\(\hept{\begin{cases}\sqrt{x-1}-3\sqrt{y+2}=2\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{cases}}\)

Đặt \(\sqrt{x-1}=a\left(a\ge0\right)\)

\(\sqrt{y+2}=b\left(b\ge0\right)\)

Khi đó hpt có dạng:

\(\hept{\begin{cases}a-3b=2\\2a+5b=15\end{cases}\Rightarrow\hept{\begin{cases}2a-6b=4\\2a+5b=15\end{cases}}\Rightarrow\hept{\begin{cases}-11b=-11\\2a+5b=15\end{cases}}\Rightarrow\hept{\begin{cases}b=1\\2a+5.1=15\end{cases}}\Rightarrow\hept{\begin{cases}b=1\\a=5\end{cases}\left(TM\right)}}\)

\(\Rightarrow\hept{\begin{cases}\sqrt{x-1}=5\\\sqrt{y+2}=1\end{cases}\Rightarrow\hept{\begin{cases}x-1=25\\y+2=1\end{cases}\Rightarrow}\hept{\begin{cases}x=26\\y=-1\end{cases}}}\)

\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\\sqrt{y+1}+\sqrt{9-x}=m\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\\sqrt{y+1}+\sqrt{9-x}=\sqrt{x+1}+\sqrt{9-y}\left(=m\right)\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\y+1+9-x+2\sqrt{\left(y+1\right).\left(9-x\right)}=x+1+9-y+2\sqrt{\left(x+1\right).\left(9-y\right)}\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\2\sqrt{\left(y+1\right).\left(9-x\right)}=2\sqrt{\left(x+1\right).\left(9-y\right)}\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\\sqrt{9y-x+9-xy}=\sqrt{9x-y+9-xy}\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\9y-x+9-xy=9x-y+9-xy\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\10x-10y=0\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-y}=m\\x=y\end{cases}}\)

<=>\(\hept{\begin{cases}\sqrt{x+1}+\sqrt{9-x}=m\\x=y\end{cases}}\)

<=>\(\hept{\begin{cases}x+1+9-x+2\sqrt{\left(x+1\right).\left(9-x\right)}=m^2\\x=y\end{cases}}\)

<=>\(\hept{\begin{cases}2\sqrt{\left(x+1\right).\left(9-x\right)}=m^2-10\\x=y\end{cases}}\)

<=>\(\hept{\begin{cases}4\left(-x^2+8x+9\right)=m^4-20m^2+100\\x=y\end{cases}}\)

<=>\(\hept{\begin{cases}-4x^2+32x+36=m^4-20m^2+100\\x=y\end{cases}}\)

<=>\(\hept{\begin{cases}-4x^2+32x-m^4+20m^2-64=0\left(1\right)\\x=y\end{cases}}\)

Xét (1)

-4x²+32x-m⁴+20m²-64=0

tính delta rồi xét trường hợp nghiệm duy nhất là ra

\(\hept{\begin{cases}m=\sqrt{x+1}+\sqrt{6-y}\left(1\right)\\m=\sqrt{6-x}-\sqrt{1+y}\left(2\right)\end{cases}}\)

ĐKXĐ : \(-1\le x\le6\); \(-1\le y\le6\)

( 1 ) - ( 2 ) , ta được :

\(\sqrt{x+1}-\sqrt{6-x}+\sqrt{6-y}+\sqrt{1+y}=0\)

\(\Leftrightarrow\sqrt{6-x}-\sqrt{x+1}=\sqrt{6-y}+\sqrt{1+y}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{6-x}\ge\sqrt{x+1}\\6-x-2\sqrt{\left(6-x\right)\left(x+1\right)}+x+1=6-y+2\sqrt{\left(6-y\right)\left(y+1\right)}+y+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le\frac{5}{2}\\-\sqrt{\left(6-x\right)\left(x+1\right)}=\sqrt{\left(6-y\right)\left(y+1\right)}\end{cases}}\)

Ta thấy VP \(\le\)0 ; VT \(\ge\)0 nên VT = VP = 0

\(\Rightarrow\orbr{\begin{cases}x=-1;y=-1\\x=-1;y=6\end{cases}}\)

với x = y = -1 thì m = \(\sqrt{7}\)

với x = -1 ; y = 6 thì m = 0

Vậy m = \(\sqrt{7}\)hoặc m = 0 thì hệ có nghiệm duy nhất

a ) \(HPT\Leftrightarrow\hept{\begin{cases}5x-y=4\left(1\right)\\3x-y=5\left(2\right)\end{cases}}\)

Lấy (1) trừ (2) :

\(\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\)

Thay \(x=-\frac{1}{2}\) vào (1) : \(y=5x-4=5.-\frac{1}{2}-4=-\frac{13}{2}\)

Vậy HPT có nghiệm \(\left(x,y\right)=\left(-\frac{1}{2},-\frac{13}{2}\right)\)

b ) \(\hept{\begin{cases}\sqrt{3}x-\sqrt{2}y=1\\\sqrt{2}x+\sqrt{3}y=\sqrt{3}\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{6}x-2y=\sqrt{2}\left(1\right)\\\sqrt{6}x+3y=3\left(2\right)\end{cases}}}\)

Lấy (2 ) -(1) thu được :

\(5y=3-\sqrt{2}\Rightarrow y=\frac{3-\sqrt{2}}{5}\)

Thay giá trị y trên vào (1) : \(x=\frac{2y+\sqrt{2}}{\sqrt{6}}=\frac{\sqrt{6}+\sqrt{3}}{5}\)

Vậy ......

Bó tay. com

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