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\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
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..................
\(a,\left\{{}\begin{matrix}2x-y=4\\x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\2x+10y=6\end{matrix}\right.\left\{{}\begin{matrix}11y=2\\2x+10y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x+10.\dfrac{2}{11}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x=\dfrac{46}{11}\end{matrix}\right.\left\{{}\begin{matrix}y=\dfrac{2}{11}\\x=\dfrac{23}{11}\end{matrix}\right.\)
\(a)\left\{{}\begin{matrix}2x-y=3\\x+2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=3\\2x+4y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-5y=5\\2x+4y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)
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 \(\left(\dfrac{29}{8};-\dfrac{33}{40}\right)\)
\(ĐK:x,y\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{2}{y}=4\\\dfrac{2}{x}+\dfrac{3}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+1=2\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\left(tm\right)\)
\(ĐK:x\ne-1;y\ne2\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y}{2-y}=-1\\\dfrac{x}{x+1}+\dfrac{2y}{2-y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0y=-2\left(vn\right)\\\dfrac{x}{x+1}+\dfrac{2y}{2-y}=2\end{matrix}\right.\Leftrightarrow x,y\in\varnothing\)
Đặt x/x+1=a
y/2-y=b
\(\Leftrightarrow\left\{{}\begin{matrix}a+2b=1\\a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-1\\a=2-b=2-\left(-1\right)=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3x+3\\y=y-2\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\varnothing\)