Câu hỏi của Kiên NT

Question

Tìm giá trị nhỏ nhất của:P = x^2 - y^2 - xy - x + y +1

Nguyen · Accepted Answer

$2P=2x^2-2y^2-2xy-2x+2y+2$

$2P=\left(x-y\right)^2+\left(1-x\right)^2+\left(y+1\right)^2$

Áp dụng BĐT Bunhiacopxki:

$\left(1^2+1^2+1^2\right)\left[\left(x-y\right)^2+\left(1-x\right)^2+\left(y+1\right)^2\right]\ge\left(x-y+1-x+y+1\right)^2$

$3.2M\ge4$

$\Leftrightarrow M\ge\dfrac{2}{3}$

Mmin$=\dfrac{2}{3}\Leftrightarrow\dfrac{1}{x-y}=\dfrac{1}{1-x}=\dfrac{1}{y+1}$

$\Leftrightarrow x=\dfrac{1}{3};y=\dfrac{-1}{3}$Câu trả lời: $2P=2x^2-2y^2-2xy-2x+2y+2$

$2P=\left(x-y\right)^2+\left(1-x\right)^2+\left(y+1\right)^2$

Áp dụng BĐT Bunhiacopxki:

$\left(1^2+1^2+1^2\right)\left[\left(x-y\right)^2+\left(1-x\right)^2+\left(y+1\right)^2\right]\ge\left(x-y+1-x+y+1\right)^2$

$3.2M\ge4$

$\Leftrightarrow M\ge\dfrac{2}{3}$

Mmin$=\dfrac{2}{3}\Leftrightarrow\dfrac{1}{x-y}=\dfrac{1}{1-x}=\dfrac{1}{y+1}$

$\Leftrightarrow x=\dfrac{1}{3};y=\dfrac{-1}{3}$