Giải bài 6. 41 trang 26 SGK Toán 8 tập 2 - Kết nối tri thức — Không quảng cáo

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)

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

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)

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