Giải bài 1 trang 25 vở thực hành Toán 8 tập 2

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) \(P – \frac{{4\left( {x – 2} \right)}}{{x + 2}} = \frac{{16}}{{x – 2}}\)

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

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

Lời giải chi tiết

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

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

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

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