Câu 9 trang 26 Sách bài tập (SBT) Toán 8 tập 1

Rút gọn các phân thức sau:

a. \(\dfrac{{14x{y^5}\left( {2x - 3y} \right)}}{{21{x^2}y{{\left( {2x - 3y} \right)}^2}}}\)

b. \(\dfrac{{8xy{{\left( {3x - 1} \right)}^3}}}{{12{x^3}\left( {1 - 3x} \right)}}\)

c. \(\dfrac{{20{x^2} - 45}}{{{{\left( {2x + 3} \right)}^2}}}\)

d.\(\dfrac{{5{x^2} - 10xy}}{{2{{\left( {2y - x} \right)}^3}}}\)

e. \(\dfrac{{80{x^3} - 125x}}{{3\left( {x - 3} \right) - \left( {x - 3} \right)\left( {8 - 4x} \right)}}\)

f. \(\dfrac{{9 - {{\left( {x + 5} \right)}^2}}}{{{x^2} + 4x + 4}}\)

g. \(\dfrac{{32x - 8{x^2} + 2{x^3}}}{{{x^3} + 64}}\)

h. \(\dfrac{{5{x^3} + 5x}}{{{x^4} - 1}}\)

i. \(\dfrac{{{x^2} + 5x + 6}}{{{x^2} + 4x + 4}}\)

Giải:

a. \(\dfrac{{14x{y^5}\left( {2x - 3y} \right)}}{{21{x^2}y{{\left( {2x - 3y} \right)}^2}}} = \dfrac{{2{y^4}}}{{3x\left( {2x - 3y} \right)}}\)

b. \(\dfrac{{8xy{{\left( {3x - 1} \right)}^3}}}{{12{x^3}\left( {1 - 3x} \right)}} = \dfrac{{ - 8xy{{\left( {3x - 1} \right)}^3}}}{{12{x^2}\left( {3x - 1} \right)}}\)

\( = \dfrac{{ - 2y{{\left( {3x - 1} \right)}^2}}}{{3x}}\)

c.  

\(\eqalign{
& {{20{x^2} - 45} \over {{{\left( {2x + 3} \right)}^2}}} = {{5\left( {4{x^2} - 9} \right)} \over {{{\left( {2x + 3} \right)}^2}}} \cr
& = {{5\left( {2x + 3} \right)\left( {2x - 3} \right)} \over {{{\left( {2x + 3} \right)}^2}}} = {{5\left( {2x - 3} \right)} \over {2x + 3}} \cr} \)

d. \(\dfrac{{5{x^2} - 10xy}}{{2{{\left( {2y - x} \right)}^3}}} = \dfrac{{ - 5x\left( {2y - x} \right)}}{{2{{\left( {2y - x} \right)}^3}}}\)

\( = \dfrac{{ - 5x}}{{2{{\left( {2y - x} \right)}^2}}}\)

e. 

\(\eqalign{
& {{80{x^3} - 125x} \over {3\left( {x - 3} \right) - \left( {x - 3} \right)\left( {8 - 4x} \right)}} \cr
& = {{5x\left( {16{x^2} - 25} \right)} \over {\left( {x - 3} \right)\left( {3 - 8 + 4x} \right)}} \cr
& = {{5x\left( {4x - 5} \right)\left( {4x + 5} \right)} \over {\left( {x - 3} \right)\left( {4x - 5} \right)}} \cr
& = {{5x\left( {4x + 5} \right)} \over {x - 3}} \cr} \)

f. 

\(\eqalign{
& {{9 - {{\left( {x + 5} \right)}^2}} \over {{x^2} + 4x + 4}} \cr
& = {{\left( {3 + x + 5} \right)\left( {3 - x - 5} \right)} \over {{{\left( {x + 2} \right)}^2}}} \cr
& = {{\left( {8 + x} \right)\left( { - 2 - x} \right)} \over {{{\left( {x + 2} \right)}^2}}} \cr
& = {{ - \left( {8 + x} \right)\left( {x + 2} \right)} \over {{{\left( {x + 2} \right)}^2}}} = {{ - \left( {8 + x} \right)} \over {x + 2}} \cr} \)

g. 

\(\eqalign{
& {{32x - 8{x^2} + 2{x^3}} \over {{x^3} + 64}} \cr
& = {{2x\left( {16 - 4x + {x^2}} \right)} \over {\left( {x + 4} \right)\left( {{x^2} - 4x + 16} \right)}} \cr
& = {{2x} \over {x + 4}} \cr} \)

h. 

\(\eqalign{
& {{5{x^3} + 5x} \over {{x^4} - 1}} = {{5x\left( {{x^2} + 1} \right)} \over {\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}} \cr
& = {{5x} \over {{x^2} - 1}} \cr} \)

i. 

\(\eqalign{
& \frac{{{x^2} + 5x + 6}}{{{x^2} + 4x + 4}} \cr
& = \frac{{{x^2} + 2x + 3x + 6}}{{{{\left( {x + 2} \right)}^2}}} \cr
& = \frac{{x\left( {x + 2} \right) + 3\left( {x + 2} \right)}}{{{{\left( {x + 2} \right)}^2}}} \cr
& = \frac{{\left( {x + 2} \right)\left( {x + 3} \right)}}{{{{\left( {x + 2} \right)}^2}}} = \frac{{x + 3}}{{x + 2}} \cr} \)

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