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

Đề bài

Tính một cách hợp lí:

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

$b)\dfrac{{19{{x}} + 8}}{{x + 1975}}.\dfrac{{2000 - x}}{{x + 1945}} + \dfrac{{19{{x}} + 8}}{{x + 1975}}.\dfrac{{2{{x}} - 25}}{{x + 1945}}$

Lời giải chi tiết

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

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