Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
15,\\
a,\\
{x^2} - 6x + 9 = {x^2} - 2.x.3 + {3^2} = {\left( {x - 3} \right)^2}\\
b,\\
\dfrac{1}{4}{a^2} + 2a{b^2} + 4{b^4} = {\left( {\dfrac{1}{2}a} \right)^2} + 2.\dfrac{1}{2}a.2{b^2} + {\left( {2{b^2}} \right)^2} = {\left( {\dfrac{1}{2}a + 2{b^2}} \right)^2}\\
c,\\
25 + 10x + {x^2} = {5^2} + 2.5.x + {x^2} = {\left( {5 + x} \right)^2}\\
d,\\
\dfrac{1}{9} - \dfrac{2}{3}{y^4} + {y^8} = {\left( {\dfrac{1}{3}} \right)^2} - 2.\dfrac{1}{3}.{y^4} + {\left( {{y^4}} \right)^2} = {\left( {\dfrac{1}{3} - {y^4}} \right)^2}\\
16,\\
a,\\
16{x^2} + 24xy + 9{y^2} = {\left( {4x} \right)^2} + 2.4x.3y + {\left( {3y} \right)^2} = {\left( {4x + 3y} \right)^2}\\
b,\\
9{x^2} - 42xy + 49{y^2} = {\left( {3x} \right)^2} - 2.3x.7y + {\left( {7y} \right)^2} = {\left( {3x - 7y} \right)^2}\\
c,\\
25{x^2} + 90x + 81 = {\left( {5x} \right)^2} + 2.5x.9 + {9^2} = {\left( {5x + 9} \right)^2}\\
d,\\
64{x^2} - 48x + 9 = {\left( {8x} \right)^2} - 2.8x.3 + {3^2} = {\left( {8x - 3} \right)^2}\\
17,\\
a,\\
{37^2} + 2.37.13 + {13^2} = {\left( {37 + 13} \right)^2} = {50^2} = 2500\\
b,\\
51,{7^2} - 2.51,7.31,7 + 31,{7^2} = {\left( {51,7 - 31,7} \right)^2} = {20^2} = 400\\
c,\\
{201^2} = {\left( {200 + 1} \right)^2} = {200^2} + 2.200.1 + {1^2}\\
= 40000 + 400 + 1\\
= 40401\\
d,\\
{199^2} = {\left( {200 - 1} \right)^2} = {200^2} - 2.200.1 + {1^2}\\
= 40000 - 400 + 1\\
= 39601\\
e,\\
37.43 = \left( {40 - 3} \right)\left( {40 + 3} \right) = {40^2} - {3^2} = 1600 - 9 = 1591\\
f,\\
20,1.19,9 = \left( {20 + 0,1} \right).\left( {20 - 0,1} \right) = {20^2} - 0,{1^2}\\
= 400 - 0,01 = 399,99\\
21,\\
a,\\
A = 9{x^2} + 42x + 49 = {\left( {3x} \right)^2} + 2.3x.7 + {7^2} = {\left( {3x + 7} \right)^2}\\
x = 1 \Rightarrow A = {\left( {3.1 + 7} \right)^2} = {10^2} = 100\\
b,\\
B = 25{x^2} - 2xy + \dfrac{1}{{25}}{y^2} = {\left( {5x} \right)^2} - 2.5x.\dfrac{1}{5}y + {\left( {\dfrac{1}{5}y} \right)^2} = {\left( {5x - \dfrac{1}{5}y} \right)^2}\\
x = - \dfrac{1}{5};y = - 5 \Rightarrow B = {\left( {5.\dfrac{{ - 1}}{5} - \dfrac{1}{5}.\left( { - 5} \right)} \right)^2} = {\left( { - 1 + 1} \right)^2} = 0
\end{array}\)
