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 \(a=\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}\) chứng minh rằng \(\sqrt[3]{a^2}=\sqrt[3]{x^2}+\sqrt[3]{y^2}\)
Cho 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)
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 :\(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\)
=> \(\sqrt{m^3+\sqrt[3]{m^6n^3}}+\sqrt{n^3+\sqrt[3]{m^3n^6}}=a\)
=> \(\sqrt{m^3+m^2n}+\sqrt{n^3+mn^2}=a\)
=> \(\sqrt{m^2\left(m+n\right)}+\sqrt{n^2\left(m+n\right)}=a\)
=> \(\sqrt{m+n}\left(m+n\right)=a\)
=> \(\left(\sqrt{m+n}\right)^3=\left(\sqrt[3]{a}\right)^3\)
=>\(\sqrt{m+n}=\sqrt[3]{a}\)
=> \(m+n=\left(\sqrt[3]{a}\right)^2\)
=> \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
Tính giải mà lười quá. Bạn cứ nhân vô là ra ah
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 \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
chúng minh rằng nếu : \(\sqrt{x^2+\sqrt[3]{x^2y^4}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\) thì \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
a.Cmr nếu \(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\) thì \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
b.Giải pt \(x^3-x^2-1=\dfrac{1}{3}\)
\(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\)
CMR:\(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)