Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{5}\end{matrix}\right.\) \(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{x+y}{xy}=\dfrac{4}{5}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4xy\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4\left(5y-5x\right)\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y=20y-20x\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y-20y+20x=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-15y+25x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-5\left(3y-5x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\3y-5x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-3y=xy\\5x=3y\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2y=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\y=\dfrac{10}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{1}{2x-3y}+\dfrac{5}{3x+y}=\dfrac{5}{8}\\\dfrac{2}{2x-3y}-\dfrac{5}{3x+y}=\dfrac{-3}{8}\end{matrix}\right.\)
Đặt \(\dfrac{1}{2x-3y}=a;\dfrac{1}{3x+y}=b\)
=> hpt <=> \(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b=\dfrac{-3}{8}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b+a+5b=\dfrac{-3}{8}+\dfrac{5}{8}=0,25\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\3a=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\a=\dfrac{1}{12}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=\dfrac{1}{12}\\b=\dfrac{13}{120}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2x-3y}=\dfrac{1}{12}\\\dfrac{1}{3x+y}=\dfrac{13}{120}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=12\\3x+y=\dfrac{120}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{516}{143}\\y=-\dfrac{228}{143}\end{matrix}\right.\)
a,\(\left\{{}\begin{matrix}-x+2y=6\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x+6y=18\left(1\right)\\10x-6y=10\left(2\right)\end{matrix}\right.\)
Cộng (1) và (2) => 7x=28
\(\Leftrightarrow\) x=4
thay x vào (1) ta có -4+2y=6
=> 2y=10
=>y=5
Vậy nghiệm của phương trình (x;y)=(4;5)
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x+\dfrac{1}{4}y=2\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\5y=15\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\varnothing\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)-xy=100\\xy-\left(x-2\right)\left(y-2\right)=64\end{matrix}\right.\)
=>xy+3x+2y+6-xy=100 và xy-xy+2x+2y-4=64
=>3x+2y=94 và 2x+2y=68
=>x=26 và x+y=34
=>x=26 và y=8
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x+3+2}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5y+20-11}{y+4}=9\end{matrix}\right.\)
=>\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+1}-\dfrac{2}{y+4}=4-3=1\\\dfrac{-2}{x+1}+\dfrac{11}{y+4}=9+5-2=12\end{matrix}\right.\)
=>x+1=18/35; y+4=9/13
=>x=-17/35; y=-43/18
\(a)\left\{{}\begin{matrix}2x-y=3\\x+2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=3\\2x+4y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-5y=5\\2x+4y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)
Vậy nghiệm hệ phương trình là (1; -1)
\(b)\left\{{}\begin{matrix}\dfrac{3}{2}x-y=\dfrac{1}{2}\\3x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2y=1\\3x-2y=1\end{matrix}\right.\Leftrightarrow0x-0y=0\left(VSN\right)\)
Vậy hệ phương trình vô số nghiệm
\(c)\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-15y-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x-3x+10y=-1\\2x-3x+15y=-12-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+15y=-16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-2x+30y=-32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}40y=-33\\-2x+30y=-32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{33}{40}\\x=\dfrac{29}{8}\end{matrix}\right.\)
Vậy nghiệm hệ phương trình là \(\left(\dfrac{29}{8};-\dfrac{33}{40}\right)\)
Giải hệ sau :
Câu a :
\(\left\{{}\begin{matrix}x+y=-1\\2x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\-x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\x=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)
Vậy ...........................
Câu b :
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) . Ta có :
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{5}\\3a+4b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=\dfrac{3}{5}\\3a+4b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-b=-\dfrac{7}{5}\\3a+4b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{7}{5}\\a=-\dfrac{6}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{7}{5}\\\dfrac{1}{y}=-\dfrac{6}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{7}\\y=-\dfrac{5}{6}\end{matrix}\right.\)
Vậy..................
\(a,\left\{{}\begin{matrix}2x-y=4\\x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\2x+10y=6\end{matrix}\right.\left\{{}\begin{matrix}11y=2\\2x+10y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x+10.\dfrac{2}{11}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x=\dfrac{46}{11}\end{matrix}\right.\left\{{}\begin{matrix}y=\dfrac{2}{11}\\x=\dfrac{23}{11}\end{matrix}\right.\)
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3x}{4}+\dfrac{7y}{3}=41\\\dfrac{5x}{2}-\dfrac{3y}{5}=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+28y=492\\25x-6y=110\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}225x+700y=12300\\225x-54y=990\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}700y+54y=12300-990\\9x+28y=492\end{matrix}\right.\)
\(\)\(\Leftrightarrow\left\{{}\begin{matrix}y=15\\x=8\end{matrix}\right.\)
Vậy (x;y) = (8;15)