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ĐKXĐ:...
Từ pt đầu:
\(\Leftrightarrow y^2+y\sqrt{y^2+1}=x-2y+\dfrac{1}{2}\)
\(\Leftrightarrow y^2+1+2y\sqrt{y^2+1}+y^2=2x-4y+2\)
\(\Leftrightarrow\left(\sqrt{y^2+1}+y\right)^2=2x-4y+2\)
\(\Leftrightarrow\sqrt{y^2+1}+y=\sqrt{2x-4y+2}\)
Thế xuống pt dưới:
\(x+\sqrt{x^2-2x+5}=1+2\sqrt{y^2+1}+2y\)
\(\Leftrightarrow\left(x-1\right)+\sqrt{\left(x-1\right)^2+4}=2y+\sqrt{\left(2y\right)^2+4}\)
Do hàm \(t+\sqrt{t^2+4}\) đồng biến
\(\Leftrightarrow x-1=2y\Rightarrow x=2y+1\)
Thế vào pt đầu:
\(\left(y+1\right)^2+y\sqrt{y^2+1}=2y+\dfrac{5}{2}\)
\(\Leftrightarrow y^2+y\sqrt{y^2+1}=\dfrac{3}{2}\)
\(\Leftrightarrow\left(\sqrt{y^2+1}+y\right)^2=4\)
\(\Leftrightarrow\sqrt{y^2+1}+y=2\)
\(\Leftrightarrow\sqrt{y^2+1}=2-y\)
\(\Leftrightarrow...\)
a)\(\left\{{}\begin{matrix}2x-3y=1\\x+2y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\cdot\left(3-2y\right)-3y=1\\x=3-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6-7y=1\\x=3-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{7}\\x=3-2\cdot\dfrac{5}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{7}\\x=\dfrac{11}{7}\end{matrix}\right.\)b) Biểu diễn lại một biến theo một biến như pt trên rồi giải, ta có :
\(\left\{{}\begin{matrix}2x+4y=5\\4x-2y=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{10}\\y=\dfrac{4}{5}\end{matrix}\right.\)
c) Cách làm tương tự như pt a ta có :
\(\left\{{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}y=\dfrac{2}{3}\\\dfrac{1}{3}x-\dfrac{3}{4}y=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{8}\\y=-\dfrac{1}{6}\end{matrix}\right.\)
d) Tương tự ta có :
\(\left\{{}\begin{matrix}0,3x-0,2y=0,5\\0,5x+0,4y=1,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)
b) ĐKXĐ: \(x,y\neq 0\).
Ta có: \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=\dfrac{1}{x}-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=\dfrac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-y=0\\xy=-1\end{matrix}\right.\\2y=x^3+1\end{matrix}\right.\).
Với x - y = 0 suy ra x = y. Do đó \(2x=x^3+1\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1=y\left(TMĐK\right)\\x=\pm\dfrac{\sqrt{5}-1}{2}=y\left(TMĐK\right)\end{matrix}\right.\).
Với xy = -1 suy ra \(y=-\dfrac{1}{x}\). Do đó \(x^3+\dfrac{2}{x}+1=0\Rightarrow x^4+x+2=0\). Phương trình vô nghiệm do \(x^4+x+2=\left(x^2-\dfrac{1}{2}\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{2}>0\).
Vậy...
Cộng vế với vế:
\(x^2+2xy+y^2+x+y=12\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+y\right)-12=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=-4\\x+y=3\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=-4\\xy=5-\left(x+y\right)=9\end{matrix}\right.\)
Theo Viet đảo, x và y là nghiệm: \(t^2-4t+9=0\) (vô nghiệm)
TH2: \(\left\{{}\begin{matrix}x+y=3\\xy=5-\left(x+y\right)=2\end{matrix}\right.\)
Theo Viet đảo, x và y là nghiệm:
\(t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)
1. \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x}+\dfrac{1}{y}=5\\x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=9\end{matrix}\right.\) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y+xy^2+x+y=5xy\\x^4y^2+x^2y^4+x^2+y^2=9x^2y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^4y^2+x^2y^4+x^2+y^2=25x^2y^2\\x^4y^2+x^2y^4+x^2+y^2=9x^2y^2\end{matrix}\right.\)\(\Leftrightarrow0=16x^2y^2\)
\(\Rightarrow\) phương trình vô nghiệm
\(x^2y+2y+x=4xy< =>xy\left(x+3\right)=4xy< =>x+3=4< =>x=1\)
Thế x=1 vào 1 trong 2 phương trình => y=1
a) \(\left\{{}\begin{matrix}5x+3y=-7\\2x-4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+3y=-7\\x-2y=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5x+3y=-7\\x=3+2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5.\left(3+2y\right)+3y=-7\\x=3+2y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13y=-22\\x=3+2y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-22}{13}\\x=3+2.\dfrac{-22}{13}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-22}{13}\\x=\dfrac{-5}{13}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm là: \(\left\{{}\begin{matrix}y=\dfrac{-22}{13}\\x=\dfrac{-5}{13}\end{matrix}\right.\).
b)\(\left\{{}\begin{matrix}7x+14y=17\\2x+4y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14x+28y=34\\14x+28y=35\end{matrix}\right.\) (vô nghiệm)
Vậy hệ phương trình vô nghiệm.
a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=-3+\dfrac{4}{y}=-3+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x+y}+6x-3y=6\\\dfrac{3}{x+y}+2x-4y=1\end{matrix}\right.\)
\(\Rightarrow4x+y=5\Rightarrow y=5-4x\)
Thế vào phương trình đầu:
\(\dfrac{1}{x+5-4x}+2x-\left(5-4x\right)=2\)
\(\Leftrightarrow\dfrac{1}{5-3x}+6x-7=0\)
\(\Leftrightarrow\left(6x-7\right)\left(5-3x\right)+1=0\)
\(\Leftrightarrow...\)