Giải Bài 3 trang 39 SGK Toán 8 tập 1 – Chân trời sáng tạo

Đề bài

Tính:

a) \(\dfrac{{4{x^2} + 2}}{{x – 2}} \cdot \dfrac{{3x + 2}}{{x – 4}} \cdot \dfrac{{4 – 2x}}{{2{x^2} + 1}}\) 

b) \(\dfrac{{x + 3}}{x} \cdot \dfrac{{x + 2}}{{{x^2} + 6x + 9}}:\dfrac{{{x^2} – 4}}{{{x^2} + 3x}}\)

Lời giải chi tiết

a) \(\dfrac{{4{x^2} + 2}}{{x – 2}} \cdot \dfrac{{3x + 2}}{{x – 4}} \cdot \dfrac{{4 – 2x}}{{2{x^2} + 1}}\)\( = \dfrac{{2\left( {2{x^2} + 1} \right)}}{{x – 2}} \cdot \dfrac{{3x + 2}}{{x – 4}} \cdot \dfrac{{ – 2\left( {x – 2} \right)}}{{2{x^2} + 1}} = \dfrac{{ – 4\left( {3x + 2} \right)}}{{x – 4}} = \dfrac{{ – 12x – 8}}{{x – 4}}\)

b) \(\dfrac{{x + 3}}{x} \cdot \dfrac{{x + 2}}{{{x^2} + 6x + 9}}:\dfrac{{{x^2} – 4}}{{{x^2} + 3x}}\)

\(\begin{array}{l} = \dfrac{{x + 3}}{x} \cdot \dfrac{{x + 2}}{{{{\left( {x + 3} \right)}^2}}} \cdot \dfrac{{{x^2} + 3x}}{{{x^2} – 4}}\\ = \dfrac{{x + 3}}{x} \cdot \dfrac{{x + 2}}{{{{\left( {x + 3} \right)}^2}}} \cdot \dfrac{{x\left( {x + 3} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\ = \dfrac{1}{{x – 2}}\end{array}\)