Lời giải: ĐK: $x,y\geq 2$
HPT \(\Rightarrow \sqrt{x+1}-\sqrt{y+1}+(\sqrt{y-2}-\sqrt{x-2})=0\)

\(\Leftrightarrow (x-y).\left[\frac{1}{\sqrt{x+1}+\sqrt{y+1}}-\frac{1}{\sqrt{y-2}+\sqrt{x-2}}\right]=0\)

\(\Leftrightarrow x-y=0\) (do dễ thấy biểu thức trong ngoặc vuông luôn âm)

\(\Leftrightarrow x=y\)

Khi đó: $\sqrt{x+1}+\sqrt{x-2}=\sqrt{m}$
$\Leftrightarrow 2x-1+2\sqrt{(x+1)(x-2)}=m$

Để hpt có nghiệm thì pt trên có nghiệm 

$\Leftrightarrow m\geq \min (2x-1+2\sqrt{(x+1)(x-2)})$

$\Leftrightarrow m\geq 2.2-1+2.0=3$

Vậy $m\geq 3$

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-2\\y\ge-3\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\\sqrt{y+3}=b\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=m\\a^2-2+b^2-3=2m-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=m\\a^2+b^2=2m\end{matrix}\right.\)

\(\Leftrightarrow a^2+\left(m-a\right)^2=2m\)

\(\Leftrightarrow2a^2-2m.a+m^2-2m=0\) (1)

Hệ đã cho có nghiệm khi và chỉ khi (1) có 2 nghiệm không âm

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta'=m^2-2\left(m^2-2m\right)\ge0\\a_1+a_2=m\ge0\\a_1a_2=\dfrac{m^2-2m}{2}\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}0\le m\le4\\m\ge0\\\left[{}\begin{matrix}m\ge2\\m\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m=0\\2\le m\le4\end{matrix}\right.\)

ĐK: \(x,y\ge0\)

\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=1\\x\sqrt{x}+y\sqrt{y}=1-3m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=1\\\left(\sqrt{x}+\sqrt{y}\right)\left(x+y-\sqrt{xy}\right)=1-3m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=1\\\left(\sqrt{x}+\sqrt{y}\right)^2-3\sqrt{xy}=1-3m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=1\\\sqrt{xy}=m\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{y}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(\Rightarrow a,b\) là nghiệm phương trình \(t^2-t+m=0\left(1\right)\)

Yêu cầu bài toán thỏa mãn khi phương trình \(\left(1\right)\) có nghiệm không âm

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=1-4m\ge0\\x_1+x_2\ge0\\x_1x_2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le\dfrac{1}{4}\\1\ge0\\m\ge0\end{matrix}\right.\Leftrightarrow0\le m\le\dfrac{1}{4}\)

ĐKXĐ : \(0\le x,y\le1\)

Ta có : 

 \(\sqrt{x}+\sqrt{1-y}=m+1;\sqrt{y}+\sqrt{1-x}=m+1\\ \Rightarrow\sqrt{x}+\sqrt{1-y}=\sqrt{y}+\sqrt{1-x}\Rightarrow\sqrt{x}-\sqrt{y}=\sqrt{1-x}-\sqrt{1-y}\)

 \(TH1:\ 1\ge x>y\ge0\Rightarrow\sqrt{x}>\sqrt{y};\sqrt{1-x}< \sqrt{1-y}\\ \Rightarrow\sqrt{x}-\sqrt{y}>0;\sqrt{1-x}-\sqrt{1-y}< 0\\ \Rightarrow\sqrt{x}-\sqrt{y}>\sqrt{1-x}-\sqrt{1-y}\left(VL\right)\)

\(TH2:\ 1\ge y>x\ge0. Tương\ tự:vôlý\)

TH3: x=y. Khi đó hệ phương trình trở thành

\(\sqrt{x}+\sqrt{1-x}=m+1\)

Áp dụng bất đẳng thức \(\sqrt{A+B}\le\sqrt{A}+\sqrt{B}\le\sqrt{2\left(A+B\right)}\) ta có:

\(1\le m+1\le\sqrt{2}\Leftrightarrow0\le m\le\sqrt{2}-1\)

Bài 2:

1.Thay m=3, ta có:

\(\left\{{}\begin{matrix}3x+2y=5\\2x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)

Bài 1:

\(\left\{{}\begin{matrix}\left|x+1\right|+\left|y-1\right|=5\\\left|x+1\right|-4y=-4\end{matrix}\right.\)

\(\Rightarrow\left|y-1\right|-4y=9\)\(\Leftrightarrow\left[{}\begin{matrix}y=-3,\left(3\right)\left(KTM\right)\left(ĐK:y\ge1\right)\\y=-1,6\left(TM\right)\left(ĐK:y< 1\right)\end{matrix}\right.\)

Thay y=-1,6 vào hpt, ta được:

\(\left\{{}\begin{matrix}\left|x+1\right|=2,4\\\left|x+1\right|=-10,4\left(vl\right)\end{matrix}\right.\)

Vậy pt vô nghiệm.

Bài 1:

(1)\(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-36}{x-3}+\dfrac{15}{y+2}=-189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)

\(\Rightarrow\dfrac{-44}{x-3}=-202\)\(\Leftrightarrow x=\dfrac{325}{101}\)\(\Rightarrow y\approx-2,3016\)

(2)\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\)\(\Leftrightarrow y=-1\)\(\Rightarrow x=26.\)

Đặt \(\left\{{}\begin{matrix}\sqrt{7x+y}=a\ge0\\\sqrt{x+y}=b\ge0\end{matrix}\right.\) \(\Rightarrow x-y=\dfrac{a^2-4b^2}{3}\)

Hệ trở thành:

\(\left\{{}\begin{matrix}a+b=6\\b+\dfrac{a^2-4b^2}{3}=m\end{matrix}\right.\)

\(\Rightarrow6-a+\dfrac{a^2-4\left(6-a\right)^2}{3}=m\)

\(\Leftrightarrow-a^2+15a-42=m\)

Với \(0\le a\le6\Rightarrow-42\le-a^2+15a-42\le12\)

\(\Rightarrow-42\le m\le12\)