Câu 1 và 3 giúp em ạ

Đáp án:

$\begin{array}{l}
1){\left( {x + 2} \right)^2} = {x^2} + 2.x.2 + {2^2}\\
 = {x^2} + 4x + 4\\
2){\left( {z - 1} \right)^2} = {z^2} - 2z + 1\\
3){\left( {2t + 3} \right)^2} = 4{t^2} + 12t + 9\\
4){\left( {3a - 1} \right)^2} = 9{a^2} - 6a + 1\\
5){\left( {4 + U} \right)^3} = {4^3} + {3.4^2}.U + 3.4.{U^2} + {U^3}\\
 = {U^3} + 12{U^2} + 48U + 64\\
6){\left( {k - 3} \right)^3} = {k^3} - 9{k^2} + 27k - 27\\
7){\left( {2m + 1} \right)^3} = 8{m^3} + 12{m^2} + 6m + 1\\
8){\left( {1 - 3b} \right)^3}\\
 = 1 - 3.1.3b + 3.1.9{b^2} - 27{b^3}\\
 = 1 - 9b + 27{b^2} - 27{b^3}\\
B3)\\
1)15{x^2} - 5x - {\left( {2x - 3} \right)^2}\\
 = 15{x^2} - 5x - \left( {4{x^2} - 12x + 9} \right)\\
 = 15{x^2} - 5x - 4{x^2} + 12x - 9\\
 = 11{x^2} + 7x - 9\\
2)\\
{\left( {3 + x} \right)^2} + {\left( {x - 2} \right)^2}\\
 = {x^2} + 6x + 9 + {x^2} - 4x + 4\\
 = 2{x^2} + 2x + 13\\
3){\left( {2x + 3} \right)^2} - \left( {3x - 1} \right)\left( {x + 4} \right)\\
 = 4{x^2} + 12x + 9 - \left( {3{x^2} + 12x - x - 4} \right)\\
 = {x^2} + x + 13\\
4){\left( {1 + x} \right)^3} - 3x\left( {2x + 1} \right)\left( {2x - 1} \right)\\
 = {x^3} + 3{x^2} + 3x + 1 - 3x\left( {4{x^2} - 1} \right)\\
 = {x^3} + 3{x^2} + 3x + 1 - 12{x^3} + 3x\\
 =  - 11{x^3} + 3{x^2} + 6x + 1\\
5){\left( {x - 4} \right)^2} + \left( {x - 3} \right)\left( {x + 3} \right)\\
 = {x^2} - 8x + 16 + {x^2} - 9\\
 = 2{x^2} - 8x + 7\\
6)\\
\left( {2x + 1} \right)\left( {4{x^2} - 2x + 1} \right) - {\left( {2x - 1} \right)^3}\\
 = {\left( {2x} \right)^3} + {1^3} - \left( {8{x^3} - 12{x^2} + 6x - 1} \right)\\
 = 8{x^3} + 1 - 8{x^3} + 12{x^2} - 6x + 1\\
 = 12{x^2} - 6x + 2\\
7){\left( {2x - 1} \right)^2} - \left( {4x - 1} \right)\left( {x + 2} \right)\\
 = 4{x^2} - 4x + 1 - \left( {4{x^2} + 8x - x - 2} \right)\\
 =  - 11x + 3\\
8){\left( {x + 5} \right)^2} + \left( {x - 1} \right)\left( {4 - x} \right)\\
 = {x^2} + 10x + 25 - {x^2} + 4x - 4 + x\\
 = 15x + 21
\end{array}$