Tháng Ba 29, 2024

Phân tích đa thức thành nhân tử: \(a)\ {{x}^{2}}+4x-{{y}^{2}}+4\) \(b)\ 4{{x}^{2}}-25-\left( 2x+7 \right)\left( 5-2x \right)\)\(c)\ {{\left( 2{{x}^{2}}-y \right)}^{2}}-64{{y}^{2}}\) \(d)\ -{{x}^{3}}+6{{x}^{2}}y-12x{{y}^{2}}+8{{y}^{3}}\) \(e)\;{x^8} – {y^8}\)

Phân tích đa thức thành nhân tử:

\(a)\ {{x}^{2}}+4x-{{y}^{2}}+4\)

\(b)\ 4{{x}^{2}}-25-\left( 2x+7 \right)\left( 5-2x \right)\)\(c)\ {{\left( 2{{x}^{2}}-y \right)}^{2}}-64{{y}^{2}}\)

\(d)\ -{{x}^{3}}+6{{x}^{2}}y-12x{{y}^{2}}+8{{y}^{3}}\)

\(e)\;{x^8} – {y^8}\)

Lời giải chi tiết:

\(a)\;{x^2} + 4x – {y^2} + 4 = \left( {{x^2} + 4x + 4} \right) – {y^2} = \left( {{x^2} + 2.2.x + {2^2}} \right) – {y^2} = {\left( {x + 2} \right)^2} – {y^2} = \left( {x – y + 2} \right)\left( {x + y + 2} \right)\)

\(\begin{array}{l}b)\;4{x^2} – 25 – \left( {2x + 7} \right)\left( {5 – 2x} \right)\\= {\left( {2x} \right)^2} – {5^2} – \left( {2x + 7} \right)\left( {5 – 2x} \right)\\= \left( {2x – 5} \right)\left( {2x + 5} \right) – \left( {2x + 7} \right)\left( {5 – 2x} \right)\\= \left( {2x – 5} \right)\left( {2x + 5} \right) + \left( {2x + 7} \right)\left( {2x – 5} \right)\\= \left( {2x – 5} \right)\left( {2x + 5 + 2x + 7} \right) = \left( {2x – 5} \right)\left( {4x + 12} \right)\end{array}\)

\(c)\;{\left( {2{x^2} – y} \right)^2} – 64{y^2} = {\left( {2{x^2} – y} \right)^2} – {\left( {8y} \right)^2} = \left( {2{x^2} – y – 8y} \right)\left( {2{x^2} – y + 8y} \right) = \left( {2{x^2} – 9y} \right)\left( {2{x^2} + 7y} \right)\)

\(d)\; – {x^3} + 6{x^2}y – 12x{y^2} + 8{y^3} = {\left( { – x} \right)^3} + 3.{x^2}.2y + 3.\left( { – x} \right).{\left( {2y} \right)^2} + {\left( {2y} \right)^3} = {\left( { – x + 2y} \right)^3} = {\left( {2y – x} \right)^3}\)

\(\begin{array}{l}e)\;{x^8} – {y^8} = {\left( {{x^4}} \right)^2} – {\left( {{y^4}} \right)^2} = \left( {{x^4} + {y^4}} \right)\left( {{x^4} – {y^4}} \right)\\ = \left( {{x^4} + {y^4}} \right)\left( {{x^2} + {y^2}} \right)\left( {{x^2} – {y^2}} \right) = \left( {{x^4} + {y^4}} \right)\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)\left( {x – y} \right)\end{array}\)