Question

[nghiemminhphuong]

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

\(P=505a+505b\)

GTNN, GTLN của P

Answer 1

[Nguyễn Linh Chi]

Ta có: \(a^2-ab+b^2=a+b\)

<=> \(a^2-a\left(b+1\right)+b^2-b=0\)

<=> \(a^2-2a.\frac{b+1}{2}+\left(\frac{b+1}{2}\right)^2-\frac{b^2}{4}-\frac{b}{2}-\frac{1}{4}+b^2-b=0\)

<=> \(\left(a-\frac{b+1}{2}\right)^2+\frac{3}{4}\left(b-1\right)^2=1\)

Ta có: \(\left(a-\frac{b+1}{2}\right)^2+\frac{3}{4}\left(b-1\right)^2=\frac{\left(a-\frac{b+1}{2}\right)^2}{1}+\frac{\left(\frac{3}{2}b-\frac{3}{2}\right)^2}{3}\)

\(\ge\frac{\left(a+b-2\right)^2}{4}\)

=> \(1\ge\frac{\left(a+b-2\right)^2}{4}\)

<=> \(\left(a+b-2\right)^2\le4\)

<=> \(-2\le a+b-2\le2\)

<=> \(0\le a+b\le4\)

mà  \(P=505a+505b=505\left(a+b\right)\)

=> \(0\le P\le2020\)

Dấu "=" xảy ra <=> \(\frac{a-\frac{b+1}{2}}{1}=\frac{\frac{3}{2}b-\frac{3}{2}}{3}\)<=> a = b 

Nếu P = 0 khi đó: a + b = 0 <=> a = b = 0 

Nếu P = 2020 <=>  a + b = 4 <=> a = b = 2

Vậy: GTNN của P = 0 đạt tại a = b = 0 

GTLN của P= 2020 đạt tại a = b = 2

Answer 2

[nghiemminhphuong]

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

\(\left(a-b\right)^2\ge0\Rightarrow\left(a+b\right)\ge ab\Rightarrow2\left(a+b\right)\ge2ab\)

\(\Leftrightarrow\left(a+b\right)^2\le2\left(a+b\right)+a^2+b^2=2\left(a+b\right)+a+b+ab\le4\left(a+b\right)\)

\(\Leftrightarrow0\le a+b\le4\Leftrightarrow0\le P\le2020\)\(D=xr\Leftrightarrow\orbr{\begin{cases}a=b=0\\a=b=2\end{cases}}\)