Sử dụng quy tắc cộng, trừ phân thức. Hướng dẫn giải Giải Bài 18 trang 41 SGK Toán 8 tập 1 – Chân trời sáng tạo – Bài tập cuối chương 1. Thực hiện các phép tính sau:…
Đề bài/câu hỏi:
Thực hiện các phép tính sau:
a) \(\dfrac{{2{x^2} – 1}}{{x – 2}} + \dfrac{{ – {x^2} – 3}}{{x – 2}}\)
b) \(\dfrac{x}{{x + y}} + \dfrac{y}{{x – y}}\)
c) \(\dfrac{1}{{x – 1}} – \dfrac{2}{{{x^2} – 1}}\)
d) \(\dfrac{{x + 2}}{{{x^2} + xy}} – \dfrac{{y – 2}}{{xy + {y^2}}}\)
e) \(\dfrac{1}{{2{x^2} – 3x}} – \dfrac{1}{{4{x^2} – 9}}\)
g) \(\dfrac{{2x}}{{9 – {x^2}}} + \dfrac{1}{{x – 3}} – \dfrac{1}{{x + 3}}\)
Hướng dẫn:
Sử dụng quy tắc cộng, trừ phân thức
Lời giải:
a)
\(\dfrac{{2{x^2} – 1}}{{x – 2}} + \dfrac{{ – {x^2} – 3}}{{x – 2}}\)
\( = \dfrac{{{x^2} – 4}}{{x – 2}}\)
\(\begin{array}{l} = \dfrac{{\left( {x – 2} \right)\left( {x + 2} \right)}}{{x – 2}}\\ = x + 2\end{array}\)
b)
\(\dfrac{x}{{x + y}} + \dfrac{y}{{x – y}}\)
\(\begin{array}{l} = \dfrac{{x(x – y)}}{{(x + y)(x – y)}} + \dfrac{{y(x + y)}}{{(x – y)(x + y)}}\\ = \dfrac{{{x^2} – xy + xy + {y^2}}}{{\left( {x – y} \right)\left( {x + y} \right)}}\\ = \dfrac{{{x^2} + {y^2}}}{{{x^2} – {y^2}}}\end{array}\)
c)
\(\dfrac{1}{{x – 1}} – \dfrac{2}{{{x^2} – 1}}\)\( = \dfrac{{x + 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} – \dfrac{2}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = \dfrac{{x – 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = \dfrac{1}{{x + 1}}\)
d)
\(\dfrac{{x + 2}}{{{x^2} + xy}} – \dfrac{{y – 2}}{{xy + {y^2}}}\)
\(\begin{array}{l} = \dfrac{{x + 2}}{{x(x + y)}} – \dfrac{{y – 2}}{{y(x + y)}}\\ = \dfrac{{\left( {x + 2} \right)y}}{{xy\left( {x + y} \right)}} – \dfrac{{\left( {y – 2} \right)x}}{{xy\left( {x + y} \right)}}\\ = \dfrac{{xy + 2y}}{{xy\left( {x + y} \right)}} – \dfrac{{xy – 2x}}{{xy\left( {x + y} \right)}}\\ = \dfrac{{2y + 2x}}{{xy\left( {x + y} \right)}}\\ = \dfrac{{2\left( {x + y} \right)}}{{xy\left( {x + y} \right)}}\\ = \dfrac{2}{{xy}}\end{array}\)
e)
\(\dfrac{1}{{2{x^2} – 3x}} – \dfrac{1}{{4{x^2} – 9}}\)
\(\begin{array}{l} = \dfrac{1}{{x\left( {2x – 3} \right)}} – \dfrac{1}{{\left( {2x – 3} \right)\left( {2x + 3} \right)}}\\ = \dfrac{{2x + 3}}{{x\left( {2x – 3} \right)\left( {2x + 3} \right)}} – \dfrac{x}{{x\left( {2x – 3} \right)\left( {2x + 3} \right)}}\\ = \dfrac{{x + 3}}{{x\left( {4{x^2} – 9} \right)}}\end{array}\)
g)
\(\dfrac{{2x}}{{9 – {x^2}}} + \dfrac{1}{{x – 3}} – \dfrac{1}{{x + 3}}\)
\(\begin{array}{l} = \dfrac{{ – 2x}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} + \dfrac{1}{{x – 3}} – \dfrac{1}{{x + 3}}\\ = \dfrac{{ – 2x}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} + \dfrac{{x + 3}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} – \dfrac{{x – 3}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\ = \dfrac{{ – 2x + 6}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\ = \dfrac{{ – 2\left( {x – 3} \right)}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\ = \dfrac{{ – 2}}{{x + 3}}\end{array}\)