Bài 1.
a) \(\left( {3xy - {x^2} + y} \right).{2 \over 3}{x^2}y \)
\(= 2{x^3}{y^2} - {2 \over 3}{x^4}y + {2 \over 3}{x^2}{y^2}\)
b) \(3{x^{n - 2}}\left( {{x^{n + 2}} - {y^{n + 2}}} \right) + {y^{n + 2}}\left( {3{x^{n - 2}} - {y^{n - 2}}} \right)\)
\(=3{x^{2n}} - 3{x^{n - 2}}.{y^{n + 2}} + 3{x^{n - 2}}.{y^{n + 2}} - {y^{2n}} \)
\(= 3{x^{2n}} - {y^{2n}}.\)
Bài 2.
a) \(x\left( {{x^3} - y} \right) + {x^2}\left( {y - {x^2}} \right) - y\left( {{x^2} - 3x} \right)\)
\(= {x^4} - xy + {x^2}y - {x^4} - {x^2}y + 3xy \)
\(= 2xy.\)
Thay \(x = {1 \over 4},y = 2012\) ta có: \(2.{1 \over 4}.2012 = 1006.\)
b) Vì \(x = 2011 \Rightarrow x + 1 = 2012\)
Do đó \({x^{10}} - 2012{x^9} + 2012{x^8} - 2012{x^7} + 2012{x^6} - ... - 2012x + 2012\)
\( = {x^{10}} - \left( {x + 1} \right){x^9} + \left( {x + 1} \right){x^8} - \left( {x + 1} \right){x^7} + \left( {x + 1} \right){x^6} - \left( {x + 1} \right)x + 2012.\)
\( = {x^{10}} - {x^{10}} - {x^9} + {x^9} + {x^8} - {x^8} - {x^7} + {x^7} + {x^6} - ... - {x^2} + x + 2012\)
\(=x + 2012 = 2011 + 2012 = 4023.\)
Bài 3. \(5x\left( {12x + 7} \right) - 3x\left( {20x - 5} \right) = - 100\)
\(\Rightarrow 60{x^2} + 35x - 60{x^2} + 15x = - 100\)
\( \Rightarrow 50x = - 100 \Rightarrow x = - 2.\)
