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ĐKXĐ:...
\(\Leftrightarrow\left(\sqrt{y}+x-4\right)\left(\sqrt{y}-x+4\right)=\sqrt{x+y}+\sqrt{x+y-9}+2\)
\(\Leftrightarrow y-\left(x-4\right)^2=\sqrt{x+y}+\sqrt{\left(y-5\right)^2}+2\)
- Với \(y\ge5\)
\(\Leftrightarrow3-\left(x-4\right)^2=\sqrt{x+y}\)
Do \(x+y=9+\left(y-5\right)^2\ge9\Rightarrow\sqrt{x+y}\ge3\)
Mà \(3-\left(x-4\right)^2\le3\)
\(\Rightarrow\) Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x-4=0\\x+y=9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\)
- Với \(0\le y< 5\)
\(\Leftrightarrow y-\left(x-4\right)^2=\sqrt{x+y}+5-y+2\)
\(\Leftrightarrow2y-\left(x-4\right)^2=\sqrt{x+y}+7\)
Ta có: \(y< 5\Rightarrow VT=2y-\left(x-4\right)^2\le2y< 10\)
\(VP=\sqrt{x+y}+7\ge3+7=10\)
\(\Rightarrow VP>VT\Rightarrow\) phương trình vô nghiệm
Vậy hệ có nghiệm duy nhất \(\left(x;y\right)=\left(4;5\right)\)
Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
a) \(\left\{{}\begin{matrix}x+2y=-1\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
Vậy..............................................................................
b) \(\left\{{}\begin{matrix}\frac{5}{x}-\frac{6}{y}=3\\\frac{4}{x}+\frac{9}{y}=7\end{matrix}\right.\)ĐKXĐ: x,y≠0
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{20}{x}-\frac{24}{y}=12\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{69}{y}=23\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=10\end{matrix}\right.\)
Vậy...................................................................................
c) \(\left\{{}\begin{matrix}3\sqrt{x+1}+\sqrt{y-1}=1\\\sqrt{x+1}-\sqrt{y-1}=-2\end{matrix}\right.\)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge-1\\y\ge1\end{matrix}\right.\)
\(\Rightarrow4\sqrt{x+1}\)\(=-1\)(vô nghiệm)
Vậy hệ pt vô nghiệm
d) Nhân 3 pt đầu rồi thu gọn
\(\left\{{}\begin{matrix}x^3-y^3-9=0\\6x^2-12x+3y^2+3y=0\end{matrix}\right.\)
\(\Rightarrow x^3-6x^2+12x-8-\left(y^3+3y^2+3y+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)^3=\left(y+1\right)^3\)
\(\Leftrightarrow x-2=y+1\Rightarrow y=x-3\)
Thế vào pt dưới:
\(2x^2+\left(x-3\right)^2-4x+x-3=0\)
\(\Leftrightarrow...\)
b/ ĐKXĐ: \(x;y\ge1\)
Trừ trên cho dưới:
\(\Rightarrow2\left(\sqrt{x^2+5}-\sqrt{y^2+5}\right)+2\left(\sqrt{x-1}-\sqrt{y-1}\right)+x^2-y^2=0\)
\(\Leftrightarrow\frac{\left(x-y\right)\left(2x+2y\right)}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\frac{2\left(x-y\right)}{\sqrt{x-1}+\sqrt{y-1}}+\left(x-y\right)\left(x+y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(\frac{2x+2y}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\frac{2}{\sqrt{x-1}+\sqrt{y-1}}+x+y\right)=0\)
\(\Leftrightarrow x-y=0\Rightarrow x=y\)
Thay vào pt đầu:
\(2\sqrt{x^2+5}=2\sqrt{x-1}+x^2\)
\(\Leftrightarrow x^2+2-2\sqrt{x^2+5}+2\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\frac{x^4-16}{x^2+2+2\sqrt{x^2+5}}+\frac{2\left(x-2\right)}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x+2\right)\left(x^2+4\right)}{x^2+2+2\sqrt{x^2+5}}+\frac{2\left(x-2\right)}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{\left(x+2\right)\left(x^2+4\right)}{x^2+2+2\sqrt{x^2+5}}+\frac{2}{\sqrt{x-1}+1}\right)=0\)
\(\Rightarrow x=y=2\)
\(\left\{{}\begin{matrix}\sqrt{2}x+2\sqrt{3}y=5\\3\sqrt{2}x-\sqrt{3}y=\frac{9}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}x+2\sqrt{3}y=5\\6\sqrt{2}x-2\sqrt{3}y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{2}x=14\\\sqrt{2}x+2\sqrt{3}y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\2\sqrt{3}y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\y=\frac{\sqrt{3}}{2}\end{matrix}\right.\)
vậy...
ĐKXĐ: \(x,y\ge0\)
Từ hệ pt trên suy ra \(\sqrt{x}+\sqrt{y+9}=\sqrt{y}+\sqrt{x+9}\Leftrightarrow x+y+9+2\sqrt{xy+9x}=x+y+9+2\sqrt{yx+9y}\Leftrightarrow2\sqrt{xy+9x}=9+2\sqrt{xy+9y}\Leftrightarrow x=y\)
Vậy hệ pt thành \(\sqrt{x}+\sqrt{x+9}=9\Leftrightarrow2x+9+2\sqrt{x^2+9x}=81\Leftrightarrow2\sqrt{x^2+9x}=72-2x\)
( đk : x \(\le\) 36, y \(\le36\))
\(\Leftrightarrow4\left(x^2+9x\right)=5184-288x+4x^2\Leftrightarrow324x=5184\Leftrightarrow x=16\) (tmđk)