Giải Bài 19 trang 41 SGK Toán 8 tập 1 – Chân trời sáng tạo

Đề bài

Thực hiện các phép tính sau:

a) \(\dfrac{{8y}}{{3{x^2}}} \cdot \dfrac{{9{x^2}}}{{4{y^2}}}\)

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

c) \(\dfrac{{2{x^2} + 4}}{{x – 3}} \cdot \dfrac{{3x + 1}}{{x – 1}}:\dfrac{{{x^2} + 2}}{{6 – 2x}}\)

d) \(\dfrac{{2{x^2}}}{{3{y^3}}}:\left( { – \dfrac{{4{x^3}}}{{21{y^2}}}} \right)\)

e) \(\dfrac{{2x + 10}}{{{x^3} – 64}}:\dfrac{{{{\left( {x + 5} \right)}^2}}}{{2x – 8}}\)

g) \(\dfrac{1}{{x + y}}\left( {\dfrac{{x + y}}{{xy}} – x – y} \right) – \dfrac{1}{{{x^2}}}:\dfrac{y}{x}\)      

Lời giải chi tiết

a)

\(\dfrac{{8y}}{{3{x^2}}} \cdot \dfrac{{9{x^2}}}{{4{y^2}}}\) \( = \dfrac{{72{x^2}y}}{{12{x^2}{y^2}}} = \dfrac{6}{y}\)            

b)

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

c)

\(\dfrac{{2{x^2} + 4}}{{x – 3}} \cdot \dfrac{{3x + 1}}{{x – 1}}:\dfrac{{{x^2} + 2}}{{6 – 2x}}\) \( = \dfrac{{2\left( {{x^2} + 2} \right)}}{{x – 3}} \cdot \dfrac{{3x + 1}}{{x – 1}} \cdot \dfrac{{6 – 2x}}{{{x^2} + 2}} = \dfrac{{2\left( {{x^2} + 2} \right)}}{{x – 3}} \cdot \dfrac{{3x + 1}}{{x – 1}} \cdot \dfrac{{ – 2\left( {x – 3} \right)}}{{{x^2} + 2}} = \dfrac{{ – 4\left( {3x + 1} \right)}}{{x – 1}}\)

d)

\(\dfrac{{2{x^2}}}{{3{y^3}}}:\left( { – \dfrac{{4{x^3}}}{{21{y^2}}}} \right)\) \( = \dfrac{{2{x^2}}}{{3{y^3}}} \cdot \dfrac{{ – 21{y^2}}}{{4{x^3}}} = \dfrac{{ – 7}}{{2xy}}\)

e)

\(\dfrac{{2x + 10}}{{{x^3} – 64}}:\dfrac{{{{\left( {x + 5} \right)}^2}}}{{2x – 8}}\) \( = \dfrac{{2x + 10}}{{{x^3} – 64}} \cdot \dfrac{{2x – 8}}{{{{\left( {x + 5} \right)}^2}}} = \dfrac{{2\left( {x + 5} \right)}}{{\left( {x – 4} \right)\left( {{x^2} + 4x + 16} \right)}} \cdot \dfrac{{2\left( {x – 4} \right)}}{{{{\left( {x + 5} \right)}^2}}} = \dfrac{4}{{\left( {{x^2} + 4x + 16} \right)\left( {x + 5} \right)}}\)

g)

\(\dfrac{1}{{x + y}}\left( {\dfrac{{x + y}}{{xy}} – x – y} \right) – \dfrac{1}{{{x^2}}}:\dfrac{y}{x}\)

\(\begin{array}{l} = \dfrac{1}{{x + y}} \cdot \left( {\dfrac{{x + y}}{{xy}} – \left( {x + y} \right)} \right) – \dfrac{1}{{{x^2}}}.\dfrac{x}{y}\\ = \dfrac{1}{{x + y}} \cdot \dfrac{{x + y}}{{xy}} – \dfrac{1}{{x + y}} \cdot \left( {x + y} \right) – \dfrac{1}{{xy}}\\ = \dfrac{1}{{xy}} – 1 – \dfrac{1}{{xy}} = -1\end{array}\)