Question

[Phạm Hữu Nam chuyên Đại số ♏ (Hội....]

tinh gia tri lon nhat cua bieu thuc

P=x.y

biet x+y =1

Answer 1

[Đặng Viết Thái]

x+y=1

<=> x=1-y

<=>P=(1-y)y=\(y-y^2\)

<=>P=\(\frac{1}{4}-\left(y^2-y+\frac{1}{4}\right)\)

<=>P=\(\frac{1}{4}-\left(y-\frac{1}{2}\right)^2\le\frac{1}{4}\)

=>Max của P=\(\frac{1}{4}\)<=>y=\(\frac{1}{2}\)

Answer 2

[Hà Hoàng Thịnh]

x+y=1

\(\Rightarrow x=1-y\)

\(\Rightarrow P=x.y=\left(1-y\right).y=y-y^2=-\left(y^2-y\right)\)

\(\Rightarrow P=-\left(y^2-2.y.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right)\)

\(\Rightarrow P=-\left(y^2-2.y.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}\)

\(\Rightarrow P=-\left(y-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vì :\(\left(y-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow-\left(y-\frac{1}{2}\right)^2\le0\)

\(\Rightarrow P\le\frac{1}{4}\)

\(\Rightarrow GTLN\)của\(P=\frac{1}{4}\)khi : \(y=\frac{1}{2}\)

\(\Rightarrow x=1-\frac{1}{2}=\frac{1}{2}\)