Làm tính cộng các phân thức sau: 5/ (2x^2y) + 3/(5xy^2)

Giải Toán 8 Bài 5: Phép cộng các phân thức đại số

Video Giải Bài 25 trang 47 Toán 8 Tập 1

Bài 25 trang 47 Toán 8 Tập 1:

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

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

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

d) $x^2+\frac{x^4+1}{1-x^2}+1;$

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

Lời giải

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

$=\frac{5.5y^2}{10x^2{y}^3}+\frac{3.2xy}{10x^2{y}^3}+\frac{x.10x^2}{10x^2{y}^3}=\frac{25y^2}{10x^2{y}^3}+\frac{6xy}{10x^2{y}^3}+\frac{10x^3}{10x^2{y}^3}=\frac{25y^2+6xy+10x^3}{10x^2{y}^3}$

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

$=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}=\frac{\left(x+1\right).x}{2x\left(x+3\right)}+\frac{\left(2x+3\right).2}{2x\left(x+3\right)}=\frac{x^2+x}{2x\left(x+3\right)}+\frac{4x+6}{2x\left(x+3\right)}=\frac{x^2+x+4x+6}{2x\left(x+3\right)}=\frac{x^2+5x+6}{2x\left(x+3\right)}=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}.$

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

$=\frac{3x+5}{x\left(x-5\right)}+\frac{25-x}{5\left(5-x\right)}=\frac{3x+5}{x\left(x-5\right)}-\frac{25-x}{5\left(x-5\right)}=\frac{\left(3x+5\right).5}{5x\left(x-5\right)}-\frac{\left(25-x\right).x}{5x\left(x-5\right)}=\frac{15x+25}{5x\left(x-5\right)}-\frac{25x-x^2}{5x\left(x-5\right)}=\frac{15x+25-25+x^2}{5x\left(x-5\right)}=\frac{15x+x^2}{5x\left(x-5\right)}=\frac{x\left(15+x\right)}{5x\left(x-5\right)}=\frac{x+15}{5\left(x-5\right)}$

d) $x^2+\frac{x^4+1}{1-x^2}+1$

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

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

$=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-2x-x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6x^2+6x+6}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{4x^2-3x+17+2x^2-2x-x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12}{x^2+x+1}.$