Giải bài 6.20 trang 19 SGK Toán 8 tập 2 – Kết nối tri thức

Đề bài

Thực hiện các phép tính:

a) \(\frac{{{x^2} – 3{\rm{x}} + 1}}{{2{{\rm{x}}^2}}} + \frac{{5{\rm{x}} – 1 – {x^2}}}{{2{{\rm{x}}^2}}}\)

b) \(\frac{y}{{x – y}} + \frac{x}{{x + y}}\)

c) \(\frac{x}{{2{\rm{x}} – 6}} + \frac{y}{{2{\rm{x}}\left( {3 – x} \right)}}\)

Lời giải chi tiết

a) \(\frac{{{x^2} – 3x + 1}}{{2{x^2}}} + \frac{{5x – 1 – {x^2}}}{{2{x^2}}} \) \( = \frac{{{x^2} – 3x + 1 + 5x – 1 – {x^2}}}{{2{x^2}}} \) \( = \frac{{2x}}{{2{x^2}}} \) \( = \frac{1}{x}\)

b) \(\frac{y}{{x – y}} + \frac{x}{{x + y}} \) \( = \frac{{y\left( {x + y} \right) + x\left( {x – y} \right)}}{{\left( {x – y} \right)\left( {x + y} \right)}} \) \( = \frac{{xy + {y^2} + {x^2} – xy}}{{{x^2} – {y^2}}} \) \( = \frac{{{x^2} + {y^2}}}{{{x^2} – {y^2}}}\)

c) \(\frac{x}{{2x – 6}} + \frac{9}{{2x\left( {3 – x} \right)}} \) \( = \frac{x}{{2\left( {x – 3} \right)}} – \frac{9}{{2x\left( {x – 3} \right)}} \) \( = \frac{{{x^2}}}{{2x\left( {x – 3} \right)}} – \frac{9}{{2x\left( {x – 3} \right)}} \) \( = \frac{{{x^2} – 9}}{{2x\left( {x – 3} \right)}} \) \( = \frac{{\left( {x – 3} \right)\left( {x + 3} \right)}}{{2x\left( {x – 3} \right)}} \) \( = \frac{{x + 3}}{{2x}}\)