Giải Bài 18 trang 41 SGK Toán 8 tập 1 – Chân trời sáng tạo

Đề bài

Thực hiện các phép tính sau:

a) $\dfrac{{2{x^2} - 1}}{{x - 2}} + \dfrac{{ - {x^2} - 3}}{{x - 2}}$                       

b) $\dfrac{x}{{x + y}} + \dfrac{y}{{x - y}}$    

c) $\dfrac{1}{{x - 1}} - \dfrac{2}{{{x^2} - 1}}$

d) $\dfrac{{x + 2}}{{{x^2} + xy}} - \dfrac{{y - 2}}{{xy + {y^2}}}$                                  

e) $\dfrac{1}{{2{x^2} - 3x}} - \dfrac{1}{{4{x^2} - 9}}$                     

g) $\dfrac{{2x}}{{9 - {x^2}}} + \dfrac{1}{{x - 3}} - \dfrac{1}{{x + 3}}$      

Lời giải chi tiết

a)

$\dfrac{{2{x^2} - 1}}{{x - 2}} + \dfrac{{ - {x^2} - 3}}{{x - 2}}$

$= \dfrac{{{x^2} - 4}}{{x - 2}}$

$\begin{array}{l} = \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{x - 2}}\\ = x + 2\end{array}$ 

b)

$\dfrac{x}{{x + y}} + \dfrac{y}{{x - y}}$

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

c)

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

d)

$\dfrac{{x + 2}}{{{x^2} + xy}} - \dfrac{{y - 2}}{{xy + {y^2}}}$          

$\begin{array}{l} = \dfrac{{x + 2}}{{x(x + y)}} - \dfrac{{y - 2}}{{y(x + y)}}\\ = \dfrac{{\left( {x + 2} \right)y}}{{xy\left( {x + y} \right)}} - \dfrac{{\left( {y - 2} \right)x}}{{xy\left( {x + y} \right)}}\\ = \dfrac{{xy + 2y}}{{xy\left( {x + y} \right)}} - \dfrac{{xy - 2x}}{{xy\left( {x + y} \right)}}\\ = \dfrac{{2y + 2x}}{{xy\left( {x + y} \right)}}\\ = \dfrac{{2\left( {x + y} \right)}}{{xy\left( {x + y} \right)}}\\ = \dfrac{2}{{xy}}\end{array}$

e)

$\dfrac{1}{{2{x^2} - 3x}} - \dfrac{1}{{4{x^2} - 9}}$    

$\begin{array}{l} = \dfrac{1}{{x\left( {2x - 3} \right)}} - \dfrac{1}{{\left( {2x - 3} \right)\left( {2x + 3} \right)}}\\ = \dfrac{{2x + 3}}{{x\left( {2x - 3} \right)\left( {2x + 3} \right)}} - \dfrac{x}{{x\left( {2x - 3} \right)\left( {2x + 3} \right)}}\\ = \dfrac{{x + 3}}{{x\left( {4{x^2} - 9} \right)}}\end{array}$

g)

$\dfrac{{2x}}{{9 - {x^2}}} + \dfrac{1}{{x - 3}} - \dfrac{1}{{x + 3}}$    

$\begin{array}{l} = \dfrac{{ - 2x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \dfrac{1}{{x - 3}} - \dfrac{1}{{x + 3}}\\ = \dfrac{{ - 2x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \dfrac{{x + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \dfrac{{x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\ = \dfrac{{ - 2x + 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\ = \dfrac{{ - 2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\ = \dfrac{{ - 2}}{{x + 3}}\end{array}$