Question

cho x>0 , y>0 , x+y =2012

a) Tìm Max \(B=\frac{2x^2+8xy+2y^2}{x^2+2xy+y^2}\)

b) Tìm Min \(C=\left(1+\frac{2012}{x}\right)^2+\left(1+\frac{2012}{y}\right)^2\)

Answer 1

\(a)\) Có \(2012=x+y\ge2\sqrt{xy}\)\(\Leftrightarrow\)\(xy\le1006^2\)

\(B=\frac{2x^2+8xy+2y^2}{x^2+2xy+y^2}=\frac{2\left(x^2+2xy+y^2\right)}{x^2+2xy+y^2}+\frac{4xy}{x^2+2xy+y^2}=2+\frac{4xy}{\left(x+y\right)^2}\)

\(\le2+\frac{4.1006^2}{2012^2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1006\)

\(b)\) \(C=\left(1+\frac{2012}{x}\right)^2+\left(1+\frac{2012}{y}\right)^2\ge\left[2+2012\left(\frac{1}{x}+\frac{1}{y}\right)\right]^2\ge\left(2+\frac{2012.4}{x+y}\right)^2\)

\(=\left(2+\frac{2012.4}{2012}\right)^2=36\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1006\)

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Answer 2

cảm ơn bạn nhiều

Answer 3

bạn ơi, mik học \(A^2+B^2\ge\left(A+B\right)^2d\text{ấu}"="\) xảy ra <=> \(A.B\ge0\) mà bạn?

Answer 4

\(C=\left(1+\frac{2012}{x}\right)^2+\left(1+\frac{2012}{y}\right)^2\ge\frac{\left[2+2012\left(\frac{1}{x}+\frac{1}{y}\right)\right]^2}{1+1}\left(Svacxo\right)\)

\(\ge\frac{\left(2+2012\cdot\frac{4}{x+y}\right)^2}{2}=\frac{\left(2+4\right)^2}{2}=\frac{36}{2}=18\)

Dấu "=" xảy ra <=> x=y=1006