Giải hệ pt: \(\left\{{}\begin{matrix}\frac{15}{x}-\frac{7}{y}=9\\\frac{4}{x}+\frac{9}{y}=35\end{matrix}\right.\)
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 , \(\left\{{}\begin{matrix}\frac{15}{x}-\frac{7}{y}=9\\\frac{4}{x}+\frac{9}{y}=35\end{matrix}\right.\)
có ái đó giúp mình với mình đang cần gấp
giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
giải hệ pt
\(\left\{{}\begin{matrix}\frac{8xy}{x^2+6xy+y^2}+\frac{17}{8}\left(\frac{y}{x}+\frac{x}{y}\right)=\frac{21}{4}\\\sqrt{x-16}+\sqrt{y-9}=7\end{matrix}\right.\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge16\\y\ge9\end{matrix}\right.\)
Từ pt thứ nhất của hệ:
\(\frac{8xy}{x^2+y^2+6xy}+\frac{17}{8}\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{21}{4}\)
\(\Leftrightarrow\frac{8}{\frac{x}{y}+\frac{y}{x}+6}+\frac{17}{8}\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{21}{4}\)
Đặt \(\frac{x}{y}+\frac{y}{x}=t\ge2\)
\(\Rightarrow\frac{8}{6+t}+\frac{17}{8}t=\frac{21}{4}\)
\(\Leftrightarrow\frac{17}{8}t^2+\frac{15}{2}t-\frac{47}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-\frac{94}{17}< 0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\frac{x}{y}+\frac{y}{x}=2\Leftrightarrow x^2+y^2=2xy\)
\(\Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x=y\)
Thay xuống pt dưới:
\(\sqrt{x-16}+\sqrt{x-9}=7\)
\(\Leftrightarrow\sqrt{x-16}-3+\sqrt{x-9}-4=0\)
\(\Leftrightarrow\frac{x-25}{\sqrt{x-16}+3}+\frac{x-25}{\sqrt{x-9}+4}=0\)
\(\Leftrightarrow...\)
giải các hệ pt sau:
a) \(\left\{{}\begin{matrix}x+2y=-1\\x-y=5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\frac{5}{x}-\frac{6}{y}=3\\\frac{4}{x}+\frac{9}{y}=7\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}3\sqrt{x+1}+\sqrt{y-1}=1\\\sqrt{x+1}-\sqrt{y-1}=-2\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\left|x-1\right|+y=5\\4x+3y=23\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}x+2y=-1\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
Vậy..............................................................................
b) \(\left\{{}\begin{matrix}\frac{5}{x}-\frac{6}{y}=3\\\frac{4}{x}+\frac{9}{y}=7\end{matrix}\right.\)ĐKXĐ: x,y≠0
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{20}{x}-\frac{24}{y}=12\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{69}{y}=23\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=10\end{matrix}\right.\)
Vậy...................................................................................
c) \(\left\{{}\begin{matrix}3\sqrt{x+1}+\sqrt{y-1}=1\\\sqrt{x+1}-\sqrt{y-1}=-2\end{matrix}\right.\)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge-1\\y\ge1\end{matrix}\right.\)
\(\Rightarrow4\sqrt{x+1}\)\(=-1\)(vô nghiệm)
Vậy hệ pt vô nghiệm
d) Nhân 3 pt đầu rồi thu gọn
Giải hệ PT
\(\left\{{}\begin{matrix}\frac{x}{4}=\frac{y}{3}=\frac{z}{9}\\7x-3y+2z=90\end{matrix}\right.\)
Ta có hệ pt :
\(\left\{{}\begin{matrix}\frac{y}{3}=\frac{x}{4}\\\frac{z}{9}=\frac{x}{4}\\7x-3y+2z=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3x}{4}\\z=\frac{9x}{4}\\7x-\frac{3.3x}{4}+\frac{2.9x}{4}=90\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3x}{4}\\z=\frac{9x}{4}\\28x-9x+18x=360\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3x}{4}\\z=\frac{9x}{4}\\37x=360\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{4}.\frac{360}{37}\\z=\frac{9}{4}.\frac{360}{37}\\x=\frac{360}{37}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{270}{37}\\z=\frac{810}{37}\\x=\frac{360}{37}\end{matrix}\right.\)
Vậy . . . . . . . . .
Giải hệ pt sau
a)\(\left\{{}\begin{matrix}\frac{x-12}{4}=\frac{y-9}{3}=z-1\\3x+5y-z=2\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{a+b}{6}=\frac{b+c}{7}=\frac{a+c}{8}\\a+b+c=14\end{matrix}\right.\)
Giải hệ phương trình :
1, \(\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{4}{x}+\frac{1}{y-2}=1\end{matrix}\right.\)
2 , \(\left\{{}\begin{matrix}\frac{2}{2x-y}-\frac{1}{x+y}=0\\\frac{3}{2x-y}-\frac{6}{x+y}=-1\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-2y\right)-15\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}2x+y=7\\-x+4y=10\end{matrix}\right.\)
1/ ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{12}{x}+\frac{3}{y-2}=3\end{matrix}\right.\) \(\Rightarrow\frac{10}{x}=-1\Rightarrow x=-10\)
\(\frac{4}{-10}+\frac{1}{y-2}=1\Rightarrow\frac{1}{y-2}=\frac{7}{5}\Rightarrow y-2=\frac{5}{7}\Rightarrow y=\frac{19}{7}\)
2/ ĐKXĐ:...
Đặt \(\left\{{}\begin{matrix}\frac{1}{2x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a-b=0\\3a-6b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{9}\\b=\frac{2}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x-y}=\frac{1}{9}\\\frac{1}{x+y}=\frac{2}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=9\\x+y=\frac{9}{2}\end{matrix}\right.\) \(\Rightarrow...\)
3/ \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-6y-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+6y=-19\end{matrix}\right.\) \(\Rightarrow...\)
4/ Bạn tự giải
giải hệ phương trình \(\left\{{}\begin{matrix}\frac{7}{x-y+2}-\frac{8}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
hệ phương trình
1 ,\(\left\{{}\begin{matrix}\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}\\2\left(x-3\right)-3\left(y+2\right)=-16\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{3}{2}\\3x-2y=5\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{x^2-y-6}{x}=x-2\\x+3y=8\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{2}{3}\\x+y=10\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}\frac{y^2+2x-8}{y}=y-3\\x+y=10\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}\frac{x+1}{y-1}=5\\3\left(2x-2\right)-4\left(3x+4\right)=5\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}2x+y=4\\\left|x-2y\right|=3\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}\frac{2x}{x+1}+\frac{y}{y+1}=3\\\frac{x}{x+1}-\frac{3y}{y+1}=-1\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}y-\left|x\right|=1\\2x-y=1\end{matrix}\right.\)
10 , \(\left\{{}\begin{matrix}\sqrt{x+3y}=\sqrt{3x-1}\\5x-y=9\end{matrix}\right.\)