Giải bài 5 trang 22 sách bài tập toán 8 – Chân trời sáng tạo

Đề bài

Tính:

a) \(x – \frac{{2x – y}}{4} + \frac{{x + 4y}}{{12}}\);

b) \(\frac{y}{x} – \frac{x}{y} – \frac{{{x^2} + {y^2}}}{{xy}}\);

c) \(\frac{4}{{x + 2}} – \frac{3}{{x – 2}} + \frac{{12}}{{{x^2} – 4}}\);

d) \(\frac{{x + y}}{{{x^2} – xy}} – \frac{{4x}}{{{x^2} – {y^2}}} – \frac{{x – y}}{{{x^2} + xy}}\).

Lời giải chi tiết

a) \(x – \frac{{2x – y}}{4} + \frac{{x + 4y}}{{12}} = \frac{{12x}}{{12}} – \frac{{3\left( {2x – y} \right)}}{{12}} + \frac{{x + 4y}}{{12}} = \frac{{12x – 6x + 3y + x + 4y}}{{12}} = \frac{{7x + 7y}}{{12}}\)

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

c) \(\frac{4}{{x + 2}} – \frac{3}{{x – 2}} + \frac{{12}}{{{x^2} – 4}} = \frac{{4\left( {x – 2} \right)}}{{\left( {x + 2} \right)\left( {x – 2} \right)}} – \frac{{3\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x – 2} \right)}} + \frac{{12}}{{\left( {x + 2} \right)\left( {x – 2} \right)}}\)

\( = \frac{{4x – 8 – 3x – 6 + 12}}{{\left( {x + 2} \right)\left( {x – 2} \right)}} = \frac{{x – 2}}{{\left( {x + 2} \right)\left( {x – 2} \right)}} = \frac{1}{{x + 2}}\)

d) \(\frac{{x + y}}{{{x^2} – xy}} – \frac{{4x}}{{{x^2} – {y^2}}} – \frac{{x – y}}{{{x^2} + xy}} = \frac{{{{\left( {x + y} \right)}^2}}}{{x\left( {x + y} \right)\left( {x – y} \right)}} – \frac{{4{x^2}}}{{x\left( {x + y} \right)\left( {x – y} \right)}} – \frac{{{{\left( {x – y} \right)}^2}}}{{x\left( {x + y} \right)\left( {x – y} \right)}}\)

\( = \frac{{{x^2} + 2xy + {y^2} – 4{x^2} – {x^2} + 2xy – {y^2}}}{{x\left( {x + y} \right)\left( {x – y} \right)}} = \frac{{4xy – 4{x^2}}}{{x\left( {x + y} \right)\left( {x – y} \right)}} = \frac{{ – 4x\left( {x – y} \right)}}{{x\left( {x + y} \right)\left( {x – y} \right)}} = \frac{{ – 4}}{{x + y}}\)