Tính: x-(2x-y/4)+(x+4y/12)

Giải SBT Toán 8 Bài 6: Cộng, trừ phân thức

Bài 5 trang 22 SBT Toán 8 Tập 1:Tính:

a)$x-\frac{2x-y}{4}+\frac{x+4y}{12}$;

b)$\frac{y}{x}-\frac{x}{y}-\frac{x^2+y^2}{xy}$;

c)$\frac{4}{x+2}-\frac{3}{x-2}+\frac{12}{x^2-4}$;

d)$\frac{x+y}{x^2-xy}-\frac{4x}{x^2-y^2}-\frac{x-y}{x^2+xy}$.

Lời giải:

a)$x-\frac{2x-y}{4}+\frac{x+4y}{12}$

$=\frac{x.12}{12}-\frac{3\left(2x-y\right)}{12}+\frac{x+4y}{12}$

$=\frac{12x-6x+3y+x+4y}{12}=\frac{7x+7y}{12}$.

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

$=\frac{y.y}{xy}-\frac{x.x}{xy}-\frac{x^2+y^2}{xy}=\frac{y^2-x^2-x^2-y^2}{xy}$

$=\frac{-2x^2}{xy}=\frac{-2x}{y}$.

c)$\frac{4}{x+2}-\frac{3}{x-2}+\frac{12}{x^2-4}=\frac{4}{x+2}-\frac{3}{x-2}+\frac{12}{\left(x-2\right)\left(x+2\right)}$

$=\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12}{\left(x+2\right)\left(x-2\right)}$

$=\frac{4x-8-3x-6+12}{\left(x+2\right)\left(x-2\right)}=\frac{x-2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x+2}$.

d)$\frac{x+y}{x^2-xy}-\frac{4x}{x^2-y^2}-\frac{x-y}{x^2+xy}$

$=\frac{x+y}{x\left(x-y\right)}-\frac{4x}{\left(x+y\right)\left(x-y\right)}-\frac{x-y}{x\left(x+y\right)}$

$=\frac{\left(x+y\right)\left(x+y\right)}{x\left(x-y\right)\left(x+y\right)}-\frac{4x^2}{x\left(x+y\right)\left(x-y\right)}-\frac{\left(x-y\right)\left(x-y\right)}{x\left(x-y\right)\left(x+y\right)}$

$=\frac{{\left(x+y\right)}^2-4x^2-{\left(x-y\right)}^2}{x\left(x+y\right)\left(x-y\right)}$

$=\frac{x^2+2xy+y^2-4x^2-\left(x^2-2xy+y^2\right)}{x\left(x+y\right)\left(x-y\right)}$

$=\frac{-3x^2+2xy+y^2-x^2+2xy-y^2}{x\left(x+y\right)\left(x-y\right)}=\frac{-4x^2+4xy}{x\left(x+y\right)\left(x-y\right)}$

$=\frac{-4x\left(x-y\right)}{x\left(x+y\right)\left(x-y\right)}=\frac{-4}{x+y}$.