Bài tập 6.18 trang 9 SBT Toán 8 Tập 2 Kết nối tri thức - KNTT

a) $\frac{5}{6x^{2}y}+\frac{7}{12x{y}^2}+\frac{11}{18xy}$

= $\frac{30y}{36x^2{y}^2}+\frac{21x}{36x^2{y}^2}+\frac{22xy}{36x^2{y}^2}$ (Mẫu thức chung là: 36x²y²)

= $\frac{30y+21x+22xy}{36x^2{y}^2}$.

b) $\frac{x^3+2x}{x^3+1}+\frac{2x}{x^2-x+1}+\frac{1}{x+1}$

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

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

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

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

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

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

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

-- Mod Toán 8 HỌC247