1. Điền vào chỗ trống : a) x^2 + 4x + …… = (x + ……)^2 b) …… – 12x + 9 = (2x – …..)^2 c) 4x^2 + …… + …… = (2x – 2y)^2 d) (x – …

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{x^2} + 4x + 4 = {\left( {x + 2} \right)^2}\\
b,\\
4{x^2} – 12x + 9 = {\left( {2x – 3} \right)^2}\\
c,\\
4{x^2} – 8xy + 4{y^2} = {\left( {2x – 2y} \right)^2}\\
d,\\
\left( {x – \dfrac{y}{2}} \right)\left( {x + \dfrac{y}{2}} \right) = {x^2} – \dfrac{{{y^2}}}{4}\\
2,\\
a,\\
A = \dfrac{{{{\left( {2t + 5} \right)}^2} + {{\left( {5t – 2} \right)}^2}}}{{4{t^2} + 4}}\\
 = \dfrac{{4{t^2} + 20t + 25 + 25{t^2} – 20t + 4}}{{4{t^2} + 4}}\\
 = \dfrac{{29{t^2} + 29}}{{4{t^2} + 4}}\\
 = \dfrac{{29\left( {{t^2} + 1} \right)}}{{4.\left( {{t^2} + 1} \right)}}\\
 = \dfrac{{29}}{4}\\
b,\\
B = 2.\left( {3 + 1} \right).\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)…..\left( {{3^{2020}} + 1} \right)\\
 = \left( {3 – 1} \right).\left( {3 + 1} \right).\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)…..\left( {{3^{2020}} + 1} \right)\\
 = \left( {{3^2} – 1} \right).\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)…..\left( {{3^{2020}} + 1} \right)\\
 = \left( {{3^4} – 1} \right)\left( {{3^4} + 1} \right)……..\left( {{3^{2020}} + 1} \right)\\
 = {3^{4040}} – 1
\end{array}\)