a, Biểu thức = x^3-3x^2+3x-1-(x^3-1)-3.(x-x^2)

                   =x^3-3x^2+3x-1-x^2+1-3x+3x^2 = 0

=> giá trị của biểu thức trên ko phụ thuộc vào biến

Ủa khai triển ra là đc mà

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

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

`=1/((x+1)(x+2))(ĐPCM)`

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

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

\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}=\dfrac{1}{\left(x+1\right)\left(x+2\right)}\left(đpcm\right)\)

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

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

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

2) \(x^4-x^2+1=0\)(1)

Đặt: t=x², khi đó:

(1)\(\Leftrightarrow t^2-t+1=0\)

\(\Leftrightarrow\left(t-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(2\right)\)

\(\Rightarrow\left(2\right)\) vô nghiệm => (1) vô nghiệm

Ta có: (x-x²+1)/(x-x²-1) - 1

= (x-x²+1)/(x-x²-1) - (x-x²-1)/(x-x²-1)

= (x-x²+1-x+x²+1)/(x-x²-1) = 2/(x-x²-1) = -2/(x²-x+1)

Ta có: x²-x+1 = x²-x+1/4+3/4 = (x - 1/2)² + 3/4 > 0 với mọi x

Nên (x-x²+1)/(x-x²-1) < 1 (đpcm)

\(x\left(x-1\right)+8\left(x-1\right)=\left(x-1\right)\left(x+8\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+8\right)=\left(x-1\right)\left(x+8\right)\)

\(\left(đpcm\right)\)