Câu 1 và 3 giúp em ạ Question Câu 1 và 3 giúp em ạ in progress 0 Môn Toán King 6 years 2020-11-28T05:18:41+00:00 2020-11-28T05:18:41+00:00 1 Answers 56 views 0
Answers ( )
Đáp án:
$\begin{array}{l}
1){\left( {x + 2} \right)^2} = {x^2} + 2.x.2 + {2^2}\\
= {x^2} + 4x + 4\\
2){\left( {z – 1} \right)^2} = {z^2} – 2z + 1\\
3){\left( {2t + 3} \right)^2} = 4{t^2} + 12t + 9\\
4){\left( {3a – 1} \right)^2} = 9{a^2} – 6a + 1\\
5){\left( {4 + U} \right)^3} = {4^3} + {3.4^2}.U + 3.4.{U^2} + {U^3}\\
= {U^3} + 12{U^2} + 48U + 64\\
6){\left( {k – 3} \right)^3} = {k^3} – 9{k^2} + 27k – 27\\
7){\left( {2m + 1} \right)^3} = 8{m^3} + 12{m^2} + 6m + 1\\
8){\left( {1 – 3b} \right)^3}\\
= 1 – 3.1.3b + 3.1.9{b^2} – 27{b^3}\\
= 1 – 9b + 27{b^2} – 27{b^3}\\
B3)\\
1)15{x^2} – 5x – {\left( {2x – 3} \right)^2}\\
= 15{x^2} – 5x – \left( {4{x^2} – 12x + 9} \right)\\
= 15{x^2} – 5x – 4{x^2} + 12x – 9\\
= 11{x^2} + 7x – 9\\
2)\\
{\left( {3 + x} \right)^2} + {\left( {x – 2} \right)^2}\\
= {x^2} + 6x + 9 + {x^2} – 4x + 4\\
= 2{x^2} + 2x + 13\\
3){\left( {2x + 3} \right)^2} – \left( {3x – 1} \right)\left( {x + 4} \right)\\
= 4{x^2} + 12x + 9 – \left( {3{x^2} + 12x – x – 4} \right)\\
= {x^2} + x + 13\\
4){\left( {1 + x} \right)^3} – 3x\left( {2x + 1} \right)\left( {2x – 1} \right)\\
= {x^3} + 3{x^2} + 3x + 1 – 3x\left( {4{x^2} – 1} \right)\\
= {x^3} + 3{x^2} + 3x + 1 – 12{x^3} + 3x\\
= – 11{x^3} + 3{x^2} + 6x + 1\\
5){\left( {x – 4} \right)^2} + \left( {x – 3} \right)\left( {x + 3} \right)\\
= {x^2} – 8x + 16 + {x^2} – 9\\
= 2{x^2} – 8x + 7\\
6)\\
\left( {2x + 1} \right)\left( {4{x^2} – 2x + 1} \right) – {\left( {2x – 1} \right)^3}\\
= {\left( {2x} \right)^3} + {1^3} – \left( {8{x^3} – 12{x^2} + 6x – 1} \right)\\
= 8{x^3} + 1 – 8{x^3} + 12{x^2} – 6x + 1\\
= 12{x^2} – 6x + 2\\
7){\left( {2x – 1} \right)^2} – \left( {4x – 1} \right)\left( {x + 2} \right)\\
= 4{x^2} – 4x + 1 – \left( {4{x^2} + 8x – x – 2} \right)\\
= – 11x + 3\\
8){\left( {x + 5} \right)^2} + \left( {x – 1} \right)\left( {4 – x} \right)\\
= {x^2} + 10x + 25 – {x^2} + 4x – 4 + x\\
= 15x + 21
\end{array}$