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Đặt 1/căn x-3=a; 1/|2y-1|=b
Theo đề, ta có; 8a+11b=5 và 4a+b=3
=>a=7/9; b=-1/9
=>|2y-1|=-9(loại)
=>Hệ vô nghiệm
Có : \(\left\{{}\begin{matrix}\sqrt{x+1}+\dfrac{2y}{y+1}=2\\2\sqrt{x+1}-\dfrac{1}{y+1}=\dfrac{3}{2}\end{matrix}\right.\)
-> \(\left\{{}\begin{matrix}2\sqrt{x+1}+\dfrac{4y}{y+1}=4\\2\sqrt{x+1}-\dfrac{1}{y+1}=\dfrac{3}{2}\end{matrix}\right.\)
-> \(\left\{{}\begin{matrix}2\sqrt{x+1}+\dfrac{4y}{y+1}-2\sqrt{x+1}+\dfrac{1}{y+1}=4-\dfrac{3}{2}\\2\sqrt{x+1}-\dfrac{1}{y+1}=\dfrac{3}{2}\end{matrix}\right.\)
-> \(\left\{{}\begin{matrix}\dfrac{4y+1}{y+1}=\dfrac{5}{2}\\\sqrt{x+1}=\dfrac{\dfrac{3}{2}+\dfrac{1}{y+1}}{2}\end{matrix}\right.\)
-> \(\left\{{}\begin{matrix}2.\left(4y+1\right)=5.\left(y+1\right)\\\sqrt{x+1}=\dfrac{\dfrac{3}{2}+\dfrac{1}{y+1}}{2}\end{matrix}\right.\)
-> \(\left\{{}\begin{matrix}8y+2=5y+5\\\sqrt{x+1}=\dfrac{\dfrac{3}{2}+\dfrac{1}{y+1}}{2}\end{matrix}\right.\)
-> \(\left\{{}\begin{matrix}3y=3->y=1\\\sqrt{x+1}=\dfrac{\dfrac{3}{2}+\dfrac{1}{1+1}}{2}\end{matrix}\right.\)
-> \(\left\{{}\begin{matrix}y=1\\\sqrt{x+1}=1\end{matrix}\right.\)
-> \(\left\{{}\begin{matrix}y=1\\x=0\end{matrix}\right.\)
Vậy .........
ĐKXĐ: x # -1/2; y # -2
\(Đặt\ \dfrac{x-1}{2x+1}=a; \dfrac{y-2}{y+2}=b \\Hệ\ tương\ đương: \\\begin{cases} a-b=1\\3a+2b=3 \end{cases} <=> \begin{cases} 3a-3b=3\\3a+2b=3 \end{cases} \\<=>\begin{cases} -5b=0\\a-b=1 \end{cases} <=>\begin{cases} b=0\\a=1 \end{cases} \\->\begin{cases} x-1=2x+1\\y-2=0 \end{cases} <=>\begin{cases} x=-2(thoả\ ĐKXĐ)\\y=2(thoả\ ĐKXĐ) \end{cases}\)
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1: Khi m=3 thì hệ phương trình (1) trở thành:
\(\left\{{}\begin{matrix}3x-2y=-1\\2x+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{13}\\y=\dfrac{5}{13}\end{matrix}\right.\)
2: Khi x=-1/2 và y=2/3 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}2\cdot\dfrac{-1}{2}+3\cdot\dfrac{2}{3}=1\\-\dfrac{1}{2}m-\dfrac{4}{3}=-1\end{matrix}\right.\Leftrightarrow m\cdot\dfrac{-1}{2}=\dfrac{1}{3}\)
hay m=-2/3
1: =>x^2+3x-4=0
=>(x+4)(x-1)=0
=>x=1 hoặc x=-4
2: =>2x-3y=1 và 3x=4y+2
=>2x-3y=1 và 3x-4y=2
=>x=2 và y=1
\(\left\{{}\begin{matrix}\dfrac{3}{x+1}-2x=-1\left(ĐK:x\ne-1\right)\\\dfrac{5}{x+1}+3y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{x+1}-6y=-3\\\dfrac{10}{x+1}+6y=22\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\dfrac{9}{x+1}+\dfrac{10}{x+1}=19\\\dfrac{3}{x+1}-2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{5}{4}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2}{x+2}+\dfrac{1}{2y-3}=2\\\dfrac{6}{x+2}-\dfrac{2}{2y-3}=1\end{matrix}\right.\left(I\right)\)
Đặt \(\left\{{}\begin{matrix}a=\dfrac{1}{x+2}\\b=\dfrac{1}{2y-3}\end{matrix}\right.\)
\(\left(I\right)\left\{{}\begin{matrix}2a+b=2\\6a-2b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+2b=4\\6a-2b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10a=5\\6a-2b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+2}=\dfrac{1}{2}\left(x\ne-2\right)\\\dfrac{1}{2y-3}=1\left(y\ne\dfrac{3}{2}\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
Vậy nghiệm hệ phương trình là \(\left(0;2\right)\)
`đk:x ne 2,y ne 1/2`
ĐẶt `a=1/(x-2),b=1/(2y-1)`
`hpt<=>` $\begin{cases}a+5b=3\\3a-b=1\\\end{cases}$
`<=>` $\begin{cases}3a+15b=9\\3a-b=1\\\end{cases}$
`<=>` $\begin{cases}16b=8\\a=3-5b\\\end{cases}$
`<=>` $\begin{cases}b=\dfrac12\\a=\dfrac12\\\end{cases}$
`<=>` $\begin{cases}x-2=2\\2y-1=2\\\end{cases}$
`<=>` $\begin{cases}x=4\\y=\dfrac32\\\end{cases}$
Đk: \(x\ne2;y\ne\dfrac{1}{2}\)
Đặt \(a=\dfrac{1}{x-2},b=\dfrac{1}{2y-1}\) (a,b khác 0)
Có hệ: \(\left\{{}\begin{matrix}a+5b=3\\3a-b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a+5b=3\\15a-5b=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}16a=8\\3a-b=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=3a-1=\dfrac{1}{2}\end{matrix}\right.\)(tm)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{1}{2}\\\dfrac{1}{2y-1}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\dfrac{3}{2}\end{matrix}\right.\)(tm)