giải hpt: \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left(y+\sqrt{xy}+x-x^2\right)=4\end{matrix}\right.\)
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Giải hệ pt
1/left{{}begin{matrix}4xsqrt{y+1}+8xleft(4x^2-4x-3right)sqrt{x+1}dfrac{x}{x+1}+x^2left(y+2right)sqrt{left(x+1right)left(y+1right)}end{matrix}right.
2/left{{}begin{matrix}xsqrt{y^2+6}+ysqrt{x^2+3}7xyxsqrt{x^2+3}+ysqrt{y^2+6}x^2+y^2+2end{matrix}right.left{{}begin{matrix}xsqrt{y^2+6}+ysqrt{x^2+3}7xyxsqrt{x^2+3}+ysqrt{y^2+6}x^2+y^2+2end{matrix}right.
3/left{{}begin{matrix}left(2x+y-1right)left(sqrt{x+3}+sqrt{xy}+sqrt{x}right)8sqrt{x}left(sqrt{x+3}+sqrt{xy}right)^2+xy2xleft(6-xright)end...
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Giải hệ pt
1/\(\left\{{}\begin{matrix}4x\sqrt{y+1}+8x=\left(4x^2-4x-3\right)\sqrt{x+1}\\\dfrac{x}{x+1}+x^2=\left(y+2\right)\sqrt{\left(x+1\right)\left(y+1\right)}\end{matrix}\right.\)
2/\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)
3/\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)
4/\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)
m.n giúp e mấy bài này vs ạ!!
giải hpt: a,\(\left\{{}\begin{matrix}x+y-\sqrt{xy}=3\\\sqrt{x+1}+\sqrt{y+1}=4\end{matrix}\right.\) b,\(\left\{{}\begin{matrix}x+y=5+\sqrt{\left(x-1\right)\left(y-1\right)}\\\sqrt{x-1}+\sqrt{y-1}=3\end{matrix}\right.\)
a.
ĐKXĐ: \(x;y\ge-1;xy\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y-3=\sqrt{xy}\\x+y+2\sqrt{xy+x+y+1}=14\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\ge0\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}u-3=\sqrt{v}\\u+2\sqrt{u+v+1}=14\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-6u+9\left(u\ge3\right)\\4\left(u+v+1\right)=\left(14-u\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\4u+4\left(u^2-6u+9\right)+4=\left(14-u\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\3u^2+8u-156=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\\left[{}\begin{matrix}u=6\\u=-\dfrac{26}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=6\\v=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=6\\xy=9\end{matrix}\right.\) \(\Rightarrow x=y=3\)
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b.
ĐKXĐ: \(x;y\ge1\)
Xét \(\sqrt{x-1}+\sqrt{y-1}=3\)
\(\Leftrightarrow x+y-2+2\sqrt{\left(x-1\right)\left(y-1\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(y-1\right)}=\dfrac{11-x-y}{2}\)
Thế vào pt đầu:
\(x+y=5+\dfrac{11-x-y}{2}\)
\(\Leftrightarrow x+y=7\Rightarrow y=7-x\)
Thế xuống pt dưới:
\(\sqrt{x-1}+\sqrt{6-x}=3\)
\(\Leftrightarrow5+2\sqrt{\left(x-1\right)\left(6-x\right)}=9\)
\(\Leftrightarrow\left(x-1\right)\left(6-x\right)=4\)
\(\Leftrightarrow...\)
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giải hệ pt \(\left\{{}\begin{matrix}x-3y+2\sqrt{xy}=4\left(\sqrt{x}-\sqrt{y}\right)\\\left(x+1\right)\left(y+\sqrt{xy}-x^2+x\right)=4\end{matrix}\right.\)
giải hpt:
1,\(\left\{{}\begin{matrix}x^2y^2-2x+y^2=0\\2x^2-4x+3+y^3=0\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\left(x^2-xy\right)\left(xy-y^2\right)=25\\\sqrt{x^2-xy}+\sqrt{xy-y^2}=3\left(x-y\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left(y+\sqrt{xy}+x-x^2\right)=4\end{matrix}\right.\)
giải HPT
a) \(\left\{{}\begin{matrix}\left(x+3\right)\left(y-5\right)=xy\\\left(2x-y\right)\left(y+15\right)=2xy\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{4x}-3y+4z^2=-2\\\sqrt{3x}+2y-3z^2=1\\-3\sqrt{x}+y+2z^2=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3y\left(1+y\right)+x^2y^2\left(2+y\right)+xy^3=30\\x^2y+x\left(1+y+y^2\right)+y=11\end{matrix}\right.\)
Ta có hpt \(\left\{{}\begin{matrix}xy+3y-5x-15=xy\\2xy+30x-y^2-15y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5x=3y-15\\6\left(3y-15\right)-y^2-15y=0\end{matrix}\right.\)
Ta có pt (2) \(\Leftrightarrow3y-y^2-80=0\Leftrightarrow y^2-3y+80=0\left(VN\right)\)
=> hpy vô nghiệm
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c) Ta có hpt \(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left(xy+x+y\right)=30\\xy\left(x+y\right)+xy+x+y=11\end{matrix}\right.\)
Đặt j\(xy\left(x+y\right)=a;xy+x+y=b\), ta có hpt
\(\left\{{}\begin{matrix}ab=30\\a+b=11\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=5;b=6\\a=6;b=5\end{matrix}\right.\)
với a=5;b=6, ta có \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}xy=1;x+y=5\\xy=5;x+y=1\end{matrix}\right.\)
đến đây thì thế y hoặc x ra pt bậc 2, còn TH còn lại bn tự giải nhé !
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b) Ta có hpt <=> \(\left\{{}\begin{matrix}2\sqrt{x}-3y+2=-4z^2\\2\sqrt{3x}+4y-2=6z^2\\-3\sqrt{x}+y-4=-2z^2\end{matrix}\right.\)
cộng 3 vế của 3 pt, ta có \(\left(2\sqrt{3}-1\right)\sqrt{x}=4\Leftrightarrow\sqrt{x}=\dfrac{4}{2\sqrt{3}-1}\Leftrightarrow x=\dfrac{16}{\left(2\sqrt{3}-1\right)^2}\)
đến đây thay căn(x)=...vào và đặt z^2=m, ta sẽ ra 1 hệ mới chỉ có 2 ẩn y và m bậc 1 , lát thế vào sẽ ra bậc 2 thì dễ rồi !
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giải hpt:
1, \(\left\{{}\begin{matrix}x^2y^2+4=2y^2\\\left(xy+2\right)\left(y-x\right)=x^3y^3\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^2+y^2-4xy\left(\dfrac{2}{x-y}-1\right)=4\left(4+xy\right)\\\sqrt{x-y}+3\sqrt{2y^2-y+1}=2y^2-x+3\end{matrix}\right.\)
giải hệ pt :
a,\(\left\{{}\begin{matrix}\sqrt{y}\left(\sqrt{x}+\sqrt{x+3}\right)=3\\\sqrt{x}+\sqrt{y}=x+1\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}x^2+x=y^2+y\\x^2+y^2=3\left(x+y\right)\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^4+y^4+x^2y^2=21\end{matrix}\right.\)
a, ĐK: \(x,y\ge0\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3\sqrt{y}}{\sqrt{x+3}-\sqrt{x}}=3\\\sqrt{x}+\sqrt{y}=x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=\sqrt{x+3}\\\sqrt{x}+\sqrt{y}=x+1\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+3}=x+1\)
\(\Leftrightarrow x+3=x^2+2x+1\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\left(l\right)\end{matrix}\right.\)
Thay \(x=1\) vào hệ phương trình đã cho ta được \(y=1\)
Vậy pt đã cho có nghiệm \(x=y=1\)
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b, \(hpt\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\left(y+\dfrac{1}{2}\right)^2\\x^2+y^2=3\left(x+y\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x+y=-1\end{matrix}\right.\\x^2+y^2=3\left(x+y\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2-3x=0\end{matrix}\right.\left(1\right)\\\left\{{}\begin{matrix}x+y=-1\\x^2+y^2=-3\end{matrix}\right.\left(vn\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=y=3\\x=y=0\end{matrix}\right.\)
Vậy ...
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c, Đặt \(\left\{{}\begin{matrix}x^2+y^2=a\\xy=b\end{matrix}\right.\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=7\\a^2-b^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=7\\a-b=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\xy=2\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)^2=9\)
\(\Rightarrow x+y=\pm3\)
TH1: \(\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=-3\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\end{matrix}\right.\)
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Giải hệ phương trình:
1. left{{}begin{matrix}x+32sqrt{left(3y-xright)left(y+1right)}sqrt{3y-2}-sqrt{dfrac{x+5}{2}}xy-2y-2end{matrix}right.
2. left{{}begin{matrix}sqrt{2y^2-7y+10-xleft(y+3right)}+sqrt{y+1}x+1sqrt{y+1}+dfrac{3}{x+1}x+2yend{matrix}right.
3. left{{}begin{matrix}sqrt{4x-y}-sqrt{3y-4x}12sqrt{3y-4x}+yleft(5x-yright)xleft(4x+yright)-1end{matrix}right.
4. left{{}begin{matrix}9sqrt{dfrac{41}{2}left(x^2+dfrac{1}{2x+y}right)}3+40xx^2+5xy+6y4y^2+9x+9end{matrix}right.
5. left{{}begin{mat...
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Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}x+3=2\sqrt{\left(3y-x\right)\left(y+1\right)}\\\sqrt{3y-2}-\sqrt{\dfrac{x+5}{2}}=xy-2y-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{2y^2-7y+10-x\left(y+3\right)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{4x-y}-\sqrt{3y-4x}=1\\2\sqrt{3y-4x}+y\left(5x-y\right)=x\left(4x+y\right)-1\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}9\sqrt{\dfrac{41}{2}\left(x^2+\dfrac{1}{2x+y}\right)}=3+40x\\x^2+5xy+6y=4y^2+9x+9\end{matrix}\right.\)
5. \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left[y+\sqrt{xy}+x\left(1-x\right)\right]=4\end{matrix}\right.\)
6. \(\left\{{}\begin{matrix}x^4-x^3+3x^2-4y-1=0\\\sqrt{\dfrac{x^2+4y^2}{2}}+\sqrt{\dfrac{x^2+2xy+4y^2}{3}}=x+2y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)