Giải câu 25 bài: Luyện tập sgk Toán 8 tập 1 Trang 46

Giải câu 25 bài: Luyện tập sgk Toán 8 tập 1 Trang 46.

Hướng dẫn: Đưa các phân thức về cùng mẫu bằng cách quy đồng rồi cộng tử với nhau.

a)$\frac{5}{2x^{2}y}+\frac{3}{5x{y}^2}+\frac{x}{y^3}$

$=\frac{5.5y^2}{2x^{2}y.5y^2}+\frac{3.2xy}{5x{y}^2.2xy}+\frac{x.10x^2}{y^3.10x^2}$

$=\frac{25y^2}{10x^2{y}^3}+\frac{6xy}{10x^2{y}^3}+\frac{10x^3}{10x^2{y}^3}$

$=\frac{25y^2+6xy+10x^3}{10x^2{y}^3}$

b)$\frac{x+1}{2x+6}+\frac{2x+3}{x\left(x+3\right)}$

$=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}$

$=\frac{x\left(x+1\right)}{2x\left(x+3\right)}+\frac{2\left(2x+3\right)}{2x\left(x+3\right)}$

$=\frac{x^2+x+4x+6}{2x\left(x+3\right)}$

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

$=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}$

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

c)$\frac{3x+5}{x^2-5x}+\frac{25-x}{25-5x}$

$=\frac{3x+5}{x^2-5x}+\frac{x-25}{5x-25}$

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

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

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

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

d)$x^2+\frac{x^4+1}{1-x^2}+1$

$=1+{\mathrm{x}}^2+\frac{x^4+1}{1-x^2}$

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

$=\frac{1-x^4+x^4+1}{1-x^2}=\frac{2}{1-x^2}$

e)$\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6}{1-x}$

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

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

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

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

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