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a) Xem lại đề
b) \(\left\{{}\begin{matrix}5x-3y=5\\2x+5y=33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5.\frac{33-5y}{2}-3y=5\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}165-25y-6y=10\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}31y=155\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=\frac{33-5.5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=4\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\frac{x}{2}-\frac{y}{3}=0\\5x+y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\\frac{x}{2}-\frac{13-5x}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\\frac{3x-26+10x}{6}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5.2\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\frac{1}{x}\\b=\frac{1}{y}\end{matrix}\right.\) \(\left(a,b\ne0\right)\)
Khi đó hệ phương trình đề cho trở thành \(\left\{{}\begin{matrix}\frac{1}{4}a+\frac{1}{3}b=2\\b-\frac{1}{2}a=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+4b=24\\-a+2b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x}=4\\\frac{1}{y}=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{4}\\y=\frac{1}{3}\end{matrix}\right.\)
KL:.......
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{4}{x+1}+\frac{1}{y}=\frac{4}{x}\\\frac{2}{x+1}+\frac{4}{x}=\frac{3}{y}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{12}{x+1}+\frac{3}{y}=\frac{12}{x}\\\frac{2}{x+1}+\frac{4}{x}=\frac{3}{y}\end{matrix}\right.\)
\(\Rightarrow\frac{14}{x+1}+\frac{4}{x}=\frac{12}{x}\Leftrightarrow\frac{14}{x+1}=\frac{8}{x}\)
Tới đây chắc bạn giải tiếp được
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
Ta có : \(\left\{{}\begin{matrix}\frac{2x-3y}{4}-\frac{x+y-1}{5}=2x-y-1\\\frac{4x+y-2}{4}=\frac{2x-y-3}{6}-\frac{x-y-1}{3}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{5\left(2x-3y\right)}{20}-\frac{4\left(x+y-1\right)}{20}=\frac{20\left(2x-y-1\right)}{20}\\\frac{3\left(4x+y-2\right)}{12}=\frac{2\left(2x-y-3\right)}{12}-\frac{4\left(x-y-1\right)}{12}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}5\left(2x-3y\right)-4\left(x+y-1\right)=20\left(2x-y-1\right)\\3\left(4x+y-2\right)=2\left(2x-y-3\right)-4\left(x-y-1\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}10x-15y-4x-4y+4=40x-20y-20\\12x+3y-6=4x-2y-6-4x+4y+4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}10x-15y-4x-4y+4-40x+20y+20=0\\12x+3y-6-4x+2y+6+4x-4y-4=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}-34x+y=-24\\12x+y=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-24+34x\\12x-24+34x=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-24+34x\\46x=28\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-\frac{76}{23}\\x=\frac{14}{23}\end{matrix}\right.\)
Vậy hệ phương trình trên có nghiệm là ( x;y ) = \(\left(\frac{14}{23};-\frac{76}{23}\right)\)
\(\left\{\begin{matrix}\frac{1\cdot3y}{4x\cdot3y}+\frac{5x}{12xy}=\frac{4\cdot4}{3xy\cdot4}\\\frac{3\cdot3y}{4x\cdot3y}-\frac{1\cdot4x}{3y\cdot4x}=\frac{-47x}{12xy}\end{matrix}\right.\)
\(\left\{\begin{matrix}\frac{3y}{12xy}+\frac{5x}{12xy}=\frac{16}{12xy}\\\frac{9y}{12xy}-\frac{4x}{12xy}=\frac{-47x}{12xy}\end{matrix}\right.\)
\(\left\{\begin{matrix}3y+5x=16\\9y-4x=-47x\end{matrix}\right.\)
\(\left\{\begin{matrix}5x+3y=16\\43x+9y=0\end{matrix}\right.\) ( nếu là toán violympic thì đến đây bạn có thể sử dụng MODE 5 bấm 1 rồi nhập vào bảng )
x=\(\frac{-12}{7}\)
y=\(\frac{172}{21}\)