[toán 9] Cm đẳng thức

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vansang02121998

$\dfrac{a^4}{c}+\dfrac{b^4}{d}=\dfrac{1}{c+d}$

$\leftrightarrow \dfrac{a^4}{c}+\dfrac{b^4}{d} = \dfrac{(a^2+b^2)^2}{c+d}$

$\leftrightarrow \dfrac{a^4d+b^4c}{cd}=\dfrac{a^4+2a^2b^2+b^4}{c+d}$

$\rightarrow (c+d)(a^4d+b^4c)=cd(a^4+2a^2b^2+b^4)$

$\leftrightarrow a^4d^2-2a^2b^2cd-b^4c^2=0$

$\leftrightarrow (a^2d-b^2c)^2=0$

$\leftrightarrow a^2d=b^2c \leftrightarrow \dfrac{a^2}{c}=\dfrac{b^2}{d}$
 
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