Xem lại giúp tớ dấu căn ở câu c và d nhé.  

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: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)

=>\(x=\sqrt{2};y=\sqrt{3}\)

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 \(\left(x,y\right)=\left(0;0\right)\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)

b.

ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)

Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:

\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)

\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)

\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)

Thay xuống pt dưới:

\(6y+y=14\Rightarrow y=2\)

\(\Rightarrow x=4\)

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\) thì pt đầu trở thành:

\(a\left(a^2-b^2+1\right)=b\)

\(\Leftrightarrow a\left(a-b\right)\left(a+b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+1\right)=0\)

\(\Leftrightarrow a=b\Rightarrow\sqrt{x+2}=\sqrt{y}\Rightarrow y=x+2\)

Thay xuống pt dưới:

\(x^2+\left(x+3\right)\left(x+3\right)=x+16\)

\(\Leftrightarrow2x^2+5x-7=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=3\\x=-\dfrac{2}{7}\left(loại\right)\end{matrix}\right.\)