Giúp em với ạ Em đang làm bài kt 15 ạ

Đáp án:

$\begin{array}{l} 1)a){\left( {x + 5} \right)^2} = {x^2} + 10x + 25\\ b){\left( {x - \dfrac{1}{3}} \right)^2} = {x^2} - \dfrac{2}{3}x + \dfrac{1}{9}\\ c){\left( {{x^2} - 3} \right)^2} = {x^4} - 6{x^2} + 9\\ d){\left( {\dfrac{x}{y} - \dfrac{y}{x}} \right)^2} = \dfrac{{{x^2}}}{{{y^2}}} - 2 + \dfrac{{{y^2}}}{{{x^2}}}\\ e)\dfrac{1}{{16}} - {x^2} = \left( {\dfrac{1}{4} - x} \right)\left( {\dfrac{1}{4} + x} \right)\\ f){\left( {x + \dfrac{1}{2}} \right)^3} = {x^3} + \dfrac{3}{2}{x^2} + \dfrac{3}{4}x + \dfrac{1}{8}\\ g){\left( {x{y^2} - \dfrac{1}{{{x^2}}}} \right)^3} = {x^3}{y^6} - 3{y^4} + \dfrac{{3{y^2}}}{{{x^3}}} - \dfrac{1}{{{x^6}}}\\ B2)a){a^2} - 6a + 9 = {\left( {a - 3} \right)^2}\\ b)\dfrac{1}{4}{x^2} + 2x{y^2} + 4{y^4} = {\left( {\dfrac{1}{2}x + 2{y^2}} \right)^2}\\ c)\dfrac{1}{9} - \dfrac{2}{3}{y^4} + {y^8} = {\left( {\dfrac{1}{3} - {y^4}} \right)^2}\\ d)8{x^6} - 12{x^4} + 6{x^2} - 1\\  = {\left( {2{x^2} - 1} \right)^3}\\ e){\left( {x - y} \right)^3} - 3{\left( {x - y} \right)^2}y + 3\left( {x - y} \right).\dfrac{{{y^2}}}{4} - \dfrac{{{y^3}}}{8}\\  = {\left( {x - y - \dfrac{y}{2}} \right)^3}\\  = {\left( {x - \dfrac{{3y}}{2}} \right)^3}\\ B3)\\ a){21^2} = {\left( {20 - 1} \right)^2} = {20^2} - 2.20 + 1 = 400 - 40 + 1 = 361\\ b){59^2} = {\left( {60 - 1} \right)^2} = {60^2} - 2.60 + 1 = 3600 - 120 + 1 = 3481\\ c)77.83 = \left( {80 - 3} \right)\left( {80 + 3} \right) = {80^2} - {3^2} = 6400 - 9 = 6301\\ d){102^3} = {\left( {100 + 2} \right)^3}\\  = {100^3} + {3.100^2}.2 + {3.100.2^2} + {2^3}\\  = 1000000 + 60000 + 1200 + 8\\  = 1061208\\ e){98^3} = {\left( {100 - 2} \right)^3}\\  = {100^3} - {3.100^2}.2 + {3.100.2^2} - {2^3}\\  = 1000000 - 60000 + 1200 - 8\\  = 941192 \end{array}$