Bài 6.41 trang 26 Toán 8 tập 2 – Kết nối tri thức: Tìm đa thức P trong các đẳng thức sau: a) P + 1/x + 2 = x/x^2 – 2x + 4

Đề bài/câu hỏi:

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

Hướng dẫn:

Áp dụng quy tắc chuyển vế trong từng bài toán

Lời giải:

a)

\(\begin{array}{l}P + \frac{1}{{x + 2}} = \frac{x}{{{x^2} – 2{\rm{x}} + 4}}\\P = \frac{x}{{{x^2} – 2{\rm{x}} + 4}} – \frac{1}{{x + 2}}\\P = \frac{{x\left( {x + 2} \right) – {x^2} + 2{\rm{x}} – 4}}{{\left( {{x^2} – 2{\rm{x}} + 4} \right)\left( {x + 2} \right)}}\\P = \frac{{{x^2} + 2{\rm{x}} – {x^2} + 2{\rm{x}} + 4}}{{{x^3} + 8}}\\P = \frac{{4{\rm{x}} – 4}}{{{x^3} + 8}}\end{array}\)

b)

\(\begin{array}{l}P – \frac{{4\left( {x – 2} \right)}}{{x + 2}} = \frac{{16}}{{x – 2}}\\P = \frac{{16}}{{x – 2}} + \frac{{4\left( {x – 2} \right)}}{{x + 2}}\\P = \frac{{16\left( {x + 2} \right) + 4\left( {x – 2} \right)\left( {x – 2} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\P = \frac{{16{\rm{x}} + 32 + 4{{\rm{x}}^2} – 16{\rm{x}} + 16}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\P = \frac{{4{{\rm{x}}^2} + 48}}{{{x^2} – 4}}\end{array}\)

c)

\(\begin{array}{l}P.\frac{{x – 2}}{{x + 3}} = \frac{{{x^2} – 4{\rm{x}} + 4}}{{{x^2} – 9}}\\ \Rightarrow P = \frac{{{x^2} – 4{\rm{x}} + 4}}{{{x^2} – 9}}.\frac{{x + 3}}{{x – 2}}\\P = \frac{{{{(x – 2)}^2}(x + 3)}}{{(x – 3)(x + 3)(x – 2)}} = \frac{{x – 2}}{{x – 3}}\end{array}\)\(\)

d)

\(\begin{array}{l}P:\frac{{{x^2} – 9}}{{2{\rm{x}} + 4}} = \frac{{{x^2} – 4}}{{{x^2} + 3{\rm{x}}}}\\ \Rightarrow P = \frac{{{x^2} – 4}}{{{x^2} + 3{\rm{x}}}}.\frac{{{x^2} – 9}}{{2{\rm{x}} + 4}}\\P = \frac{{(x – 2)(x + 2)(x – 3)(x + 3)}}{{2{\rm{x}}(x + 3)(x + 2)}}\\P = \frac{{(x – 2)(x – 3)}}{{2{\rm{x}}}}\end{array}\)