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ĐK: \(x\ge0,y\ge0\)
\(\left\{{}\begin{matrix}x\sqrt{y}+y\sqrt{x}=12\\x\sqrt{x}+y\sqrt{y}=28\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)=12\\\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)=28\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)=12\\\left(\sqrt{x}+\sqrt{y}\right)\left[\left(\sqrt{x}+\sqrt{y}\right)^2-3\sqrt{xy}\right]=28\end{matrix}\right.\)(1)
Đặt \(a=\sqrt{x}+\sqrt{y}\left(a\ge0\right)\)
b=\(\sqrt{xy}\left(b\ge0\right)\)
Vậy (1)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}ab=12\\a\left(a^2-3b\right)=28\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}ab=12\left(3\right)\\a^3-3ab=28\left(4\right)\end{matrix}\right.\)
Lấy (3) thay vào (4) ta được \(a^3-3.12=28\Leftrightarrow a^3=64\Leftrightarrow a=4\Leftrightarrow b=3\)
Vậy ta có \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=4\\\sqrt{xy}=3\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=9\\y=1\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\y=9\end{matrix}\right.\end{matrix}\right.\)
Vậy (x;y)={(9;1);(1;9)}
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2b+ab^2=30\\a^3+b^3=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a^2b+3ab^2=90\\a^3+b^3=35\end{matrix}\right.\)
\(\Rightarrow\left(a+b\right)^3=125\Rightarrow a+b=5\)
Cũng từ \(a^2b+ab^2=30\Rightarrow ab\left(a+b\right)=30\Rightarrow ab=\dfrac{30}{a+b}=6\)
Theo Viet đảo, a và b là nghiệm của:
\(t^2-5t+6=0\Rightarrow\left[{}\begin{matrix}t=2\\t=3\end{matrix}\right.\)
\(\Rightarrow\left(a;b\right)=...\Rightarrow x;y\)
giải hệ pt \(\left\{{}\begin{matrix}\sqrt{2}x-\sqrt{3}y=1\\x+\sqrt{3}y=\sqrt{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{2}x-\sqrt{3}y=1\left(1\right)\\x+\sqrt{3}y=\sqrt{2}\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)+\left(2\right):\)
\(\sqrt{2}x+x-\sqrt{3}y+\sqrt{3}y=1+\sqrt{2}\)
\(\Rightarrow\sqrt{2}x+x-\sqrt{2}-1=0\)
\(\Rightarrow x\left(1+\sqrt{2}\right)-\left(1+\sqrt{2}\right)=0\)
\(\Rightarrow\left(1+\sqrt{2}\right)\left(x-1\right)=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Thay \(x=1\) vào \(\left(2\right):1+\sqrt{3}y=\sqrt{2}\)
\(\Rightarrow\sqrt{3}y=\sqrt{2}-1\)
\(\Rightarrow y=\dfrac{\sqrt{2}-1}{\sqrt{3}}\)
Vậy hệ pt có nghiệm duy nhất \( \left(x;y\right)=\left(1;\dfrac{\sqrt{2}-1}{\sqrt{3}}\right)\)
\(\left\{{}\begin{matrix}\sqrt{2}x-\sqrt{3}y=1\\x+\sqrt{3}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{2}+1\right)x=1+\sqrt{2}\\x+\sqrt{3}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+\sqrt{2}}{\sqrt{2}+1}=1\\x+\sqrt{3}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\1+\sqrt{3}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{\sqrt{2}-1}{\sqrt{3}}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất \(\left(x;y\right)=\left(1;\dfrac{\sqrt{2}-1}{\sqrt{3}}\right)\)
ĐKXĐ:...
\(\Leftrightarrow\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}+\frac{4}{\sqrt{y-1}}+\sqrt{y-1}=28\)
Ta có:
\(VT\ge2\sqrt{\frac{36.4\sqrt{x-2}}{\sqrt{x-2}}}+2\sqrt{\frac{4\sqrt{y-1}}{\sqrt{y-1}}}=28\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\frac{9}{\sqrt{x-2}}=\sqrt{x-2}\\\frac{4}{\sqrt{y-1}}=\sqrt{y-1}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=11\\y=5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2}=a\left(a>0\right)\\\sqrt{y-1}=b\left(b>0\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{36}{a}+\dfrac{4}{b}=28-4a-b\)
\(\Leftrightarrow\left(\dfrac{36}{a}+4a\right)+\left(\dfrac{4}{b}+b\right)=28\)
\(VT\ge2\sqrt{\dfrac{36}{a}\times4a}+2\sqrt{\dfrac{4}{b}\times b}=28\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{36}{a}=4a\\\dfrac{4}{b}=b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\) \(\left(a,b>0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2}=3\\\sqrt{y-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=5\end{matrix}\right.\) (n)
Vậy . . . >3<
\(\hept{\begin{cases}\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)=12\\\left(\sqrt{x}+\sqrt{y}\right)\left(x+y-\sqrt{xy}\right)=28\end{cases}}\)
\(\Rightarrow\sqrt{x}+\sqrt{y}=\frac{12}{\sqrt{xy}}\)
\(\Rightarrow\frac{12}{\sqrt{xy}}\left(x+y-\sqrt{xy}\right)=28\)
\(\Leftrightarrow\frac{x+y-\sqrt{xy}}{\sqrt{xy}}=\frac{7}{3}\)
\(\Leftrightarrow\frac{x+y}{\sqrt{xy}}=\frac{4}{3}\)
tc \(x+y\ge2\sqrt{xy}\)
\(\Rightarrow\frac{x+y}{\sqrt{xy}}\ge2>\frac{4}{3}\)=>pt vô nghiệm
Lời giải:
Đặt \(\left(\sqrt{x},\sqrt{y}\right)=\left(a,b\right)\)
Khi đó hệ phương trình chuyển về: \(\hept{\begin{cases}ab\left(a+b\right)=12\\a^3+b^3=28\end{cases}}\Leftrightarrow\hept{\begin{cases}ab\left(a+b\right)=12\\\left(a+b\right)^3-3ab\left(a+b\right)=28\end{cases}}\)
Lấy 3 lần PT (1) +PT (2) thu được: \(\left(a+b\right)^3=28+36=64\Rightarrow a+b=4\)
Mà \(ab\left(a+b\right)=12\Rightarrow ab=3\)
Khi đó, áp dụng định lý Viete đảo thì \(a,b\) là nghiệm của pt: \(x^2-4x+3=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
Hay \(\left(a,b\right)=\left(1,3\right)\) và hoán vị hay \(\left(x,y\right)=\left(1,9\right)\) và hoán vị.