Ta có : 

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) vì \(x^2+y^2=1\)

\(\Rightarrow\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)

\(\Leftrightarrow\frac{x^4.b+y^4.a}{ab}=\frac{\left(x^2+y^2\right)^2}{ab}\)

\(\Leftrightarrow\left(x^4.b+y^4.a\right)\left(a+b\right)=ab\left(x^2+y^2\right)^2\)

\(\Rightarrow x^4ab+x^4b^2+a^2y^4+aby^4\)

\(=ab\left(x^2+y^2\right)\left(x^2+y^2\right)\)

\(\Rightarrow ab\left(x^4+x^2y^2+x^2y^2+y^4\right)\)

\(\Rightarrow abx^4+abx^2y^2+abx^2y^2+abx^2y^2+aby^4\)

\(\Rightarrow b^2x^4+a^2y^4\)

\(=2abx^2y^2\)

\(\Rightarrow\left(bx^2\right)^2+\left(ay^2\right)^2-ax^2.by^2-ax^2-by^2=0\)

\(\Rightarrow\left[\left(bx^2\right)^2-ax^2.by^2\right]+\left[\left(ay^2\right)^2-ax^2.by^2\right]=0\)

\(bx^2\left(bx^2-ay^2\right)+ay^2\left(ay^2-bx^2\right)=0\)

\(bx^2\left(bx^2-ay^2\right)-ay^2\left(bx^2-ay^2\right)\)

\(\left(bx^2-ay^2\right)^2=0\)

\(bx^2-ay^2=0\)

\(bx^2=ay^2\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}\)

Mà \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\Rightarrow x^2.\frac{x^2}{a}+y.\frac{y^2}{b}=\frac{x^2+y^2}{a+b}\)

\(\Rightarrow\frac{x^2}{a}\left(x^2+y^2\right)=\frac{x^2+y^2}{a+b}\)

\(\Rightarrow\frac{x^2}{a}=\frac{1}{a+b}\Rightarrow\frac{y^2}{b}=\frac{x^2}{a}=\frac{1}{a+b}\)

Ta có :

\(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{a^{1002}}=\left(\frac{x^2}{a}\right)^{1002}+\left(\frac{y^2}{b}\right)^{1002}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}< đpcm>\)

Hok tốt 

P/s : _Làm bừa nên chắc k đúng đâu - - _M bt a hok ngu thek nào r mak (:

_E cóa thý a hok ngu âu >: ?

_Với cả giải vợi lak đầy đủ roy hả ?

_Thank nhìu nhìu <<<: 

\(\frac{x^4}{a}=\frac{y^4}{b}=\frac{1}{a+b}=\frac{x^4+y^4}{a+b}\Rightarrow x^4+y^4=1.\)

Mà \(x^2+y^2=1\)=>\(x^4+y^4=x^2+y^2=1.\)

Nếu x =0 => y =1 => a =0 vô lí 

Xem lại đề  dc ko ( hay mình làm sai?)

đề đúng r bạn

đây là bài tổng quát nè bạn, áp dụng bài này nhé ^_^

https://olm.vn/hoi-dap/question/1123004.html

\(x^2+y^2=1\)\(\Leftrightarrow\)\(\left(x^2+y^2\right)^2=1\) \(\left(1\right)\)

Thay \(\left(1\right)\) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) ta được : 

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)

\(\Leftrightarrow\)\(\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)

\(\Leftrightarrow\)\(\left(a+b\right)\left(x^4b+y^4a\right)=ab\left(x^4+2x^2y^2+y^4\right)\)

\(\Leftrightarrow\)\(x^4ab+y^4a^2+x^4b^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)

\(\Leftrightarrow\)\(x^4b^2+y^4a^2=2x^2y^2ab\)

\(\Leftrightarrow\)\(x^4b^2-2x^2y^2ab+y^4a^2=0\)

\(\Leftrightarrow\)\(\left(x^2b\right)^2-2.x^2b.y^2a+\left(y^2a\right)^2=0\)

\(\Leftrightarrow\)\(\left(x^2b-y^2a\right)=0\)

\(\Leftrightarrow\)\(x^2b-y^2a=0\)

\(\Leftrightarrow\)\(x^2b=y^2a\)

\(\Leftrightarrow\)\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\) ( thay \(x^2+y^2=1\) ) 

\(\Leftrightarrow\)\(\left(\frac{x^2}{a}\right)^{1002}=\left(\frac{y^2}{b}\right)^{1002}=\left(\frac{1}{a+b}\right)^{1002}\)

\(\Leftrightarrow\)\(\frac{x^{2004}}{a^{1002}}=\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}\)

Do đó : 

\(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\) ( đpcm ) 

Chúc bạn học tốt ~ 

Ta có:

\(x^2+y^2=1\Rightarrow\left(x^2+y^2\right)^2=1\)(1)

Thay (1) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)ta có:

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)

\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+2x^2y^2+y^4\right).ab\)

\(\Leftrightarrow x^4ab+x^4b^2+y^4a^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)

\(\Leftrightarrow x^4b^2+y^4a^2=2x^2y^2ab\)

\(\Leftrightarrow\left(x^2b\right)^2-2x^2y^2ab+\left(y^2a\right)^2=0\)

\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\)

\(\Leftrightarrow x^2b-y^2a=0\)

\(\Leftrightarrow x^2b=y^2a\)

\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)

\(\Rightarrow\left(\frac{x^2}{a}\right)^{1002}=\left(\frac{y^2}{b}\right)^{1002}=\left(\frac{1}{a+b}\right)^{1002}\)

\(\Rightarrow\frac{x^{2004}}{a^{1002}}=\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}\)

\(\Rightarrow\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\left(đpcm\right)\)

Chúc bạn học tốt!

Các bợn làm nhanh dùm mk nha. Bài kiểm tra sáng mai mình nộp rồi. Ai nhanh nhất mình tick cho nha

sai đề nha bn : là \(\dfrac{2}{\left(a+b\right)^{1002}}\) mới đúng

+ \(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{1}{a+b}=\dfrac{x^2+y^2}{a+b}\)

\(\Rightarrow\dfrac{bx^4+ay^4}{ab}=\dfrac{x^2+y^2}{a+b}\)

\(\Rightarrow\left(bx^4+ay^4\right)\left(a+b\right)=ab\left(x^2+y^2\right)\)\(\Rightarrow abx^4+aby^4+a^2y^4+b^2x^4=abx^2+aby^2\)

\(\Rightarrow a^2y^4+b^2x^4=abx^2\left(1-x^2\right)+aby^2\left(1-y^2\right)\)

\(\Rightarrow a^2y^4+b^2x^4=abx^2y^2+abx^2y^2\)

\(\Rightarrow\left(ay^2\right)^2+\left(bx^2\right)^2-2abx^2y^2=0\)

\(\Rightarrow\left(ay^2-bx^2\right)^2=0\)

\(\Rightarrow ay^2-bx^2=0\Rightarrow ay^2=bx^2\)

\(\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\) ( tính chất dãy tỉ số bằng nhau )

\(\Rightarrow\dfrac{x^{2004}}{a^{1002}}=\dfrac{y^{2002}}{b^{1002}}=\dfrac{1}{\left(a+b\right)^{1002}}\)

\(\Rightarrow\dfrac{x^{2004}}{a^{1002}}+\dfrac{y^{2004}}{b^{1002}}=\dfrac{2}{\left(a+b\right)^{1002}}\) ( đpcm )