Bài tập 6.32 trang 13 SBT Toán 8 Tập 2 Kết nối tri thức - KNTT

a) Điều kiện xác định của P là: $\left\{\begin{array}{l}x-1\ne 0\\ 1-x^3\ne 0\\ x+1\ne 0\\ x^2+2x+1={\left(x+1\right)}^2\ne 0\end{array}\right.hay\left\{\begin{array}{l}x\ne 1\\ x\ne -1\end{array}\right.$.

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

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

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

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

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

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

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

c) Thay $x=\frac{1}{2}$ vào P sau khi rút gọn ta có: $P=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3$

-- Mod Toán 8 HỌC247