[toán 9] chúng minh

C

chonhoi110

Đặt $\sqrt[3]{x^2}=u ; \sqrt[3]{y^2}=v$

~> $\sqrt{u^3+u^2v}+\sqrt{v^3+v^2u}=a$

~> $u^3+v^3+u^2v+v^2u+2\sqrt{(u^3+u^2v)(v^3+v^2u)}=a^2$

<~> $(u+v)(u^2+v^2)+2uv(u+v)=a^2$

<~> $(u+v)^3=a^2$ ~> $\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}$
 
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