Question

Giá trị của phân thức \( P = \frac{5x^2 - 10xy + 5y^2}{2x^2 - 2y^2} \) tại \( x = -\frac{1}{2}; y = -3 \) là:

Biết \(\frac{x^2 - x + 2}{x^2 - x + 1} - M = \frac{x}{x^2 - x + 1}\). Tính giá trị của \( M \) tại \( x = 1 \).

Rút gọn biểu thức \(\left( \frac{1}{x^2 + x} - \frac{2-x}{x+1} \right) : \left( \frac{1}{x} + x - 2 \right)\)

Answer 1

a)P= \(\dfrac{5\left(x^2-2xy+y^2\right)}{2\left(x^2-y^2\right)}\)=\(\dfrac{5\left(x-y\right)^2}{2\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{5\left(x-y\right)}{2\left(x+y\right)}\)

Với x = \(\dfrac{-1}{2}\);y = -3

⇒P = \(\dfrac{5\left(\left(\dfrac{-1}{2}\right)-\left(-3\right)\right)}{2\left(\left(\dfrac{-1}{2}\right)+\left(-3\right)\right)}\)=\(\dfrac{5\left(\dfrac{-1+3.2}{2}\right)}{2\left(\dfrac{-1+2.\left(-3\right)}{2}\right)}\)=\(\dfrac{5.\dfrac{5}{2}}{2.\dfrac{-7}{2}}\)=\(\dfrac{25}{2}\):(-7)=\(\dfrac{-25}{14}\)

b)M = \(\dfrac{x^2-x+2}{x^2-x+1}\) - \(\dfrac{x}{x^2-x+1}\)= \(\dfrac{x^2-2x+2}{x^2-x+1}\)

Với x = 1

⇒M=\(\dfrac{1^2-2.1+2}{1^2-1+1}\)=\(\dfrac{1}{1}\)=1

c) \(\left(\dfrac{1}{x\left(x+1\right)}+\dfrac{x-2}{x+1}\right):\left(\dfrac{1+x^2-2x}{x}\right)\)

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

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

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