Giải mục 2 trang 41, 42 SGK Toán 8 tập 1 – Cánh diều

LT4

Thực hiện phép tính:

\(a)\dfrac{{4{\rm{x}} + 3y}}{{{x^2} – {y^2}}} – \dfrac{{3{\rm{x}} + 4y}}{{{x^2} – {y^2}}}\)

\(b)\dfrac{{2{\rm{x}}y – 3{y^2}}}{{{x^2} – 3{\rm{x}}y}} – \dfrac{x}{{3{\rm{x}} – 9y}}\)

Phương pháp giải:

Áp dụng quy tắc trừ hai phân thức cùng mẫu, trừ hai phân thức khác mẫu để thực hiện phép tính.

Lời giải chi tiết:

\(a)\dfrac{{4{\rm{x}} + 3y}}{{{x^2} – {y^2}}} – \dfrac{{3{\rm{x}} + 4y}}{{{x^2} – {y^2}}} = \dfrac{{\left( {{\rm{4x}} + 3y} \right) – \left( {3{\rm{x}} + 4y} \right)}}{{{x^2} – {y^2}}} = \dfrac{{4{\rm{x}} + 3y – 3{\rm{x}} – 4y}}{{{x^2} – {y^2}}} = \dfrac{{x – y}}{{{x^2} – {y^2}}} = \dfrac{{x – y}}{{\left( {x – y} \right)\left( {x + y} \right)}} = \dfrac{1}{{x + y}}\)

\(\begin{array}{l}b)\dfrac{{2{\rm{x}}y – 3{y^2}}}{{{x^2} – 3{\rm{x}}y}} – \dfrac{x}{{3{\rm{x}} – 9y}}\\ = \dfrac{{2{\rm{x}}y – 3{y^2}}}{{x\left( {x – 3y} \right)}} – \dfrac{{{x^2}}}{{3\left( {x – 3y} \right)}}\\ = \dfrac{{3\left( {2{\rm{x}}y – 3{y^2}} \right)}}{{3{\rm{x}}\left( {x – 3y} \right)}} – \dfrac{{{x^2}}}{{3{\rm{x}}\left( {x – 3y} \right)}}\\ = \dfrac{{6{\rm{x}}y – 9{y^2} – {x^2}}}{{3{\rm{x}}\left( {x – 3y} \right)}} = \dfrac{{ – \left( {{x^2} – 6{\rm{x}}y + 9{y^2}} \right)}}{{3{\rm{x}}\left( {x – 3y} \right)}} = \dfrac{{ – {{\left( {x – 3y} \right)}^2}}}{{3{\rm{x}}\left( {x – 3y} \right)}} = \dfrac{{ – \left( {x – 3y} \right)}}{{3{\rm{x}}}}\end{array}\)

LT5

Tính một cách hợp lí:

\(\dfrac{{x – 5y}}{{2{\rm{x}} – 3y}} – \dfrac{{24{\rm{x}}y}}{{4{{\rm{x}}^2} – 9{y^2}}} – \dfrac{{x + 8y}}{{3y – 2{\rm{x}}}}\)

Phương pháp giải:

Áp dụng phân thức đối để tính hợp lí.

Lời giải chi tiết:

Ta có:

\(\begin{array}{l}\dfrac{{x – 5y}}{{2{\rm{x}} – 3y}} – \dfrac{{24{\rm{x}}y}}{{4{{\rm{x}}^2} – 9{y^2}}} – \dfrac{{x + 8y}}{{3y – 2{\rm{x}}}}\\ = \dfrac{{x – 5y}}{{2{\rm{x}} – 3y}} – \dfrac{{24{\rm{x}}y}}{{{{\left( {2{\rm{x}}} \right)}^2} – {{\left( {3y} \right)}^2}}} + \left( { – \dfrac{{x + 8y}}{{3y – 2{\rm{x}}}}} \right)\\ = \dfrac{{x – 5y}}{{2{\rm{x}} – 3y}} – \dfrac{{24{\rm{x}}y}}{{\left( {2{\rm{x}} – 3y} \right)\left( {2{\rm{x}} + 3y} \right)}} + \dfrac{{x + 8y}}{{2{\rm{x}} – 3y}}\\ = \dfrac{{x – 5y}}{{2{\rm{x}} – 3y}} + \dfrac{{x + 8y}}{{2{\rm{x}} – 3y}} – \dfrac{{24{\rm{x}}y}}{{\left( {2{\rm{x}} – 3y} \right)\left( {2{\rm{x}} + 3y} \right)}}\\ = \dfrac{{2{\rm{x}} + 3y}}{{2{\rm{x}} – 3y}} – \dfrac{{24{\rm{x}}y}}{{\left( {2{\rm{x}} – 3y} \right)\left( {2{\rm{x}} + 3y} \right)}}\\ = \dfrac{{{{\left( {2{\rm{x}} + 3y} \right)}^2}}}{{\left( {2{\rm{x}} – 3y} \right)\left( {2{\rm{x}} + 3y} \right)}} – \dfrac{{24{\rm{x}}y}}{{\left( {2{\rm{x}} – 3y} \right)\left( {2{\rm{x}} + 3y} \right)}}\\ = \dfrac{{4{{\rm{x}}^2} + 12{\rm{x}}y + 9{y^2} – 24{\rm{x}}y}}{{\left( {2{\rm{x}} – 3y} \right)\left( {2{\rm{x}} + 3y} \right)}}\\ = \dfrac{{4{{\rm{x}}^2} – 12{\rm{x}}y + 9{y^2}}}{{\left( {2{\rm{x}} – 3y} \right)\left( {2{\rm{x}} + 3y} \right)}} = \dfrac{{{{\left( {2{\rm{x}} – 3y} \right)}^2}}}{{\left( {2{\rm{x}} – 3y} \right)\left( {2{\rm{x}} + 3y} \right)}} = \dfrac{{2{\rm{x}} – 3y}}{{2{\rm{x}} + 3y}}\end{array}\)