\(\frac{2}{x^2-x+1}-\frac{1}{x+1}=\frac{2x+2-x^2+x-1}{x^3+1}=\frac{3x+1-x^2}{x^3+1}\\ \)

\(\Rightarrow3x+1-x^2=2x-1\Rightarrow\left(x+1\right)\left(x-2\right)=0\Rightarrow x=2\)

\(\frac{2}{x^2-x+1}=\frac{1}{x+1}+\frac{2x-1}{x^3+1}\)

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

ĐKXĐ : x\(\ne\)-1

MTC : (x+1)(x^2-x+1)

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

2x+2=x^2-x+1+2x-1

-x^2+2x-2x+x=-2+1-1

x-x^2=-2

x(1-x)=-2

.....................

\(A=\frac{\left(x+1\right)^2+8}{7-\left(y+1\right)^2}\) => không có GTNN cũng chẳng có LN

\(A-1=\frac{1}{1.2}+\frac{1}{2.3}..+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}\)\(=\frac{99}{100}\)

\(A=1+\frac{99}{100}=\frac{199}{100}\)

=1+1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/98-1/99+1/99-1/100
=1+1/2+1/2-1/100

=199/100

\(a\sqrt{b-1}+b\sqrt{a-1}\Leftrightarrow\sqrt{a}\sqrt{ab-a}+\sqrt{b}\sqrt{ab-b}\)

\(\le\sqrt{\left(a+b\right)\left(2ab-a-b\right)}\le\frac{a+b-a-b+2ab}{2}=ab\)

BĐT đc chứng minh

\(x=\sqrt{a-1};y=\sqrt{b-1}\) bỏ căn đi viết cho dẽ nhìn

\(x^2=a-1;y^2=b-1\Leftrightarrow\left(x^2+1\right)y+\left(y^2+1\right)x\le\left(x^2+1\right)\left(y^2+1\right)\)

\(\Leftrightarrow\left(x^2+1\right)\left(y^2-2y+1\right)+\left(y^2+1\right)\left(x^2-2x+1\right)\ge0\)

\(\Leftrightarrow\left(x^2+1\right)\left(y-1\right)^2+\left(y^2+1\right)\left(x-1\right)^2\ge0\)Đúng với mọi x,y => dpcm

Đẳng thức khi x=y=1=> a=b=2