ĐKXĐ: \(x\ne1;x\ne-1\)

\(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}\) \(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}-\dfrac{4x}{2\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x^2+2x+1-4x}{2\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{x-1}{2\left(x+1\right)}\)

\(\dfrac{x+10}{x-2}+\dfrac{x-18}{x+2}+\dfrac{x+2}{x^2-4}=\dfrac{\left(x+10\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-18\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+12x+20+x^2-16x-36+x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2-3x-14}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x^2+4x\right)-\left(7x+14\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x\left(x+2\right)-7\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x-7\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-7}{x-2}\)

Ta có:

(2x – 3) . (x² – 5x + 1)

= 2x. (x² – 5x + 1)  + (-3). (x² – 5x + 1)

= 2x . x² + 2x . (-5x) + 2x . 1 + (-3).x² + (-3).(-5x) + (-3). 1

= 2x³ + (-10x² ) + 2x + (-3x²) + 15x + (-3)

= 2x³ + (-10x² + -3x²) + (2x + 15x) + (-3)

ĐKXĐ: \(x\ne\dfrac{2}{5}\)

\(\dfrac{2x}{5x-2}+\dfrac{4}{5x-2}\) \(=\dfrac{2x+4}{5x-2}\)

Chọn B

a) `(x^3-x^2)/(x^3-2x^2+x)`

`=(x^2(x-1))/(x(x-1)(x-1))`

`=x/(x-1)`

`=>` 2 phân thức bằng nhau.

b) `(x^2+2x+1)/(2x^2-2)`

`=((x+1)(x+1))/(2(x+1)(x-1))`

`=(x+1)/(2(x-1))`

`=(x+1)/(2x-2)`

`=>` 2 phân thức bằng nhau

a) Ta có: \(\dfrac{x^3-x^2}{x^3-2x^2+x}\)

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

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

b) Ta có: \(\dfrac{x^2+2x+1}{2x^2-2}\)

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

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

1) \(\Delta'=1-m>0\forall m< 1\)

Vậy phương trình đã cho luôn có hai nghiệm phân biệt

2) Do a = 1; c = -1 nên a và c trái dấu

Do đó phương trình luôn có hai nghiệm phân biệt

Theo Viét, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1x_2=-1\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_2+x_1}{x_1x_2}=\dfrac{-2}{-1}=2\)