\(\left\{{}\begin{matrix}x+2y=5m-1\\-2x+y=2\end{matrix}\right.< =>\left\{{}\begin{matrix}2x+4y=10m-2\\-2x+y=2\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}5y=10m\\-2x+y=2\end{matrix}\right.< =>\left\{{}\begin{matrix}y=2m\\x=m-1\end{matrix}\right.\)

=>\(\sqrt{x}+\sqrt{y}=\sqrt{2}\left(1\right)\) 

=>\(\sqrt{m-1}+\sqrt{2m}=\sqrt{2}\) (\(m\ge1\))

\(< =>\left(\sqrt{m-1}\right)^2=|\left(\sqrt{2}-\sqrt{2m}\right)^2|\)

<=>\(m-1=\left[\sqrt{2}.\left(1-\sqrt{m}\right)\right]^2< =>m-1=|2.\left(1-\sqrt{m}\right)^2|\)

<=>\(m-1=|2\left(1-2\sqrt{m}+m\right)|=\left|2-4\sqrt{m}+2m\right|\)

với \(\left|2-4\sqrt{m}+2m\right|=2-4\sqrt{m}+2m< =>m\le1\)

ta có pt:

<=>\(m-1-2+4\sqrt{m}-2m=0\)

\(< =>-m+4\sqrt{m}-3=0< =>-\left(m-4\sqrt{m}+3\right)=0\)

<=>\(m-4\sqrt{m}+3=0< =>\left(\sqrt{m}-3\right)\left(\sqrt{m}-1\right)=0\)

<=>\(\left[{}\begin{matrix}\sqrt{m}-3=0\\\sqrt{m}-1=0\end{matrix}\right.< =>\left[{}\begin{matrix}m=9\left(loai\right)\\m=1\left(TM\right)\end{matrix}\right.\) 

nếu \(|2-4\sqrt{m}+2m|=-2+4\sqrt{m}-2m< =>m\ge1\)

=>\(-2+4\sqrt{m}-2m=m-1< =>3m-4\sqrt{m}+1=0\)

<=>\(3\left(m-2.\dfrac{2}{3}\sqrt{m}+\dfrac{1}{3}\right)=3\left(m-2.\dfrac{2}{3}\sqrt{m}+\dfrac{4}{9}-\dfrac{4}{9}+\dfrac{1}{3}\right)=0\)

<=>\(\left(\sqrt{m}-1\right)\left(\sqrt{m}-\dfrac{1}{3}\right)=0\)=>\(\left[{}\begin{matrix}\sqrt{m}-1=0\\\sqrt{m}-\dfrac{1}{3}=0\end{matrix}\right.< =>\left\{{}\begin{matrix}m=1\left(TM\right)\\m=\dfrac{1}{3}\left(loai\right)\end{matrix}\right.\)

vậy m=1 thì pt đã cho có 2 nghiệm (x,y) thỏa mãn

\(\sqrt{x}+\sqrt{y}=\sqrt{2}\)

chỗ cuối sửa thành x=1/9 (loại ) hộ :((

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

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

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