Bài 3: Phép cộng và phép trừ đa thức

\(\begin{array}{l}A + B + C = (4{{{x}}^2} - 5{{x}}y + 3{y^2}) + (3{{{x}}^2} + 2{{x}}y + {y^2}) + ( - {x^2} + 3{{x}}y + 2{y^2})\\ = 4{{{x}}^2} - 5{{x}}y + 3{y^2} + 3{{{x}}^2} + 2{{x}}y + {y^2} - {x^2} + 3{{x}}y + 2{y^2}\\ = \left( {4{{{x}}^2} + 3{{{x}}^2} - {x^2}} \right) + \left( { - 5{{x}}y + 2{{x}}y + 3{{x}}y} \right) + \left( {3{y^2} + {y^2} + 2{y^2}} \right)\\ = 6{{{x}}^2} + 6{y^2}\end{array}\)

\(\begin{array}{l}C - A - B = \left( { - {x^2} + 3{{x}}y + 2{y^2}} \right) - \left( {4{{{x}}^2} - 5{{x}}y + 3{y^2}} \right) - \left( {3{{{x}}^2} + 2{{x}}y + {y^2}} \right) \\ = - {x^2} + 3{{x}}y + 2{y^2} - 4{{{x}}^2} + 5{{x}}y - 3{y^2} - 3{{{x}}^2} - 2{{x}}y - {y^2} \\ = \left( { - 4{{{x}}^2} - 3{{{x}}^2} - {x^2}} \right) + \left( {5{{x}}y - 2{{x}}y + 3{{x}}y} \right) + \left( { - 3{y^2} - {y^2} + 2{y^2}} \right)\\ =  - 8{{{x}}^2}{{ + 6x}}y - 2{y^2}\end{array}\)

Ta có:

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

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

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

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

\(M = {x^2} + 12{{x}}y - {y^2}\)

\(\begin{array}{l}N = {x^3} + {x^2}y - 2{{{x}}^2} - xy - {y^2} + 3y + x - 1\\ = \left( {{x^3} + {x^2}y - 2{{{x}}^2}} \right) + \left( { - xy - {y^2} + 2y} \right) + y + x - 1\\ = {x^2}\left( {x + y - 2} \right) - y\left( {x + y - 2} \right) + \left( {x + y - 2} \right) + 1\\ = {x^2}.0 - y.0 + 0 + 1 = 1\end{array}\)

Ta có:

\(\begin{array}{l}M = x - \left( {y - z} \right) - 2{{x}} + y + z - \left( {2 - x - y} \right)\\ = x - y + z - 2{{x}} + y + z - 2 + x + y\\ = y + 2{{z}} - 2\\N = x - \left[ {x - \left( {y - 2{{z}}} \right) - 2{{z}}} \right]\\ = x - \left( {x - y + 2{{z}} - 2{{z}}} \right) = x - x + y = y\\ \text{suy ra } M - N = y + 2{{z}} - 2 - y = 2{{z}} - 2\end{array}\)