Giải bài 1 trang 49 SGK Toán 8 tập 1 - Cánh diều

Đề bài

Thực hiện phép tính:

$a){\displaystyle \frac{x}{xy+y^2}}-{\displaystyle \frac{y}{x^2+xy}}$                            

$b){\displaystyle \frac{x^2+4}{x^2-4}}-{\displaystyle \frac{x}{x+2}}-{\displaystyle \frac{x}{2-x}}$

$c){\displaystyle \frac{a^2+ab}{b-a}}:{\displaystyle \frac{a+b}{2{\mathrm{a}}^2-2b^2}}$                          

$d)\left({\displaystyle \frac{2\mathrm{x}+1}{2\mathrm{x}-1}}-{\displaystyle \frac{2\mathrm{x}-1}{2\mathrm{x}+1}}\right):{\displaystyle \frac{4\mathrm{x}}{10\mathrm{x}-5}}$

Lời giải chi tiết

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

$\begin{array}{l}b){\displaystyle \frac{x^2+4}{x^2-4}}-{\displaystyle \frac{x}{x+2}}-{\displaystyle \frac{x}{2-x}}\\ ={\displaystyle \frac{x^2+4}{\left(x-2\right)\left(x+2\right)}}-{\displaystyle \frac{x}{x+2}}+{\displaystyle \frac{x}{x-2}}\\ ={\displaystyle \frac{x^2+4-x\left(x-2\right)+x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}}\\ ={\displaystyle \frac{x^2+4-x^2+2\mathrm{x}+x^2+2\mathrm{x}}{\left(x-2\right)\left(x+2\right)}}={\displaystyle \frac{x^2+4\mathrm{x}+4}{\left(x-2\right)\left(x+2\right)}}={\displaystyle \frac{{\left(x+2\right)}^2}{\left(x-2\right)\left(x+2\right)}}={\displaystyle \frac{x+2}{x-2}}\end{array}$

$\begin{array}{l}c){\displaystyle \frac{a^2+ab}{b-a}}:{\displaystyle \frac{a+b}{2{\mathrm{a}}^2-2b^2}}\\ ={\displaystyle \frac{a\left(a+b\right)}{b-a}}.{\displaystyle \frac{2{\mathrm{a}}^2-2b^2}{a+b}}\\ ={\displaystyle \frac{a\left(a+b\right).2.\left(a^2-b^2\right)}{-\left(a-b\right).\left(a+b\right)}}\\ ={\displaystyle \frac{a\left(a+b\right).2.\left(a-b\right).\left(a+b\right)}{-\left(a-b\right)\left(a+b\right)}}=-2\mathrm{a}\left(a+b\right)\end{array}$

$\begin{array}{l}d)\left({\displaystyle \frac{2\mathrm{x}+1}{2\mathrm{x}-1}}-{\displaystyle \frac{2\mathrm{x}-1}{2\mathrm{x}+1}}\right):{\displaystyle \frac{4\mathrm{x}}{10\mathrm{x}-5}}\\ ={\displaystyle \frac{{\left(2\mathrm{x}+1\right)}^2-{\left(2\mathrm{x}-1\right)}^2}{\left(2\mathrm{x}+1\right)\left(2\mathrm{x}-1\right)}}.{\displaystyle \frac{10x-5}{4\mathrm{x}}}\\ ={\displaystyle \frac{\left(2\mathrm{x}+1+2\mathrm{x}-1\right)\left(2\mathrm{x}+1-2\mathrm{x}+1\right)}{\left(2\mathrm{x}+1\right)\left(2\mathrm{x}-1\right)}}.{\displaystyle \frac{5.\left(2\mathrm{x}-1\right)}{4\mathrm{x}}}\\ ={\displaystyle \frac{4\mathrm{x}.2.5\left(2\mathrm{x}-1\right)}{\left(2\mathrm{x}+1\right)\left(2\mathrm{x}-1\right).4\mathrm{x}}}={\displaystyle \frac{10}{2\mathrm{x}+1}}\end{array}$