Cho biết \({x^3} = 2p + 1\)  trong đó \(x\) là số tự nhiên, \(p\) là số nguyên tố. Tìm \(x.\)

\(3{y^2} + 2xy - {x^2}\)         

\(3{y^2} + 2xy + {x^2}\)

\(3{y^2} - 2xy - {x^2}\)          

\(3{y^2} + 2xy\)

\({\left( {{x^4} - 2} \right)^2} - {\left( {2{x^2}} \right)^2}\).

\({\left( {{x^4} + 4} \right)^2} - {\left( {4{x^2}} \right)^2}\).

\({\left( {{x^4} + 2} \right)^2} - {\left( {4{x^2}} \right)^2}\).

\({\left( {{x^4} + 2} \right)^2} - {\left( {2{x^2}} \right)^2}\).

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

\({\left( {3x - 2y} \right)^2} - {\left( {2x - 3y} \right)^2} = \left( {5x - y} \right)\left( {x - 5y} \right)\).

\({\left( {3x - 2y} \right)^2} - {\left( {2x - 3y} \right)^2} = \left( {x - y} \right)\left( {x + y} \right)\).

\({\left( {3x - 2y} \right)^2} - {\left( {2x - 3y} \right)^2} = 5\left( {x - y} \right)\left( {x - 5y} \right)\).

\(2x - 3;5x - 5\).

\(2x - 3; - 5x + 1\).

\( - 5x + 1;2x - 3\).      

\(2x - 3; - 5x - 5\).

\( - \dfrac{{11}}{7}\)  

\(\dfrac{7}{{11}}\)

\(1\)

\(\dfrac{{11}}{7}\)

\(2\).

\(3\).

\(1\).

\(4\).

\({\left( {{x^2}} \right)^3} + {3^3}\).

\({\left( {{x^2}} \right)^3} - {3^3}\).

\({\left( {{x^2}} \right)^3} + {9^3}\).

\({\left( {{x^2}} \right)^3} - {9^3}\).