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Áp dụng hằng đẳng thức để làm phép chia:
a) \(({x^2} + 2xy + {y^2}):(x + y)\) ;
b) \((4{x^2} – 4x + 1):(1 – 2x)\) ;
c) \((25{x^2} – 4{y^2}):(5x – 2y)\) ;
d) \((27{x^3} + 8):(3x + 2)\(.
\(\eqalign{ & a)\,\,\left( {{x^2} + 2xy + {y^2}} \right):\left( {x + y} \right) = {{{x^2} + 2xy + {y^2}} \over {x + y}} = {{{{\left( {x + y} \right)}^2}} \over {x + y}} = x + y \cr & b)\,\,\left( {4{x^2} – 4x + 1} \right):\left( {1 – 2x} \right) = {{4{x^2} – 4x + 1} \over {1 – 2x}} = {{{{\left( {1 – 2x} \right)}^2}} \over {1 – 2x}} = 1 – 2x \cr & c)\,\,\left( {25{x^2} – 4{y^2}} \right):\left( {5x – 2y} \right) = {{25{x^2} – 4{y^2}} \over {5x – 2y}} = {{\left( {5x – 2y} \right)\left( {5x + 2y} \right)} \over {5x – 2y}} = 5x + 2y \cr & d)\,\,\left( {27{x^3} + 8} \right):\left( {3x + 2} \right) = {{27{x^3} + 8} \over {3x + 2}} = {{{{\left( {3x} \right)}^3} + {2^3}} \over {3x + 2}} = {{\left( {3x + 2} \right)\left( {9{x^2} – 6x + 4} \right)} \over {3x + 2}} = 9{x^2} – 6x + 4 \cr} \)