Giải bài 25 trang 47 - Sách giáo khoa Toán 8 tập 1

Đề bài

Làm tính cộng các phân thức sau:

Hướng dẫn giải

a) MTC = \(10x^2y^3\)

Ta có: \({\dfrac{5}{2{x^2}y}} + {\dfrac{3}{5x{y^2}}} + {\dfrac{x}{{y^3}}} = {\dfrac{25{y^2}} {10{x^2}{y^3}}} + {\dfrac{6xy}{10{x^2}{y^3}}} + {\dfrac{10x^3}{10{x^2}{y^3}}}\)

            \(= {\dfrac{25{y^2} + 6xy + 10{x^3}} {10{x^2}{y^3}}}\)

b) Ta có : \(2x + 6 = 2 ( x+ 3 )\)

MTC = \(2x ( x+ 3 )\)

Do đó : \({\dfrac{x + 1}{2x + 6}} + {\dfrac{2x + 3}{x( {x + 3} )}}\) = \(= {\dfrac{x + 1}{2\left( {x + 3} \right)}} + {\dfrac{2x + 3}{x\left( {x + 3} \right)}}\)

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

\(= {\dfrac{{x^2} + 5x + 6}{2x\left( {x + 3} \right)}} = {\dfrac{\left( {x + 2} \right) \left( {x + 3} \right)}{2x\left( {x + 3} \right)}}\) \(\dfrac{x+2}{2x}\)

c) Ta có : \(x^2 - 5x = x(x-5)\)

               \(25 - 5x = 5(5-x) = -5 ( x-5 )\)

MTC = \(5x ( x-5)\)

Do đó : \({\dfrac{3x + 5} {{x^2} - 5x}} + {\dfrac{25 - x} {25 - 5x}} = {\dfrac{3x + 5} {x(x-5)}} + {\dfrac{25-x} {-5(x - 5)}}\)

\(= {\dfrac{3x + 5}{x\left( {x - 5} \right)}} + {\dfrac{x - 25}{5\left( {x - 5} \right)}} = {\dfrac{5\left( {3x + 5} \right)}{5x\left( {x - 5} \right)}} + {\dfrac{x\left( {x - 25} \right)}{5x\left( {x - 5} \right)}}\)

\(= {\dfrac{15x + 25 + {x^2} - 25x}{5x\left( {x - 5} \right)}} = {\dfrac{{x^2} - 10x + 25} {5x\left( {x - 5} \right)}}\)

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

d) MTC = \(1-x^2\)

Ta có : \({x^2} + {\dfrac{{x^4} + 1}{1 - {x^2}}} + 1\) \(= {x^2} + 1 + {\dfrac{{x^4} + 1}{1 - {x^2}}}\)

 \(= {\dfrac{\left( {1 + {x^2}} \right)\left( {1 - {x^2}} \right)}{1 - {x^2}}} + {\dfrac{{x^4} + 1}{1 - {x^2}}}\)\(= {\dfrac{1 - {x^4} + {x^4} + 1}{1 - {x^2}}} = {\dfrac{2} {1 - {x^2}}}\)

e) MTC = \((x-1)(x^2 + x + 1 )\)

Do đó : \({\dfrac{4{x^2} - 3x + 17}{{x^3} - 1}} + {\dfrac{2x - 1}{{x^2} + x + 1}} + {\dfrac6 {1 - x}}\)

\(= {\dfrac{4{x^2} - 3x + 17}{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + {\dfrac{2x - 1} {{x^2} + x + 1}} + {\dfrac{ - 6} {x - 1}}\)

\(= {\dfrac{4{x^2} - 3x + 17} {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + {\dfrac{\left( {2x - 1} \right)\left( {x - 1} \right)}{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} +{\dfrac{ - 6\left( {{x^2} + x + 1} \right)}{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\)

\(= {\dfrac{4{x^2} - 3x + 17 + 2{x^2} - 3x + 1 - 6{x^2} - 6x - 6} {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\)

\(= {\dfrac{ - 12x + 12} {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = {\dfrac{ - 12\left( {x - 1} \right)}{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = {\dfrac{ - 12} {{x^2} + x + 1}}\)