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a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)
=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75
=>x=7; y=5
b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)
=>4x+9y=8 và -8x+3y=5
=>x=-1/4; y=1
c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)
=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5
=>2x-3y=-5,5 và 3x-2y=-4,5
=>x=-1/2; y=3/2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
\(e,\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy\in\left\{2;-3\right\}\end{matrix}\right.\)
Vì \(\frac{x}{y}=2>0\Rightarrow xy>0\Rightarrow xy=2\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\left(h\right)\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
\(a,\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\frac{1}{y}\right)^2-\frac{x}{y}=3\\\left(x+\frac{1}{y}\right)+\frac{x}{y}=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{y}=a\\\frac{x}{y}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-b=3\\a+b=3\end{matrix}\right.\)
Làm nốt nha
\(\left\{{}\begin{matrix}x+3\sqrt{xy+x-y^{2-y}}=5y+4\left(1\right)\\\sqrt{4y^2-x-2}+\sqrt{y-1}=x-1\left(2\right)\end{matrix}\right.\)
ĐK: x\(\ge1,y\ge1\),x\(\ge y\)
(1)\(\Leftrightarrow\left(x-y\right)+3\sqrt{x\left(y+1\right)-y\left(y+1\right)}-4y-4=0\Leftrightarrow\left(x-y\right)+3\sqrt{\left(x-y\right)\left(y+1\right)}-4\left(y+1\right)=0\left(3\right)\)
Chia 2 vế của (3) cho y+1>0 thì (3) và đặt t=\(\sqrt{\dfrac{x-y}{y+1}}\)(t\(\ge0\))
Vậy (3)\(\Leftrightarrow t^2+3t-4=0\Leftrightarrow t^2-t+4t-4=0\Leftrightarrow t\left(t-1\right)+4\left(t-4\right)=0\Leftrightarrow\left(t-1\right)\left(t+4\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}t-1=0\\t+4=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}t=1\left(tm\right)\\t=-4\left(ktm\right)\end{matrix}\right.\)
Ta có t=1\(\Leftrightarrow\sqrt{\dfrac{x-y}{y+1}}=1\Leftrightarrow x-y=y+1\Leftrightarrow x=2y+1\)
Thay vào phương trình (2)\(\Leftrightarrow\sqrt{4y^2-\left(2y+1\right)-2}+\sqrt{y-1}=2y+1-1\Leftrightarrow\sqrt{4y^2-2y-3}+\sqrt{y-1}=2y\Leftrightarrow\left(\sqrt{4y^2-2y-3}-3\right)+\left(\sqrt{y-1}-1\right)=2\left(y-2\right)\Leftrightarrow\dfrac{4y^2-2y-12}{\sqrt{4y^2-2y-3}+3}+\dfrac{y-2}{\sqrt{y-1}+1}-2\left(y-2\right)=0\Leftrightarrow\dfrac{2\left(y-2\right)\left(2y+3\right)}{\sqrt{4y^2-2y-3}+3}+\dfrac{y-2}{\sqrt{y-1}+1}-2\left(y-2\right)=0\Leftrightarrow\left(y-2\right)\left[\dfrac{2\left(2y+3\right)}{\sqrt{4y^2-2y-3}+3}+\dfrac{1}{\sqrt{y-1}+1}-2\right]=0\Leftrightarrow\)\(\left[{}\begin{matrix}y-2=0\left(4\right)\\\dfrac{2\left(2y+3\right)}{\sqrt{4y^2-2y-3}+3}+\dfrac{1}{\sqrt{y-1}+1}-2=0\left(5\right)\end{matrix}\right.\)
(4)\(\Leftrightarrow y=2\Leftrightarrow x=5\left(tm\right)\)
(5)\(\Leftrightarrow\dfrac{2\left(2y+3\right)}{\sqrt{4y^2-2y-3}+3}=2y+3-\sqrt{y+1}< 2y+3\Rightarrow\dfrac{2\left(2y+3\right)}{\sqrt{4y^2-2y-3}+3}\ge2\Leftrightarrow\)VT của (5)>2\(\Rightarrow\) vô nghiệm
Vậy (x;y)=(5;2)