\(x^{2010}+y^{2010}=x^{2011}+y^{2011}=x^{2012}+y^{2012}\)

\(\Leftrightarrow x^{2010}+x^{2012}-2x^{2011}+y^{2010}+y^{2012}-2y^{2011}=0\)

\(\Leftrightarrow x^{2010}\left(x^2-2x+1\right)+y^{2010}\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow x^{2010}\left(x-1\right)^2+y^{2010}\left(y-1\right)^2=0\)

\(x^{2010};y^{2010}>0\Leftrightarrow x=y=1.\Rightarrow x^{2016}+y^{2016}=2\)

\(x^{2010}+y^{2010}=x^{2011}+y^{2011}=x^{2012}+y^{2012}\)

\(\Leftrightarrow x^{2010}+x^{2012}-2x^{2011}+y^{2010}+y^{2012}-2y^{2011}=0\)

\(\Leftrightarrow x^{2010}\left(x^2-2x+1\right)+y^{2010}\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow x^{2010}\left(x-1\right)^2+y^{2010}\left(y-1\right)^2=0\)

\(x^{2010};y^{2010}>0\Leftrightarrow x=y=1.\Rightarrow x^{2016}+y^{2016}=2\)

...................................................................................................................

theo đầu bài ta có\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\)=>\(3x^2+3y^2=10xy\)

A=\(\dfrac{x-y}{x+y}\)

=>\(A^2=\left(\dfrac{x-y}{x+y}\right)^2=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}=\dfrac{3x^2-6xy+3y^2}{3x^2+6xy+3y^2}=\dfrac{10xy-6xy}{10xy+6xy}=\dfrac{4xy}{16xy}=\dfrac{1}{4}\)

=>A=\(\sqrt{\dfrac{1}{4}}=\dfrac{-1}{2}hoặc\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\) (cộng trừ căn 1/4 nhé)

vì y>x>0=> A=-1/2

1) Vì a, b là số nguyên tố và a - 1 chia hết cho b nên a là số nguyên tố lẻ >=3 và b =2( vì a -1 chẵn)

b³ - 1 = 7 chia hết cho a, nên a =7. Vậy a = b² + b + 1( 7 = 2² + 2 + 1)

cũng bằng 3

$\text{Ta coˊ :}\genfrac{}{}{0ex}{}{x}{xy+x+1}+\genfrac{}{}{0ex}{}{y}{yz+y+1}+\genfrac{}{}{0ex}{}{z}{xz+z+1}$

$=\genfrac{}{}{0ex}{}{x}{xy+x+1}+\genfrac{}{}{0ex}{}{xy}{xyz+xy+x}+\genfrac{}{}{0ex}{}{xyz}{x^{2}yz+xyz+xy}$

$=\genfrac{}{}{0ex}{}{x}{xy+x+1}+\genfrac{}{}{0ex}{}{xy}{xy+x+1}+\genfrac{}{}{0ex}{}{1}{xy+x+1}\left(\text{Vıˋ }xyz=1\right)$

$=\genfrac{}{}{0ex}{}{x+xy+1}{xy+x+1}$

$=1$

Áp dụng BĐT \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)

\(\Rightarrow P\ge\frac{1}{2}\left(2x+\frac{1}{x}+2y+\frac{1}{y}\right)^2=\frac{1}{2}\left[2\left(x+y\right)+\frac{1}{x}+\frac{1}{y}\right]^2\)

\(\Rightarrow P\ge\frac{1}{2}\left[2\left(x+y\right)+\frac{4}{x+y}\right]^2=18\)

\(\Rightarrow P_{min}=18\) khi \(x=y=\frac{1}{2}\)

_Solution:

Prove with Cauchy-Schwarz inequality engel form, we have:

\(A=\frac{1}{x^3+3xy^2}+\frac{1}{y^3+3x^2y}\ge\frac{4}{x^3+y^3+3xy^2+3x^2y}\)

\(A\ge\frac{4}{\left(x+y\right)^3}\)

Other way: \(x+y\le1\Rightarrow\left(x+y\right)^3\le1\Rightarrow\frac{1}{\left(x+y\right)^3}\ge1\)

\(\Rightarrow A\ge4\) (proof)

We have ''='' \(\Leftrightarrow x=y=\frac{1}{2}\).