Cộng trừ đa thức

Ta có: $3y\left( {{x^2} - xy} \right) - 7{x^2}\left( {y + xy} \right)$$= 3{x^2}y - 3x{y^2} - 7{x^2}y - 7{x^3}y$$= \left( {3{x^2}y - 7{x^2}y} \right) - 3x{y^2} - 7{x^3}y$$=  - 4{x^2}y - 3x{y^2} - 7{x^3}y$

Ta có:

$\dfrac{1}{5}xy\left( {x + y} \right) - 2\left( {y{x^2} - x{y^2}} \right)$$= \dfrac{1}{5}{x^2}y + \dfrac{1}{5}x{y^2} - 2{x^2}y + 2x{y^2}$$= \left( {\dfrac{1}{5}x{y^2} + 2x{y^2}} \right) + \left( {\dfrac{1}{5}{x^2}y - 2{x^2}y} \right)$

$= \dfrac{{11}}{5}x{y^2} - \dfrac{9}{5}{x^2}y$

Ta có $4{x^3}yz - 4x{y^2}{z^2} - yz\left( {xyz + {x^3}} \right)$$= 4{x^3}yz - 4x{y^2}{z^2} - x{y^2}{z^2} - {x^3}yz$$= \left( {4{x^3}yz - {x^3}yz} \right) + \left( { - 4x{y^2}{z^2} - x{y^2}{z^2}} \right)$

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

Ta có $A + B + C = 4{x^2} - 5xy + 3{y^2} + \left( {3{x^2} + 2xy + {y^2}} \right) + \left( { - {x^2} + 3xy + 2{y^2}} \right)$

$= 4{x^2} - 5xy + 3{y^2} + 3{x^2} + 2xy + {y^2} - {x^2} + 3xy + 2{y^2}$

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

$= 6{x^2} + 6{y^2}$

Ta có $A - B - C = 4{x^2} - 5xy + 3{y^2} - \left( {3{x^2} + 2xy + {y^2}} \right) - \left( { - {x^2} + 3xy + 2{y^2}} \right)$

$= 4{x^2} - 5xy + 3{y^2} - 3{x^2} - 2xy - {y^2} + {x^2} - 3xy - 2{y^2}$

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

$= 2{x^2} - 10xy$

Ta có $C - A - B =  - {x^2} + 3xy + 2{y^2} - \left( {4{x^2} - 5xy + 3{y^2}} \right) - \left( {3{x^2} + 2xy + {y^2}} \right)$

$=  - {x^2} + 3xy + 2{y^2} - 4{x^2} + 5xy - 3{y^2} - 3{x^2} - 2xy - {y^2}$

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

$=  - 8{x^2} + 6xy - 2{y^2}$

Ta có $M + \left( {5{x^2} - 2xy} \right) = 6{x^2} + 10xy - {y^2}$$\Rightarrow M = 6{x^2} + 10xy - {y^2} - \left( {5{x^2} - 2xy} \right)$

$\Rightarrow M = 6{x^2} + 10xy - {y^2} - 5{x^2} + 2xy$

$\Rightarrow M = \left( {6{x^2} - 5{x^2}} \right) + \left( {10xy + 2xy} \right) - {y^2}$

$\Rightarrow M = {x^2} + 12xy - {y^2}$

Ta có $M - \left( {3xy - 4{y^2}} \right) = {x^2} - 7xy + 8{y^2}$

$\Rightarrow M = {x^2} - 7xy + 8{y^2} + 3xy - 4{y^2}$

$\Rightarrow M = {x^2} + \left( { - 7xy + 3xy} \right) + \left( {8{y^2} - 4{y^2}} \right)$

$\Rightarrow M = {x^2} - 4xy + 4{y^2}$

Ta có $\left( {25{x^2}y - 10x{y^2} + {y^3}} \right) - A = 12{x^2}y - 2{y^3}$$\Rightarrow A = \left( {25{x^2}y - 10x{y^2} + {y^3}} \right) - \left( {12{x^2}y - 2{y^3}} \right)$

$\Rightarrow A = 25{x^2}y - 10x{y^2} + {y^3} - 12{x^2}y + 2{y^3}$

$\Rightarrow A = \left( {25{x^2}y - 12{x^2}y} \right) + \left( {{y^3} + 2{y^3}} \right) - 10x{y^2}$

$\Rightarrow A = 13{x^2}y + 3{y^3} - 10x{y^2}$

Ta có $B + 3x{y^2} + 3x{z^2} - 3xyz - 8{y^2}{z^2} + 10 = 0$$\Rightarrow B =  - \left( {3x{y^2} + 3x{z^2} - 3xyz - 8{y^2}{z^2} + 10} \right)$

$\Rightarrow B =  - 3x{y^2} - 3x{z^2} + 3xyz + 8{y^2}{z^2} - 10$

Thay $x =  - 1;y =  - 1$ vào biểu thức C  ta được

$C = \left( { - 1} \right)\left( { - 1} \right) + {\left( { - 1} \right)^2}{\left( { - 1} \right)^2} + {\left( { - 1} \right)^3}{\left( { - 1} \right)^3} + ... + {\left( { - 1} \right)^{100}}{\left( { - 1} \right)^{100}}$

$C = \underbrace {1 + 1 + 1 + ... + 1 + 1}_{100\,\,{\rm{so}}\,\,1}$ $= 100.1 = 100$

Thay $x =  - 1;y =  - 1$ vào biểu thức P  ta được

$P = a{\left( { - 1} \right)^4}{\left( { - 1} \right)^4} + b{\left( { - 1} \right)^3}.\left( { - 1} \right) + c\left( { - 1} \right)\left( { - 1} \right)$$= a + b + c = 2020$

Vậy $P = 2020.$

Ta có $N = {x^3} + {x^2}y - 2{x^2} - xy - {y^2} + 3y + x - 1$$= \left( {{x^3} + {x^2}y - 2{x^2}} \right) + \left( { - xy - {y^2} + 2y} \right) + y + x - 1$

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

$= {x^2}.0 - y.0 + 0 + 1 = 1$

Vậy $N = 1.$

Ta có:

$\begin{array}{l}M = x - \left( {y - z} \right) - 2x + y + z - \left( {2 - x - y} \right) = x - y + z - 2x + y + z - 2 + x + y\\ = (x - 2x + x) + ( - y + y + y) + (z + z) - 2\\ = y + 2z - 2\end{array}$

$N = x - \left[ {x - \left( {y - 2z} \right) - 2z} \right] = x - (x - y + 2z - 2z) = x - (x - y) = x - x + y = y$

Khi đó $M - N = y + 2z - 2 - y = 2z - 2$

Ta có: $12\left( {12x + 15y} \right) = 12(3.4x + 3.5y) = 12.3(4x + 5y) = 12P$.

Thay $x = y = 1;z =  - 1$ vào biểu thức P ta được:

$\begin{array}{l}P = 1.1.( - 1) + {1^2}{.1^2}.{( - 1)^2} + {1^3}{.1^3}.{( - 1)^3} + {1^4}{.1^4}.{( - 1)^4} + ... + {1^{2019}}{.1^{2019}}.{( - 1)^{2019}} + {1^{2020}}{.1^{2020}}.{( - 1)^{2020}}\\ = ( - 1) + 1 + ( - 1) + 1 + ... + ( - 1) + 1 = 0\end{array}$.