Ta có x + y= 3 => x= 3 - y

=> (3 - y)^2 + y^2 \(\ge\)5

Giải bất phương trình trên, ta được: y \(\ge\)2

Chỉ biết giải đến đó, min P= 33 thì phải

cảm ơn bn , tôi nghĩ ra rồi

bn ra dc \(y\ge2\)thì thay vào \(x^2+y^2\ge5\) ra dc \(x\ge1\)

khi đó min P = 1+16+6.4.1=41 khi và chỉ khi x=1 và y=2

tks bn 

đoán xem

Trả lời :

2 + 27 + 2019 = 2048

Hok tốt

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2 + 27 + 2019 =2048

Ta có:\(\frac{x-3}{2011}+\frac{x-2}{2012}=\frac{x-2012}{2}+\frac{x-2011}{3}\)

\(\Rightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-2}{2012}-1\right)=\left(\frac{x-2012}{2}-1\right)+\left(\frac{x-2011}{3}-1\right)\)

\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}=\frac{x-2014}{2}+\frac{x-2014}{3}\)

\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}-\frac{x-2014}{2}-\frac{x-2014}{3}=0\)

\(\Rightarrow\left(x-2014\right).\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\right)\)

\(\Rightarrow x-2014=0\)( vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\ne0\))

\(\Rightarrow x=2014\)

Vậy x= 2014.

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