Bậc của đa thức \(P = {x^6} - {x^4}{y^3} + 5{x^2}{y^2} + {y^4} - 1\) là

a) \(A = - 2x{y^5}\left( { - {x^2}{y^4}} \right)\left( {6{x^2}y} \right)\)

\[ = 2{x^3}{y^9} \cdot 6{x^2}y\]

\[ = 12{x^5}{y^{10}}\].

b) \[B = \left( {15{x^5}{y^3} - 10{x^3}{y^5} + 25{x^4}{y^4}} \right):5{x^2}{y^2}\]

\[ = 15{x^5}{y^3}:\left( {5{x^2}{y^2}} \right) - 10{x^3}{y^5}:\left( {5{x^2}{y^2}} \right) + 25{x^4}{y^4}:\left( {5{x^2}{y^2}} \right)\]

\[ = 3{x^3}y - 2x{y^3} + 5{x^2}{y^2}\].

c) \(C = 3{x^2}y\left( {2{x^2} - y} \right) - 4{x^2}\left( {{x^2}y - {y^2}} \right)\).

\[{\rm{ = }}\left( {{\rm{6}}{x^4}y - 3{x^2}{y^2}} \right) - \left( {4{x^4}y - 4{x^2}{y^2}} \right)\]

\[{\rm{ = 6}}{x^4}y - 3{x^2}{y^2} - 4{x^4}y + 4{x^2}{y^2}\]

\[{\rm{ = }}\left( {{\rm{6}}{x^4}y - 4{x^4}y} \right) + \left( {4{x^2}{y^2} - 3{x^2}{y^2}} \right)\]

\[{\rm{ = 2}}{x^4}y + {x^2}{y^2}\].

a) \(3\left( {2x - 3} \right) + 2\left( {2 - x} \right) = - 3\)

\(6x - 9 + 4 - 2x = - 3\)

\(6x - 2x = 9 - 4 - 3\)

\(4x = 2\)

\(x = \frac{1}{2}\).

Vậy \(x = \frac{1}{2}\).

b) \(3x\left( {2x + 3} \right) - \left( {2x + 5} \right)\left( {3x - 2} \right) = 8\)

\(6{x^2} + 9x - \left( {6{x^2} - 4x + 15x - 10} \right) = 8\)

\(6{x^2} + 9x - 6{x^2} + 4x - 15x + 10 = 8\)

\(\left( {6{x^2} - 6{x^2}} \right) + \left( {9x + 4x - 15x} \right) = 8 - 10\)

\( - 2x = - 2\)

\(x = 1\)

Vậy \(x = 1\).