\(5a^2+5b^2+8ab-2a+2b+2=0\)

\(\Leftrightarrow4a^2+4b^2+8ab+a^2-2a+1+b^2-2b+1=0\)

\(\Leftrightarrow\left(2a+2b\right)^2+\left(a-1\right)^2+\left(b+1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}2a+2b=0\\a-1=0\\b+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}a\cdot1+2\left(-1\right)=0\left(tm\right)\\a=1\\b=-1\end{cases}}}\)

Thay a, b vào B ta được :

\(B=\left(1-1\right)^{2018}+\left(1-2\right)^{2019}+\left(-1+1\right)^{2020}\)

\(B=0^{2018}+\left(-1\right)^{2019}+0^{2020}\)

\(B=-1\)

Dòng 2 là \(+2b\)nhé mình bấm lộn :)

P/s: ko chắc 

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

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

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

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

\(P=\frac{1}{x^2+x+1}\left[x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right]\)

\(P=\frac{1}{x^2+x+1}\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\)

\(P=\frac{1}{x^2+x+1}\cdot\left(x-\frac{1}{2}\right)^2+\frac{1}{x^2+x+1}\cdot\frac{3}{4}\)

Vì \(\frac{1}{x^2+x+1}\cdot\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow P\ge\frac{1}{x^2+x+1}\cdot\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{1}{x^2+x+1}\cdot\left(x-\frac{1}{2}\right)^2\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)

Vậy...

dễ hơn nè

Ta thấy x² + x + 1 > 0

Ta có : 2 ( x - 1 )² \(\ge\)0 \(\Rightarrow\)2x² - 4x + 2 \(\ge\)0 \(\Rightarrow\)3 ( x² - x + 1 ) \(\ge\)x² + x + 1

\(\Rightarrow\frac{x^2-x+1}{x^2+x+1}\ge\frac{1}{3}\) . Dấu " = " xảy ra  \(\Leftrightarrow\)x = 1 

I'm sorry. she's  you can call her after dinner

The five of us are very intelligent

He sent me a pen yesterday

Can you try on this shirt ?

He brought some money for him

My bike broke down on my way to school yesterday

If your car break down you can ask him to repair it

ĐKXĐ: \(x\ne\left\{-1;-\frac{1}{2}\right\}\)

\(\Leftrightarrow\left(\frac{x^2-4x+1}{x+1}+1\right)+\left(\frac{x^2-5x+1}{2x+1}+1\right)=0\)

\(\Leftrightarrow\frac{x^2-3x+2}{x+1}+\frac{x^2-3x+2}{2x+1}=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(\frac{1}{x+1}+\frac{1}{2x+1}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right).\frac{3x+2}{\left(x+1\right)\left(2x+1\right)}=0\)

Tập nghiệm: \(S=\left\{1;2;-\frac{2}{3}\right\}\)