a: Thay \(y=\dfrac{1}{3}\) vào (d3), ta được:

\(\dfrac{-2}{3}x+\dfrac{5}{3}=\dfrac{1}{3}\)

\(\Leftrightarrow x=2\)

Thay x=2 và \(y=\dfrac{1}{3}\) vào (d), ta được:

\(2\left(m-2\right)+m+7=\dfrac{1}{3}\)

\(\Leftrightarrow3m=\dfrac{1}{3}-3=\dfrac{-8}{3}\)

hay \(m=-\dfrac{8}{9}\)

(d₃₎//(d₄₎\(\Leftrightarrow\left\{{}\begin{matrix}m^2+6m=7\\2n+7\ne-n^2-9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m^2+6m-7=0\\n^2+2n+16\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m-1\right)\left(m+7\right)=0\\\left(n+1\right)^2+15\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m-1=0\\m+7=0\end{matrix}\right.\\\left(n+1\right)^2+15\ne0\left(luônđúng\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-7\end{matrix}\right.\)

\(\left(d3\right)\equiv\left(d4\right)\Leftrightarrow\left\{{}\begin{matrix}m^2+6m=7\\2n+7=-n^2-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2+6m-7=0\\n^2+2n+16=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m-1\right)\left(m+7\right)=0\\\left(n+1\right)^2+15=0\left(vôlí\right)\end{matrix}\right.\)

1.

a, \(\left\{{}\begin{matrix}2x-3y=3\\-4x=3x-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3y=3\\-4x-3x=13\end{matrix}\right.\)\(\left\{{}\begin{matrix}-4x+6y=-6\\-4x-3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9y=-19\\-4x+6y=-6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=-\dfrac{19}{9}\end{matrix}\right.\)

b, \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=3\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=9\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=2\\\dfrac{3}{x}+\dfrac{3}{y}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\left(TM\right)\\y=\dfrac{1}{2}\left(TM\right)\end{matrix}\right.\)

c, \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{5}{y}=1\\\dfrac{2}{x}+\dfrac{1}{y}=3\end{matrix}\right.\left(x,y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{5}{y}=1\\\dfrac{10}{x}+\dfrac{5}{y}=15\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{x}=16\\\dfrac{10}{x}+\dfrac{5}{y}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{16}\left(TM\right)\\y=\dfrac{13}{7}\left(TM\right)\end{matrix}\right.\)

d, \(\left\{{}\begin{matrix}\sqrt{x+1}-3\sqrt{y-1}=-4\\2\sqrt{x+1}-\sqrt{y-1}=2\end{matrix}\right.\left(x\ge-1,y\ge1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x+1}-6\sqrt{y-1}=-8\\2\sqrt{x+1}-\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-5\sqrt{y-1}=-10\\2\sqrt{x+1}-6\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y-1}=2\\2\sqrt{x+1}-6\sqrt{y-1}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(TM\right)\\y=5\left(TM\right)\end{matrix}\right.\)

2: a: Khi m=-1 thì hệ sẽ là:

-x-y=2 và 4x+y=5

=>3x=7 và x+y=-2

=>x=7/3 và y=-2-7/3=-6/3-7/3=-13/3

b: Để hệ có nghiệm duy nhất thì \(\dfrac{m}{4}< >\dfrac{-1}{-m}=\dfrac{1}{m}\)

=>m^2<>4

=>m<>2; m<>-2

c: Để hệ có vô số nghiệm thì \(\dfrac{m}{4}=\dfrac{-1}{-m}=\dfrac{2}{m+6}\)

=>m^2=4 và m^2+6m=8

=>\(m\in\varnothing\)

d: Để hệ vô nghiệm thì m/4=-1/-m<>2/m+6

=>m=2 hoặc m=-2

Đề thi tuyển sinh THPT Hoàng Văn Thụ, Hòa Bình, 2013-2014

Giải:

PT hoành độ giao điểm là (m+1)m=x²

<=> x²-(m+1)x+m=0

\(\Delta=\left(m+1\right)^2-4m=m^2-2m+1=\left(m-1\right)^2,m\ne1\)

\(\sqrt{\Delta}=m-1\)

\(x_1=\frac{m+1+m-1}{1}=2m\)

\(\Rightarrow y_1=\left(2m\right)^2-\left(m+1\right)2m+m=4m^2-2m^2-2m+m=2m^2-m\)

\(x_2=\frac{m+1-m+1}{1}=2\)

\(\Rightarrow y_2=4-\left(m+1\right)\cdot2+m=4-2m-2+m=2-m\)

=> A(2m;2m²-m)

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