Question

Giúp mik bài này vs,sẽ follow ai giải đc ạ.Cảm ơnnnnn

Answer 1

a) \(\dfrac{x^2+2}{x^3-1}+\dfrac{x+1}{x^2+x+1}+\dfrac{1}{1-x}\)

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

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

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

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

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

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

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

Answer 2

Answer 3

a)ĐKXĐ:x khác 1

C=x^2+2/(x-1)(x^2+x+1) +(x+1)(x-1)/(x-1)(x^2+x+1) - (x^2+x+1)/(x-1)(x^2+x+1)

C=x^2+2+x^2-1-x^2-x-1/(x-1)(x^2+x+1)

C=x^2-x/(x-1)(x^2+x+1)

C=x(x-1)/(x-1)(x^2+x+1)

C=x/x^2+x+1

b)Xét C-1/3=x/x^2+x+1 -x^2+x+1/x^2+x+1

                  =-(x^2+1)/x^2+x+1

                  =-(x^2+1)/(x+1/2)^2+3/4

Vì x^2+1/(x+1/2)^2+3/4 >0

=>C-1/3=-(x^2+1)/(x+1/2)^2+3/4<0

=>C<1/3