Gọi \(x\) là giá trị thỏa mãn \(5\left( {3x + 5} \right) - 4\left( {2x - 3} \right)\)\( = 5x + 3\left( {2x - 12} \right) + 1.\) Khi đó

$5{x^3}{y^3}$

$ - 5{x^3}{y^3}$

$ - {x^3}{y^3}$

${x^3}{y^2}$

${x^2} - 2xy + {y^2}$

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

${x^2} - {y^2}$

${x^2} + 2xy + {y^2}$

\( - 2{x^4}{y^5}\)

\(\dfrac{1}{2}{x^5}{y^4}\)

\(2{x^5}{y^4}\)

\( - 2{x^5}{y^4}\)

\(\dfrac{1}{{81}}y\)

\( - \dfrac{1}{{27}}y\)

\(\dfrac{1}{{81}}xy\)

\( - \dfrac{1}{{81}}y\)

\(24{x^5} + 20{x^4} + 12{x^3} - 4{x^2}\)

\( - 24{x^5} - 20{x^4} + 12{x^3} + 1\)                

\( - 24{x^5} - 20{x^4} + 12{x^3} - 4{x^2}\)

\( - 24{x^5} - 20{x^4} - 12{x^3} + 4{x^2}\)

\(4{x^2} + 12x + 9\)

\(4{x^2} - 9\)

\(2{x^2} - 3\)

\(4{x^2} + 9\)

\(P = 4{x^2} - 6x.\) Với  \(x =  - \dfrac{3}{2}\) thì \(P = 18\) .

\(P = 4{x^2} + 6x.\) Với  \(x =  - \dfrac{3}{2}\) thì \(P = 0\) .

\(P = 4{x^2} - 6x.\) Với  \(x =  - \dfrac{3}{2}\) thì \(P =  - 18\) .

\(P = 4{x^2} + 6x.\) Với  \(x =  - \dfrac{3}{2}\) thì \(P = 18\) .