Question

[Nobody]

a) Chứng minh a² + b² + 4a - 6b +13 > 0

b) Tìm GTNN A= a² + b² -2a +10b -5

Answer 1

[Huyen Trang]

a) Ta có: \(a^2+b^2+4a-6b+13\)

\(=\left(a^2+4a+4\right)+\left(b^2-6b+9\right)\)

\(=\left(a+2\right)^2+\left(b-3\right)^2\ge0\left(\forall x,y\right)\)

=> đpcm

b) Ta có: 

\(A=a^2+b^2-2a+10b-5\)

\(A=\left(a^2-2a+1\right)+\left(b^2+10b+25\right)-31\)

\(A=\left(a-1\right)^2+\left(b+5\right)^2-31\ge-31\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+5\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}a=1\\b=-5\end{cases}}\)

Vậy \(Min_A=-31\Leftrightarrow\hept{\begin{cases}a=1\\b=-5\end{cases}}\)

Answer 2

[Xyz OLM]

a) Ta có : a² + b² + 4a - 6b + 13 = (a² + 4a + 4) + (b² - 6b + 9) = (a + 2)² + (b - 3)² \(\ge\)0\(\forall\)x;y

b) Ta có A = a² + b² - 2a + 10b - 5 = (a² - 2a + 1) + (b² + 10b + 25) - 31 = (a - 1)² + (b + 5)² - 31 \(\ge\)-31

Dấu "=" xảy  ra <=> \(\hept{\begin{cases}a-1=0\\b+5=0\end{cases}}\Rightarrow\hept{\begin{cases}a=1\\b=-5\end{cases}}\)

Vậy Min A = -31 <=> a = 1 ; b = -5