{x^2}{y^2}\)\( = \left( {3{x^5}{y^2}:2{x^2}{y^2}} \right) + \left( {4{x^3}{y^2}:2{x^2}{y^2}} \right) - \left( {8{x^2}{y^2}:2{x^2}{y^2}} \right)\)

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

Ta có: \(24{x^4}{y^3}:12{x^3}{y^3} \)\(= \left( {24:12} \right).\left( {{x^4}:{x^3}} \right).\left( {{y^3}:{y^3}} \right) \)\(= 2x\) nên A sai.

+ \(18{x^6}{y^5}:\left( { - 9{x^3}{y^3}} \right) \)\(= \left( {18:\left( { - 9} \right)} \right).\left( {{x^6}:{x^3}} \right).\left( {{y^5}:{y^3}} \right) \)\(=  - 2{x^3}{y^2}\) nên B sai.

+ \(40{x^5}{y^2}:\left( { - 2{x^4}{y^2}} \right) \)\(= \left( {40:\left( { - 2} \right)} \right).\left( {{x^5}:{x^4}} \right).\left( {{y^2}:{y^2}} \right) \)\(=  - 20x\) nên C đúng.

+ \(9{a^3}{b^4}{x^4}:3{a^2}{b^2}{x^2} \)\( = \left( {9:3} \right).\left( {{a^3}:{a^2}} \right).\left( {{b^4}:{b^2}} \right).\left( {{x^4}:{x^2}} \right) \)\(= 3a{b^2}{x^2}\) nên D sai.

Ta có: \({\left( {3x - y} \right)^7}:{\left( {y - 3x} \right)^2} = {\left( {3x - y} \right)^7}:{\left( {3x - y} \right)^2} = {\left( {3x - y} \right)^5}\)  nên A sai.

+ \({\left( {x - y} \right)^5}:{\left( {x - y} \right)^2} = {\left( {x - y} \right)^{5 - 2}} = {\left( {x - y} \right)^3}\) nên B đúng.

+ \({\left( {2x - 3y} \right)^9}:{\left( {2x - 3y} \right)^6} = {\left( {2x - 3y} \right)^{9 - 6}} = {\left( {2x - 3y} \right)^3}\) nên C đúng.

+ ${\left( {x - 2y} \right)^{50}}:{\left( {x - 2y} \right)^{21}} = {\left( {x - 2y} \right)^{50 - 21}} = {\left( {x - 2y} \right)^{29}}$ nên D đúng.