Giải bài 3 trang 48 SGK Toán 8 tập 1 – Cánh diều

Đề bài

Tính một cách hợp lí:

\(a)\dfrac{{{x^2} – 49}}{{{x^2} + 5}}.\left( {\dfrac{{{x^2} + 5}}{{x – 7}} – \dfrac{{{x^2} + 5}}{{x + 7}}} \right)\)

\(b)\dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{2000 – x}}{{x + 1945}} + \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{2{\rm{x}} – 25}}{{x + 1945}}\)

Lời giải chi tiết

\(\begin{array}{l}a)\dfrac{{{x^2} – 49}}{{{x^2} + 5}}.\left( {\dfrac{{{x^2} + 5}}{{x – 7}} – \dfrac{{{x^2} + 5}}{{x + 7}}} \right)\\ = \dfrac{{\left( {x – 7} \right)\left( {x + 7} \right)}}{{{x^2} + 5}}.\dfrac{{{x^2} + 5}}{{x – 7}} – \dfrac{{\left( {x – 7} \right)\left( {x + 7} \right)}}{{{x^2} + 5}}.\dfrac{{{x^2} + 5}}{{x + 7}}\\ = x + 7 – \left( {x – 7} \right) = 14\end{array}\)

\(\begin{array}{l}b)\dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{2000 – x}}{{x + 1945}} + \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{2{\rm{x}} – 25}}{{x + 1945}}\\ = \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\left( {\dfrac{{2000 – x}}{{x + 1945}} + \dfrac{{2{\rm{x}} – 25}}{{x + 1945}}} \right)\\ = \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{2000 – x + 2{\rm{x}} – 25}}{{x + 1945}}\\ = \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{x + 1975}}{{x + 1945}} = \dfrac{{19{\rm{x}} + 8}}{{x + 1945}}\end{array}\)