chúng minh rằng nếu : \(\sqrt{x^2+\sqrt[3]{x^2y^4}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\) thì \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
Những câu hỏi liên quan
cho \(a=\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}\) chứng minh rằng \(\sqrt[3]{a^2}=\sqrt[3]{x^2}+\sqrt[3]{y^2}\)
Cho \(\sqrt{x^2+\sqrt[3]{x^4}y^2}+\sqrt{y^2+\sqrt[3]{x^2y^4}=a}\)
Chứng minh rằng: \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
Chứng minh nếu \(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\) thì \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
Cho P=\(P=\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}\). Chứng minh rằng: \(\sqrt[3]{P^2}=\sqrt[3]{x^2}+\sqrt[3]{y^2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x^2}=a\ge0\\\sqrt[3]{y^2}=b\ge0\end{matrix}\right.\)
\(P=\sqrt{a^3+a^2b}+\sqrt{b^3+ab^2}=\sqrt{a^2\left(a+b\right)}+\sqrt{b^2\left(a+b\right)}\)
\(=a\sqrt{a+b}+b\sqrt{a+b}=\left(a+b\right)\sqrt{a+b}\)
\(\Rightarrow P^2=\left(a+b\right)^2\left(a+b\right)=\left(a+b\right)^3\)
\(\Rightarrow\sqrt[3]{P^2}=a+b=\sqrt[3]{x^2}+\sqrt[3]{y^2}\) (đpcm)
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Rút gọn:
Adfrac{sqrt[3]{x^4}+sqrt[3]{x^2y^2}+sqrt[3]{y^4}}{sqrt[3]{x^2}+sqrt[3]{xy}+sqrt[3]{y^2}}
Bdfrac{sqrt[3]{xy}left(sqrt[3]{y^2}-sqrt[3]{x^2}right)+left(sqrt[3]{x^4}-sqrt[3]{y^4}right)}{sqrt[3]{x^4}+sqrt[3]{x^2y^2}-sqrt[3]{x^3y}}.sqrt[3]{x^2}
Cleft(dfrac{xsqrt[3]{x}-2xsqrt[3]{y}+sqrt[3]{x^2y^2}}{sqrt[3]{x^2}-sqrt[3]{xy}}+dfrac{sqrt[3]{x^2y}-sqrt[3]{xy^2}}{sqrt[3]{x}-sqrt[3]{y}}right).dfrac{1}{sqrt[3]{x^2}}
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Rút gọn:
\(A=\dfrac{\sqrt[3]{x^4}+\sqrt[3]{x^2y^2}+\sqrt[3]{y^4}}{\sqrt[3]{x^2}+\sqrt[3]{xy}+\sqrt[3]{y^2}}\)
\(B=\dfrac{\sqrt[3]{xy}\left(\sqrt[3]{y^2}-\sqrt[3]{x^2}\right)+\left(\sqrt[3]{x^4}-\sqrt[3]{y^4}\right)}{\sqrt[3]{x^4}+\sqrt[3]{x^2y^2}-\sqrt[3]{x^3y}}.\sqrt[3]{x^2}\)
\(C=\left(\dfrac{x\sqrt[3]{x}-2x\sqrt[3]{y}+\sqrt[3]{x^2y^2}}{\sqrt[3]{x^2}-\sqrt[3]{xy}}+\dfrac{\sqrt[3]{x^2y}-\sqrt[3]{xy^2}}{\sqrt[3]{x}-\sqrt[3]{y}}\right).\dfrac{1}{\sqrt[3]{x^2}}\)
7 tháng 9 2016 lúc 20:50
cmr nếu \(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\)
thì \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
Đặt \(m=\sqrt[3]{x^2}\)và \(n=\sqrt[3]{y^2}\)
=> m3 = x2 và n3 = y2
Ta có :\(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\)
=> \(\sqrt{m^3+\sqrt[3]{m^6n^3}}+\sqrt{n^3+\sqrt[3]{m^3n^6}}=a\)
=> \(\sqrt{m^3+m^2n}+\sqrt{n^3+mn^2}=a\)
=> \(\sqrt{m^2\left(m+n\right)}+\sqrt{n^2\left(m+n\right)}=a\)
=> \(\sqrt{m+n}\left(m+n\right)=a\)
=> \(\left(\sqrt{m+n}\right)^3=\left(\sqrt[3]{a}\right)^3\)
=>\(\sqrt{m+n}=\sqrt[3]{a}\)
=> \(m+n=\left(\sqrt[3]{a}\right)^2\)
=> \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
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Tính giải mà lười quá. Bạn cứ nhân vô là ra ah
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cme nếu \(\sqrt{x^2+\sqrt[3]{x^4y^2}}\)\(+\sqrt{y^2+\sqrt[3]{x^2y^4}}\)=a
thì \(\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{a}\)
CMR: Nếu \(\sqrt{x^2+\sqrt[3]{x^2y^4}}+\sqrt{y^2+\sqrt[3]{x^4y^2}=a}\)
Thì \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
CMR nếu : \(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\)
thì \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)