Cho \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)và \(x^2+y^2=1\)
Chứng minh : \(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\)
Cho \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) và \(x^2+y^2=1\) Chứng minh rằng: \(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{2}{\left(a+b\right)^{102}}\)
Cho:
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) và \(x^2+y^2=1\)
CMR:
\(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\)
cho \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)và \(x^2+y^2=1\). CMR: \(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{2}{\left(a+b\right)^{102}}\)
Cho \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) và \(x^2+y^2=1\) CMR : \(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{2}{\left(a+b\right)^{102}}\)
\(\frac{x^4}{a}\)=\(\frac{y^4}{b}\)=\(\frac{1}{a+b}\)và x2+y2=1
CMR:\(\frac{x^{2004}}{a^{1002}}\)+\(\frac{y^{2004}}{b^{1002}}\)=\(\frac{2}{\left(a+b\right)^{1002}}\)
\(\frac{x^4}{a}=\frac{y^4}{b}=\frac{1}{a+b}=\frac{x^4+y^4}{a+b}\Rightarrow x^4+y^4=1.\)
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?)
Cho \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)và \(x^2+y^2=1\)
Chứng minh \(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{2002}}=\frac{2}{\left(a+b\right)^{102}}\)
đây là bài tổng quát nè bạn, áp dụng bài này nhé ^_^
https://olm.vn/hoi-dap/question/1123004.html
Cho \(\frac{x^4}{a}\)+ \(\frac{y^4}{b}\)= \(\frac{1}{a+b}\) và \(x^2\) + \(y^2\)= 1
CMR: \(\frac{x^{2004}}{a^{1002}}\) + \(\frac{y^{2004}}{b^{1002}}\) = \(\frac{2}{\left(a+b\right)^{1002}}\)
\(x^2+y^2=1\)\(\Leftrightarrow\)\(\left(x^2+y^2\right)^2=1\) \(\left(1\right)\)
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 ~
Cho \(\frac{x^{\text{4}}}{a}+\frac{y^{\text{4}}}{b}=\frac{1}{a+b};x^2+y^2=1\)
Chứng minh rằng:\(\frac{x^{200\text{4}}}{a^{1002}}+\frac{y^{200\text{4}}}{b^{1002}}=\frac{2}{\left(a+b\right)^{102}}\)
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!
Cho \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b};x^2+y^2=1\). Chứng minh:
a) bx2 = ay2
b) \(\frac{x^{2008}}{a^{2004}}+\frac{y^{2008}}{b^{2004}}=\frac{2}{\left(a+b\right)^{1004}}\)
Em vào câu hỏi tương tự tham khảo:
a) Ta có: \(x^2+y^2=1\Leftrightarrow x^4+2x^2y^2+y^4=1\)
Khi đó: \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
<=> \(\left(a+b\right)\left(\frac{x^4}{a}+\frac{y^4}{b}\right)=x^4+2x^2y^2+y^4\)
<=> \(\frac{b}{a}x^4+\frac{a}{b}y^4=2x^2y^2\)
<=> \(\frac{x^4}{a^2}+\frac{y^4}{b^2}-\frac{2x^2y^2}{ab}=0\)
<=> \(\left(\frac{x^2}{a}-\frac{y^2}{b}\right)^2=0\)
a) \(\frac{x^2}{a}=\frac{y^2}{b}\Leftrightarrow bx^2=ay^2\)
b) \(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)( dãy tỉ số bằng nhau)
Khi đó: \(\frac{x^{2008}}{a^{1004}}+\frac{y^{2008}}{b^{1004}}=2\frac{x^{2008}}{a^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\)