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\(\Leftrightarrow\left\{{}\begin{matrix}1+\frac{2}{x-1}+3-\frac{6}{y+2}=7\\\frac{2}{x-1}-\frac{5}{y+2}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x-1}-\frac{6}{y+2}=3\\\frac{2}{x-1}-\frac{5}{y+2}=4\end{matrix}\right.\)
\(\Rightarrow\frac{1}{y+2}=1\Rightarrow y+2=1\Rightarrow y=-1\)
\(\Rightarrow\frac{2}{x-1}=3+\frac{6}{y+2}=9\Rightarrow x-1=\frac{2}{9}\Rightarrow x=\frac{11}{9}\)
\(\begin{cases} 4xy +4(x^2+y^2)+\frac{3}{(x+y)^2}=7\\2x+\frac{1}{x+y}=1\end{cases}\)
\(\begin{cases} 4xy +8(x^2+y^2)+\frac{5}{(x+y)^2}=13\\2x+\frac{1}{x+y}=1\end{cases}\)
Giải hpt:
\(\begin{cases}2^{\sqrt{x^2-y}}+2^{\frac{-1}{\sqrt{x^2-y}}}=\frac{5}{2}\\4x\sqrt{y}-\sqrt{x\left(y+9\right)}=4y-x-2\end{cases}\)
giúp mình bài này với mình đang cần gấp
1/ tìm nghiệm pt/bpt sau
a/ x2 + \(\sqrt[3]{x^4-x^2}\) =2x +1
b/ \(\begin{cases}2x^2-y^2=1\\x^2+xy=2\end{cases}\)
2/ hệ có nghiệm duy nhất khi a=
\(\begin{cases}x+y=6\\x^2+y^2=a\end{cases}\)
hệ phương trình
1, \(\left\{{}\begin{matrix}3x=6\\x-3y=2\end{matrix}\right.\)
2,\(\left\{{}\begin{matrix}3x+5y=15\\2y=-7\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}7x-2y=1\\3x+y=6\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}3\left(x+y\right)+9=2\left(x-y\right)\\2\left(x+y\right)=3\left(x-y\right)+11\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}3\left(x+y\right)+5\left(x-y\right)=12\\-5\left(x+y\right)+2\left(x-y\right)=11\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}2\left(3x-2\right)-4=5\left(3y+2\right)\\4\left(3x-2\right)+7\left(3y+2\right)=-2\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=\frac{4}{5}\\\frac{1}{x}-\frac{1}{y}=\frac{1}{5}\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}\frac{15}{x}-\frac{7}{y}=9\\\frac{4}{x}+\frac{9}{y}=35\end{matrix}\right.\)
GIẢI HỆ PHƯƠNG TRÌNH
(GIẢI GIÚP EM VS MỌI NGƯỜI)
1, \(\begin{cases} x(y+z)=8 \\ y(x+z)=18\\ z(x+y)=20 \end{cases}\)
2, \(\begin{cases} \dfrac{xy}{x+y} =\dfrac{8}{3}\\ \dfrac{yz}{z+y} =\dfrac{12}{5}\\ \dfrac{xz}{x+z} =\dfrac{24}{7} \end{cases} \)
3, \(\begin{cases} x^{2} + 2yz=x\\ y^{2} + 2xz=y\\ z^{2} + 2xy=z\\ \end{cases}\)
4, \(\begin{cases} \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} =2\\ \dfrac{2}{xy} -\dfrac{1}{z^{2}} =4 \end{cases} \)
Giải hệ phương trình:
\(\begin{cases} a^2 - b^2 =-2\\a-b=\frac{3}{2}ab \end{cases}\)
Giải các hệ
\(\left\{{}\begin{matrix}\sqrt{x+y}+\sqrt{2x+y+2}=7\\3x+2y=23\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=\frac{-5}{4}\\x^4+y^2+xy\left(1+2x\right)=\frac{-5}{4}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x^2+1\right)+y\left(x+y\right)=7y\\\left(x^2+1\right)\left(x+y-2\right)=-y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x\left(x+y+1\right)=3\\\left(x+y\right)^2-\frac{5}{x^2}=-1\end{matrix}\right.\)
Giải các hệ phương trình
\(\left\{{}\begin{matrix}\frac{1}{x+1}+\frac{1}{y}=\frac{1}{3}\\\frac{1}{\left(x+1\right)^2}-\frac{1}{y^2}=\frac{1}{4}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+m\right)^2-y^2+y\left(x+m\right)=11\\x+2y=7-m\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+a\right)^2+2\left(y-a\right)^2-\left(x+a\right)\left(y-a\right)=2\\x+y=2\end{matrix}\right.\)