Điền đơn thức vào chỗ trống: \(12{x^3}{y^2}{z^2} - 18{x^2}{y^2}{z^4} = ....\left( {2x - 3{z^2}} \right)\)

\(1 - 2x\)

\(x - {x^2}\)

\({x^2} - x\)

\({x^2} + x\)

\(12{x^8} + 21{x^5}\)

\(12{x^8} + 21{x^6}\)

\( - 12{x^8} + 21{x^5}\)

\( - 12{x^8} - 21{x^5}\)

\(A = 4\)

$A =  - 4$

\(A = 19\)

\(A =  - 19\)

\({\left( {4x - \dfrac{1}{2}} \right)^2}\)

\({\left( {x - \dfrac{1}{2}} \right)^2}\)

\({\left( {4x + \dfrac{1}{2}} \right)^2}\)

\({\left( {x + \dfrac{1}{2}} \right)^2}\)

\({\left( {A + B} \right)^2} = {A^2} + 2AB + {B^2}\)

\({\left( {A - B} \right)^3} = {A^3} - 3{A^2}B - 3A{B^2} + {B^3}\)

\({A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)\)

\({A^3} - {B^3} = \left( {A - B} \right)\left( {{A^2} + AB + {B^2}} \right)\)

5

6

7

8

\( - 16x - 3\)

\( - 3\)

\( - 16x\)

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