giai he phuong trinh \(\left\{{}\begin{matrix}\sqrt{x+1}+\sqrt{y-1}=2+\sqrt{6}\\x+y=5+2\sqrt{6}\end{matrix}\right.\)
Những câu hỏi liên quan
Giai he phuong trinh:
a) \(\left\{{}\begin{matrix}x^2-y^2=1\\4x^2-5xy=2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x+\sqrt{y+2018}=1\\\sqrt{x+2018}+y=1\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x+y=\sqrt{4z-1}\\y+z=\sqrt{4x-1}\\z+x=\sqrt{4y-1}\end{matrix}\right.\)
Giai phuong trinh va he phuong trinh:
a) \(\sqrt{x^2+6}=x-2\sqrt{x^2-1}\)
b) \(x^2+3x+1=\left(x+3\right).\sqrt{x^2+1}\)
c) \(\left\{{}\begin{matrix}x^2+y^2=11\\x+xy+y=3+4\sqrt{2}\end{matrix}\right.\)
Giai he phuong trinh:
a) \(\left\{{}\begin{matrix}x+\sqrt{y+2018}=1\\\sqrt{x+2018}+y=1\end{matrix}\right.\)
Giai he phuong trinh:
a) \(\left\{{}\begin{matrix}5x+3y=31\\\sqrt{\dfrac{x+2}{y-3}}+\sqrt{\dfrac{y-3}{x+2}}=2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y-12}-\dfrac{x}{y}=2\end{matrix}\right.\)
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!!
a)\(\left\{{}\begin{matrix}2x+\left|y\right|=3\\x-y=6\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\sqrt{3}x+y=\sqrt{2}\\\sqrt{3}x-\sqrt{2}y=-1\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}2\sqrt{x+3}+\sqrt{y^2-4y+4}=2\\\sqrt{x+3}-3\left|2-y\right|=1\end{matrix}\right.\)
a, Với y >= 0
hpt có dạng \(\left\{{}\begin{matrix}2x+y=3\\x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=9\\y=x-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\)(ktmđk)
Với y < 0 hpt có dạng
\(\left\{{}\begin{matrix}2x-y=3\\x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-3-6=-9\end{matrix}\right.\)(tm)
b, bạn tự làm
c, đk : x>= 3
\(\left\{{}\begin{matrix}2\sqrt{x+3}+\left|y-2\right|=2\\\sqrt{x+3}-3\left|y-2\right|=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x+3}+\left|y-2\right|=2\\2\sqrt{x+3}-6\left|y-2\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7\left|y-2\right|=1\\2\sqrt{x+3}+\left|y-2\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y-2=\dfrac{1}{7}\\y-2=-\dfrac{1}{7}\end{matrix}\right.\\2\sqrt{x+3}+\left|y-2\right|=2\end{matrix}\right.\)
bạn tự giải nốt nhé
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giải hệ phuong trình :\(\left\{{}\begin{matrix}\left(\sqrt{x^2+1}+x\right)\left(\sqrt{y^2+1}-y\right)=1\\3\sqrt{x+2y-2}+x\sqrt{x-2y+6}=10\end{matrix}\right.\)
\(\left(\sqrt{x^2+1}+x\right)\left(\sqrt{y^2+1}-y\right)=1\)
\(\Leftrightarrow\sqrt{x^2+1}+x=\sqrt{y^2+1}+y\) (1)
Tương tự ta có: \(\sqrt{y^2+1}-y=\sqrt{x^2+1}-x\) (2)
Cộng vế (1) và (2) \(\Rightarrow x-y=y-x\Rightarrow x=y\)
Thế xuống dưới:
\(3\sqrt{3x-2}+x\sqrt{6-x}=10\)
Đặt \(\sqrt{6-x}=a\Rightarrow\left\{{}\begin{matrix}0\le a\le\frac{4\sqrt{3}}{3}\\x=6-a^2\end{matrix}\right.\)
\(\Rightarrow a^3-6a+10-3\sqrt{16-3a^2}=0\)
\(\Leftrightarrow\left(a^3-3a-2\right)+3\left(4-a-\sqrt{16-3a^2}\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+1\right)^2+\frac{12a\left(a-2\right)}{4-a+\sqrt{16-a^2}}=0\)
\(\Leftrightarrow\left(a-2\right)\left[\left(a+1\right)^2+\frac{12a}{4-a+\sqrt{16-a^2}}\right]=0\)
\(\Leftrightarrow a=2\Leftrightarrow...\)
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\(\left\{{}\begin{matrix}y^3-x^3-x^2-y^2-xy^2+x^2y=6x-6y+6\\y\sqrt{x+3}+\left(y+6\right)\sqrt{x+10}=y^2+4x\end{matrix}\right.\)
giai he
tim m de he phuong trinh va phuong trinh co nghiem
\(a,\sqrt{x^2+3x+2m}=\sqrt{4x-x^2}\)
b, \(\left\{{}\begin{matrix}x+y+1=x\\x^2+y^2=m\end{matrix}\right.\)