Đề bài

Tìm đa thức P trong các đẳng thức sau:

🥂a) \(P + \frac{1}{{x + 2}} = \frac{x}{{{x^2} - 2{\rm{x}} + 4}}\)

🎉b) \(P - \frac{{4\left( {x - 2} \right)}}{{x + 2}} = \frac{{16}}{{x - 2}}\)

ꦗc) \(P.\frac{{x - 2}}{{x + 3}} = \frac{{{x^2} - 4{\rm{x}} + 4}}{{{x^2} - 9}}\)

ꩲd) \(P:\frac{{{x^2} - 9}}{{2{\rm{x}} + 4}} = \frac{{{x^2} - 4}}{{{x^2} + 3{\rm{x}}}}\)

Lời giải chi tiết

🥀a)  \(P = \frac{x}{{{x^2} - 2{\rm{x}} + 4}} - \frac{1}{{x + 2}}\)\( = \frac{{x\left( {x + 2} \right) - {x^2} + 2{\rm{x}} - 4}}{{\left( {{x^2} - 2{\rm{x}} + 4} \right)\left( {x + 2} \right)}}\)\( = \frac{{4{\rm{x}} - 4}}{{{x^3} + 8}}\).

ꦍb) \(P = \frac{{16}}{{x - 2}} + \frac{{4\left( {x - 2} \right)}}{{x + 2}}\)\( = \frac{{16\left( {x + 2} \right) + 4\left( {x - 2} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\)\( = \frac{{4{{\rm{x}}^2} + 48}}{{{x^2} - 4}}\).

ဣc) \(P = \frac{{{x^2} - 4{\rm{x}} + 4}}{{{x^2} - 9}}:\frac{{x - 2}}{{x + 3}} = \frac{{{{(x - 2)}^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\frac{{x + 3}}{{x - 2}}\)\( = \frac{{x - 2}}{{x - 3}}\).

🍃d) \(P = \frac{{{x^2} - 4}}{{{x^2} + 3{\rm{x}}}}.\frac{{{x^2} - 9}}{{2{\rm{x}} + 4}} = \frac{{(x - 2)(x + 2)}}{{x(x + 3)}}.\frac{{(x - 3)(x + 3)}}{{2(x + 2)}} = \frac{{(x - 2)(x - 3)}}{{2x}}\).