\(\left(2x+1\right)^2-2\left(2x+1\right)\left(3-x\right)+\left(3-x\right)^2\)

\(=\left[\left(2x+1\right)-\left(3-x\right)\right]^2\)

\(=\left(3x-2\right)^2\)

p/s: chúc bạn học tốt

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

Vậy a=1 đó

1)We have: \(a-b=8\)

\(\Rightarrow\left(a-b\right)^2=64\)

\(\Rightarrow a^2-2ab+b^2=64\)

\(\Rightarrow a^2+2ab+b^2-4ab=64\)

\(\Rightarrow\left(a+b\right)^2=64+4ab=64+4\cdot10=64+40=104\)

Hence: \(\left(a+b\right)^2=104\)

2)We have: \(a+b=8\)

\(\Rightarrow\left(a+b\right)^2=64\)

\(\Rightarrow a^2+2ab+b^2=64\)

\(\Rightarrow a^2-2ab+b^2+4ab=64\)

\(\Rightarrow\left(a-b\right)^2=64-4ab=64-4\cdot10=64-40=24\)

Hence \(\left(a-b\right)^2=24\)

áp dụng cosi a^2+1>=2a tương tự và cộng vế tương ứng suy ra đpcm

\(a^2+b^2+2\ge2\left(a+b\right)\)

\(\Leftrightarrow a^2+b^2+2-2\left(a+b\right)\ge0\)

\(\Leftrightarrow a^2+b^2+2-2a-2b\ge0\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2\ge0\)( luôn đúng )

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}b-1=0\\b-1=0\end{cases}}\)\(\Leftrightarrow a=b=1\)

Vậy ...