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1) Ta có: \(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}-3\sqrt{y}=15\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11\sqrt{x}=33\\3\sqrt{x}-\sqrt{y}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)
2) Ta có: \(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2\sqrt{x+3}+4\sqrt{y+1}=-4\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y+1}=0\\\sqrt{x+3}-2\sqrt{y+1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y+1}=0\\\sqrt{x+3}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\x+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
4. Đk: \(x,y\ge0\)
\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}=1\\\sqrt{y}+\sqrt{x+1}=1\end{matrix}\right.\left(1\right)\)
Ta có: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}\ge0+1=1\\\sqrt{y}+\sqrt{x+1}\ge0+1=1\end{matrix}\right.\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=0,\sqrt{x+1}=1\\\sqrt{y}=0,\sqrt{y+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)<tmđk>
Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(0;0\right)\)
\(2,ĐK:x\ge4;y\ge1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-4}=a\\\sqrt{y-1}=b\end{matrix}\right.\left(a,b\ge0\right)\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\a^2+b^2=58\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2ab+58=16\\a^2+b^2=58\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}ab=-21\\a+b=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=4-b\\b^2-4b-21=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}b=7\Rightarrow a=-3\\b=-3\Rightarrow a=7\end{matrix}\right.\left(loại\right)\)
Vậy hệ vô nghiệm
\(1,\\ \forall x=0\\ HPT\Leftrightarrow1=19\left(\text{vô lí}\right)\\ \forall x\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x^3}+y^3=19\\\dfrac{y}{x^2}+\dfrac{y^2}{x}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{1}{x}+y\right)^3-3\cdot\dfrac{y}{x}\left(\dfrac{1}{x}+y\right)=19\\\dfrac{y}{x}\left(\dfrac{1}{x}+y\right)=-6\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}+y=a\\\dfrac{y}{x}=b\end{matrix}\right.\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}a^3-3ab=19\\ab=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+y=1\\\dfrac{y}{x}=-6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}1+xy=x\\y=-6x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3};y=-2\\x=-\dfrac{1}{2};y=3\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\dfrac{1}{3};-2\right);\left(-\dfrac{1}{2};3\right)\)
\(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y+2\right)=\left(y-1\right)\left(x-\text{4}\right)\\\left(2x+3\right)\left(2y+1\right)=\left(y-1\right)\left(4x+1\right)\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)
\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)
Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))
a, ĐK: \(x,y\ge0\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3\sqrt{y}}{\sqrt{x+3}-\sqrt{x}}=3\\\sqrt{x}+\sqrt{y}=x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=\sqrt{x+3}\\\sqrt{x}+\sqrt{y}=x+1\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+3}=x+1\)
\(\Leftrightarrow x+3=x^2+2x+1\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\left(l\right)\end{matrix}\right.\)
Thay \(x=1\) vào hệ phương trình đã cho ta được \(y=1\)
Vậy pt đã cho có nghiệm \(x=y=1\)
b, \(hpt\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\left(y+\dfrac{1}{2}\right)^2\\x^2+y^2=3\left(x+y\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x+y=-1\end{matrix}\right.\\x^2+y^2=3\left(x+y\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2-3x=0\end{matrix}\right.\left(1\right)\\\left\{{}\begin{matrix}x+y=-1\\x^2+y^2=-3\end{matrix}\right.\left(vn\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=y=3\\x=y=0\end{matrix}\right.\)
Vậy ...
