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

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

`a)` Thay `m=\sqrt{3}+1` vào hệ ptr có:

`{(\sqrt{3}x-2y=1),(3x+(\sqrt{3}+1)y=1):}`

`<=>{(3x-2\sqrt{3}y=\sqrt{3}),(3x+(\sqrt{3}+1)y=1):}`

`<=>{((3\sqrt{3}+1)y=1-\sqrt{3}),(\sqrt{3}x-2y=1):}`

`<=>{(y=[-5+2\sqrt{3}]/13),(\sqrt{3}x-2[-5+2\sqrt{3}]/13=1):}`

`<=>{(x=[4+\sqrt{3}]/13),(y=[-5+2\sqrt{3}]/13):}`

`b){((m-1)x-2y=1),(3x+my=1):}`

`<=>{(x=[1-my]/3),((m-1)[1-my]/3-2y=1):}`

`<=>{(x=[1-my]/3),(m-m^2y-1+my-6y=3):}`

`<=>{(x=[1-my]/3),((-m^2+m-6)y=4-m):}`

`<=>{(x=[1-my]/3),(y=[4-m]/[-m^2+m-6]):}`

   Mà `-m^2+m-6` luôn `ne 0`

   `=>AA m` thì đều tìm được `1` giá trị `y` từ đó tìm được `x`

 `=>AA m` thì hệ ptr có `1` nghiệm duy nhất

`c){((m-1)x-2y=1),(3x+my=1):}`

`<=>{(x=[1-my]/3),(y=[4-m]/[-m^2+m-6]):}`

`<=>{(x=(1-m[4-m]/[-m^2+m-6]):3),(y=[4-m]/[-m^2+m-6]):}`

`<=>{(x=[-m^2+m-6-4m+m^2]/[-3m^2+3m-18]),(y=[4-m]/[-m^2+m-6]):}`

`<=>{(x=[-3m-6]/[3(-m^2+m-6)]),(y=[4-m]/[-m^2+m-6]):}`

Ta có: `x-y=[-3m-6]/[3(-m^2+m-6)]-[4-m]/[-m^2+m-6]`

                `=[-3m-6-12+3m]/[-3(m^2-m+6)]`

                `=[-18]/[-3(m^2-m+6)]=6/[(m-1/2)^2+23/4]`

Vì `(m-1/2)^2+23/4 >= 23/4`

`<=>6/[(m-1/2)^2+23/4] <= 24/23`

Hay `x-y <= 24/23`

Dấu "`=`" xảy ra `<=>m-1/2=0<=>m=1/2`

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

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

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

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

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

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

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

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 :((

a: Khi m=căn 2 thì hệ sẽ là:

2x-y=căn 2+1 và x+y*căn 2=2

=>\(\left\{{}\begin{matrix}2x-y=\sqrt{2}+1\\2x+2y\sqrt{2}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-y-2y\sqrt{2}=\sqrt{2}-3\\2x-y=\sqrt{2}+1\end{matrix}\right.\)

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

b: Để hệ có nghiệm thì 2/1<>-1/m

=>-1/m<>2

=>m<>-1/2

Ta có: \(\left\{{}\begin{matrix}\left(\sqrt{3}-\sqrt{2}\right)x+y=\sqrt{2}\\x+\left(\sqrt{3}+\sqrt{2}\right)y=\sqrt{6}\end{matrix}\right.\)

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

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

Vậy: Hệ phương trình vô nghiệm