Question

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

a) Tìm x để C có nghĩa

b) Rút gọn C

c) Tính C khi \(x^2\)-x=0

d) Tìm x để C =\(\dfrac{-1}{2}\) Giúp mình nha cảm ơn mọi nhười nhiều

Answer 1

a) C có nghĩa ⇔\(\left\{{}\begin{matrix}2x-2\ne0\\2x^2-2\ne0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

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

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

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

= \(\dfrac{1}{2\left(x+1\right)}\)

c) Ta có   x²-x=0 ⇒ \(\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Thay x=0 vào C= \(\dfrac{1}{2\left(x+1\right)}\)  ⇒ C= \(\dfrac{1}{2}\)

Thay x= 1  vào C = \(\dfrac{1}{2\left(x+1\right)}\)  ⇒ C= \(\dfrac{1}{4}\)

d)  C= \(\dfrac{1}{2\left(x+1\right)}\)= \(\dfrac{-1}{2}\)

⇔-2(x+1)=2 ⇔ x=-2