Giải bài 6.23 trang 19 SGK Toán 8 tập 2 – Kết nối tri thức

Đề bài

Thực hiện các phép tính sau;

\(a)\frac{{{x^2} + 4{\rm{x}} + 4}}{{{x^2} – 4}} + \frac{x}{{2 – x}} + \frac{{4 – x}}{{5{\rm{x}} – 10}}\)

\(b)\frac{x}{{{x^2} + 1}} – \left( {\frac{3}{{x + 6}} + \frac{{x – 2}}{{x + 4}}} \right) + \left[ {\frac{3}{{x + 6}} – \left( {\frac{1}{{{x^2} + 1}} – \frac{{x – 2}}{{x + 4}}} \right)} \right]\)

Lời giải chi tiết

\(\begin{array}{l}a)\frac{{{x^2} + 4{\rm{x}} + 4}}{{{x^2} – 4}} + \frac{x}{{2 – x}} + \frac{{4 – x}}{{5{\rm{x}} – 10}}\\ = \frac{{{{\left( {x + 2} \right)}^2}}}{{\left( {x + 2} \right)\left( {x – 2} \right)}} – \frac{x}{{x – 2}} + \frac{{4 – x}}{{5\left( {x – 2} \right)}}\\ = \frac{{x + 2}}{{x – 2}} – \frac{x}{{x – 2}} + \frac{{4 – x}}{{5\left( {x – 2} \right)}}\\ = \frac{{5\left( {x + 2} \right) – 5x + 4 – x}}{{5\left( {x – 2} \right)}} = \frac{{ – x + 14}}{{5\left( {x – 2} \right)}}\end{array}\)

\(\begin{array}{l}b)\frac{x}{{{x^2} + 1}} – \left( {\frac{3}{{x + 6}} + \frac{{x – 2}}{{x + 4}}} \right) + \left[ {\frac{3}{{x + 6}} – \left( {\frac{1}{{{x^2} + 1}} – \frac{{x – 2}}{{x + 4}}} \right)} \right]\\ = \frac{x}{{{x^2} + 1}} – \frac{3}{{x + 6}} – \frac{{x – 2}}{{x + 4}} + \frac{3}{{x + 6}} – \frac{1}{{{x^2} + 1}} + \frac{{x – 2}}{{x + 4}}\\ = \frac{x}{{{x^2} + 1}} – \frac{1}{{{x^2} + 1}} = \frac{{x – 1}}{{{x^2} + 1}}\end{array}\)