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)
Chứng minh răng nếu: \(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=3\)thì \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{9}\)
cho \(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\)
Cmr \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
cho \(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\)
CMR: \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
\(Cho\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\)
Tính giá trị của \(P=\sqrt[3]{x^2}+\sqrt[3]{y^2}\)theo a
cmr nếu \(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\)
thì \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
Giải hệ phương trình:
a,\(\left\{{}\begin{matrix}\sqrt{x+y}\left(\sqrt{y}+1\right)=\sqrt{x^2+y^2}+2\\x\sqrt{y-1}+y\sqrt{x-1}=\dfrac{x^2+4y-4}{2}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}x^3+2y^2=x^2y+2xy\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
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 \(\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)
Ai giải được bài nào thì giúp mình vs
1/ \(\hept{\begin{cases}x^3-3x^2y-4x^2+4y^3+16xy=16y^2\\\sqrt{x-2y}+\sqrt{x+y}=2\sqrt{3}\end{cases}}\)
2/\(\hept{\begin{cases}\sqrt{x^2+xy+2y^2}+\sqrt{xy}=3y\\\sqrt{x-1}+\sqrt{y-1}+x+y=6\end{cases}}\)
3/\(\hept{\begin{cases}\sqrt{x+y}+\sqrt{x+3}=\frac{1}{3}\left(y-3\right)\\\sqrt{x+y}+\sqrt{x}=x+3\end{cases}}\)
