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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 \(\left(x,y\right)=\left(0;0\right)\)
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 \(\left(x;y\right)=\left(1;\dfrac{\sqrt{2}-1}{\sqrt{3}}\right)\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(1-\sqrt{3}\right)x+2y=1-\sqrt{3}\\\left(1-\sqrt{3}\right)x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\sqrt{3}\\x=1+\left(1+\sqrt{3}\right)\cdot\left(-\sqrt{3}\right)=-2-\sqrt{3}\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}-x-\sqrt{2}y=\sqrt{3}\\x+\sqrt{2}y=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\in R\\x=-\sqrt{3}-y\sqrt{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3-y^3-9=0\\6x^2-12x+3y^2+3y=0\end{matrix}\right.\)
\(\Rightarrow x^3-6x^2+12x-8-\left(y^3+3y^2+3y+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)^3=\left(y+1\right)^3\)
\(\Leftrightarrow x-2=y+1\Rightarrow y=x-3\)
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:
\(2\sqrt{x^2+5}=2\sqrt{x-1}+x^2\)
\(\Leftrightarrow x^2+2-2\sqrt{x^2+5}+2\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\frac{x^4-16}{x^2+2+2\sqrt{x^2+5}}+\frac{2\left(x-2\right)}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x+2\right)\left(x^2+4\right)}{x^2+2+2\sqrt{x^2+5}}+\frac{2\left(x-2\right)}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{\left(x+2\right)\left(x^2+4\right)}{x^2+2+2\sqrt{x^2+5}}+\frac{2}{\sqrt{x-1}+1}\right)=0\)
\(\Rightarrow x=y=2\)
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
Câu 1 \(\left\{{}\begin{matrix}2x+2y+2xy=10\left(1\right)\\x^2+y^2=5\left(2\right)\end{matrix}\right.\)
=>2.(2) - (1)=\(\left(x-1\right)^2+\left(y-1\right)^2+\left(x-y\right)^2=0\)
<=>\(\left\{{}\begin{matrix}x-1=0\\y-1=0\\x-y=0\end{matrix}\right.\) =>x=y=1
Câu 2 dùng vi-et đảo
Câu 3 rút x=y+1 từ pt trên rồi thế xuống dưới
Câu 4 lấy pt trên cộng pt dưới rồi xét dấu GTTĐ
Ta có: \(\left\{{}\begin{matrix}5\sqrt{x-1}-\sqrt{y+2}=19\\2\sqrt{x-1}+\sqrt{y+2}=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{x-1}-2\sqrt{x-1}=19\\2\sqrt{x-1}-\sqrt{y+2}=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}=19\\2\sqrt{x-1}-\sqrt{y+2}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=\frac{19}{3}\\2\sqrt{x-1}-\sqrt{y+2}=9\end{matrix}\right.\)
Đến đây thì mình bí rồi
P/S: Không chắc có đúng hay không nữa