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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 <<<:
\(x^2+y^2=1\)\(\Leftrightarrow\)\(\left(x^2+y^2\right)^2=1\) \(\left(1\right)\)
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!
\(\frac{x}{2004}+\frac{x}{1002}+\frac{667x}{668}-2004=0\)
\(\Leftrightarrow\frac{x}{2004}+\frac{x}{1002}+\frac{667x}{668}=2004\)
\(\Rightarrow\frac{x+2x+2001x}{2004}=2004\)
\(\Rightarrow\frac{2004x}{2004}=2004\)
\(\Rightarrow x=2004\)
A) ta có \(\frac{X}{2}=\frac{Y}{3}\)=>\(\frac{X}{8}=\frac{Y}{12}\)(1)
\(\frac{Y}{4}=\frac{Z}{5}\)=>\(\frac{Y}{12}=\frac{Z}{15}\)(2)
Từ (1)và (2)=>\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\) và x-y-z=28
đến đây tự làm
c) \(\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)
\(\Rightarrow\left(x-\frac{1}{5}\right)^{2004}=0\) và \(\left(y+0,4\right)^{100}=0\) và \(\left(z-3\right)^{678}=0\)
+) \(\left(x-\frac{1}{5}\right)^{2004}=0\Rightarrow x-\frac{1}{5}=0\Rightarrow x=\frac{1}{5}\)
+) \(\left(y+0,4\right)^{100}=0\Rightarrow y+0,4=0\Rightarrow y=-0,4\)
+) \(\left(z-3\right)^{678}=0\Rightarrow z-3=0\Rightarrow z=3\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\frac{1}{5};-0,4;3\right)\)
\(\frac{x^4}{a}=\frac{y^4}{b}=\frac{1}{a+b}=\frac{x^4+y^4}{a+b}\Rightarrow x^4+y^4=1.\)
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