\(\Leftrightarrow\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)
\(\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+y^4+2x^2y^2}{a+b}\Leftrightarrow\left(a+b\right)\left(x^4b+y^4a\right)=ab\left(x^4+y^4+2x^2y^2\right)\)
\(\Leftrightarrow x^4ab+y^4a^2+x^4b^2+y^4ab=x^4ab+y^4ab+2x^2y^2ab\)
\(\Leftrightarrow y^4a^2+x^4b^2=2x^2y^2ab\Leftrightarrow\left(x^2b-y^2a\right)^2=0\Leftrightarrow\frac{x^2}{a}=\frac{y^2}{b}\)
\(\Rightarrow\left(\frac{x^2}{a}\right)^{1001}=\left(\frac{y^2}{b}\right)^{1001}\Leftrightarrow\frac{x^{2002}}{a^{1001}}=\frac{y^{2002}}{b^{2011}}\)
Mà: \(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\Leftrightarrow\left(\frac{x^2}{a}\right)^{1001}=\frac{1}{\left(a+b\right)^{1001}}\)
\(\Rightarrow\frac{x^{2002}}{a^{1001}}+\frac{y^{2002}}{b^{1001}}=\frac{2}{\left(a+b\right)^{1001}}\left(đpcm\right)\)
\(x^2+y^2=1\Rightarrow y^2=1-x^2\)
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\Leftrightarrow\frac{b.x^4+a.y^4}{ab}=\frac{1}{a+b}\)
\(\Leftrightarrow bx^4+ay^4=\frac{ab}{a+b}\Leftrightarrow bx^4+a\left(1-x^2\right)^2-\frac{ab}{a+b}=0\)
\(\Leftrightarrow bx^4+a\left(x^4-2x^2+1\right)-\frac{ab}{a+b}=0\)
\(\Leftrightarrow\left(a+b\right)x^4-2ax^2+a-\frac{ab}{a+b}=0\)
\(\Leftrightarrow\left(a+b\right)x^4-2ax^2+\frac{a^2}{a+b}=0\Leftrightarrow\left(a+b\right)\left[x^4-2.x.\frac{a}{a+b}+\left(\frac{a}{a+b}\right)^2\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(x^2-\frac{a}{a+b}\right)=0\Rightarrow x^2=\frac{a}{a+b}\) (do \(a+b\ne0\))
\(\Rightarrow y^2=1-x^2=\frac{b}{a+b}\)
\(\Rightarrow\) \(\frac{x^2}{a}=\frac{a}{a\left(a+b\right)}=\frac{1}{a+b}\) ; \(\frac{y^2}{b}=\frac{b}{b\left(a+b\right)}=\frac{1}{a+b}\)
Thay vào bài toán:
\(\frac{x^{2002}}{a^{1001}}+\frac{y^{2002}}{b^{1001}}=\left(\frac{x^2}{a}\right)^{1001}+\left(\frac{y^2}{b}\right)^{1001}=\left(\frac{1}{a+b}\right)^{1001}+\left(\frac{1}{a+b}\right)^{1001}=\frac{2}{\left(a+b\right)^{1001}}\)
x2+y2=1⇒y2=1−x2
x4a+y4b=1a+b⇔b.x4+a.y4ab=1a+b
⇔bx4+ay4=aba+b⇔bx4+a(1−x2)2−aba+b=0
⇔bx4+a(x4−2x2+1)−aba+b=0
⇔(a+b)x4−2ax2+a−aba+b=0
⇔(a+b)x4−2ax2+a2a+b=0⇔(a+b)[x4−2.x.aa+b+(aa+b)2]=0
⇔(a+b)(x2−aa+b)=0⇒x2=aa+b (do a+b≠0)
⇒y2=1−x2=ba+b
⇒ x2a=aa(a+b)=1a+b ; y2b=bb(a+b)=1a+b
Thay
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}=\frac{x^2+y^2}{a+b}\)
\(\Rightarrow\frac{x^4b+y^4a}{ab}=\frac{x^2+y^2}{a+b}\)
\(\Rightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^2+y^2\right)ab\)
\(\Rightarrow x^4ab+y^4ab+x^4b^2+y^4a^2=x^2ab+y^2ab\)
\(\Rightarrow x^4b^2+y^4a^2=x^2ab-x^4ab+y^2ab-y^4ab\)
\(\Rightarrow x^4b^2+y^4a^2-x^2ab\left(1-x^2\right)-y^2ab\left(1-y^2\right)=0\)
\(\Rightarrow\left(x^2b\right)^2+\left(y^2a\right)^2-2x^2y^2ab=0\) ( do \(x^2+y^2=1\) )
\(\Rightarrow\left(x^2b-y^2a\right)=0\)
\(\Rightarrow x^2b=y^2a\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\frac{x^{2002}}{a^{1001}}=\frac{y^{2002}}{b^{1001}}=\frac{1}{\left(a+b\right)^{1001}}\)
\(\Rightarrow\frac{x^{2002}}{a^{1001}}+\frac{y^{2002}}{b^{1001}}=\frac{2}{\left(a+b\right)^{1001}}\)