Question

Cho biểu thức A = \(\left[\dfrac{\left(x-2\right)\left(x+1\right)}{x-1}-\left(x+2\right)\right]\)\(\dfrac{x^2-2x+1}{2}\)

a) Tìm điều kiện xác định của A và rút gọn A

b) Tìm x để A = -2

c) Tìm giá trị nhỏ nhất của A

Answer 1

a,\(A=\left[\dfrac{\left(x-2\right)\left(x+1\right)}{x-1}-\left(x+2\right)\right].\dfrac{x^2-2x+1}{2}\)

A xác định \(\Leftrightarrow x-1\ne0\)

\(\Leftrightarrow x\ne1\)

Rút gọn:

\(A=\left[\dfrac{\left(x-2\right)\left(x+1\right)}{x-1}-\left(x+2\right)\right].\dfrac{\left(x-1\right)^2}{2}\)

\(A=\left[\dfrac{\left(x-2\right)\left(x+1\right)}{x-1}-\dfrac{x+2}{1}\right].\dfrac{\left(x-1\right)^2}{2}\)

\(A=\left[\dfrac{\left(x-2\right)\left(x+1\right)}{x-1}-\dfrac{\left(x+2\right)\left(x-1\right)}{x-1}\right].\dfrac{\left(x-1\right)^2}{2}\)

\(A=\left(\dfrac{x^2-x-2-x^2-x+2}{x-1}\right).\dfrac{\left(x-1\right)^2}{2}\)

\(A=\dfrac{-2}{x-1}.\dfrac{\left(x-1\right)^2}{2}\)

\(A=\dfrac{-2.\left(x-1\right)^2}{\left(x-1\right).2}\)

\(A=-x+1\)

b,\(A=-2\Leftrightarrow-x+1=-2\)

\(\Leftrightarrow-x=-3\)

\(\Leftrightarrow x=3\)