Question

[hahaha]

Cho x+y=1

Tìm GTNN của A=x²+y²

help me, please

Answer 1

[Nhật Minh]

\(A=\frac{1}{2}\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\frac{1}{2}\left(x+y\right)^2=\frac{1}{2}\)

Min A= 1/2  khi x = y =1/2

Answer 2

[Hoàng Phúc]

Vì x+y=1

=>y=1-x

Ta có: \(A=x^2+y^2=x^2+\left(1-x\right)^2=x^2+1\left(1-x\right)-x\left(1-x\right)=x^2+1-x-x+x^2\)

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

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

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

Vì \(2\left(x-\frac{1}{2}\right)^2>=0\) với mọi x

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

Dấu "=" xảy ra <=>\(x=\frac{1}{2}\);mà x+y=1=>\(y=\frac{1}{2}\)

Khi đó GTNN của A=x²+y² là 1/2 tại \(x=y=\frac{1}{2}\)