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=>2x-2y=8 và 2x+3y=5m+3
=>-5y=8-5m-3=-5m+5 và x-y=4
=>y=m-1 và x=4+m-1=m+3
x^2+y^2-4=(m+3)^2+(m-1)^2-4
=m^2+6m+9+m^2-2m+1-4
=2m^2+4m+6
=2(m^2+2m+3)
=2(m^2+2m+1+2)
=2[(m+1)^2+2]>=4
=>A<=2019/4
Dấu = xảy ra khi m=-1
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Để hệ có nghiệm duy nhất thì \(\frac{m+1}{1}<>\frac{-1}{m-1}\)
=>\(\left(m+1\right)\left(m-1\right)<>-1\)
=>\(m^2-1<>-1\)
=>\(m^2<>0\)
=>m<>0
\(\begin{cases}\left(m+1\right)x-y=m+1\\ x+\left(m-1\right)y=2\end{cases}\)
=>\(\begin{cases}x\left(m+1\right)-y=m+1\\ x\left(m+1\right)+\left(m^2-1\right)y=2\left(m+1\right)\end{cases}\)
=>\(\begin{cases}x\left(m+1\right)+\left(m^2-1\right)y-x\left(m+1\right)+y=2m+2-m-1\\ x\left(m+1\right)-y=m+1\end{cases}\)
=>\(\begin{cases}m^2\cdot y=m+1\\ x\left(m+1\right)=m+1+y\end{cases}\Rightarrow\begin{cases}y=\frac{m+1}{m^2}\\ x\left(m+1\right)=m+1+\frac{m+1}{m^2}\end{cases}\)
=>\(\begin{cases}y=\frac{m+1}{m^2}\\ x=1+\frac{1}{m^2}=\frac{m^2+1}{m^2}\end{cases}\)
\(S=x+y=\frac{m^2+1+m+1}{m^2}=1+\frac{1}{m}+\frac{2}{m^2}\)
\(=2\left(\frac{1}{m^2}+\frac{1}{2m}+\frac12\right)=2\left(\frac{1}{m^2}+2\cdot\frac{1}{m}\cdot\frac14+\frac{1}{16}+\frac{7}{16}\right)\)
\(=2\left(\frac{1}{m}+\frac14\right)^2+\frac78\ge\frac78\forall m\)
Dấu '=' xảy ra khi \(\frac{1}{m}+\frac14=0\)
=>m=-4
\(\left\{{}\begin{matrix}3x-y=2m-1\\x+2y=3m+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y+x+2y=4m-2+3m+2\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\m+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\2y=2m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m+1\end{matrix}\right.\)
\(x^2+y^2+3\\ =m^2+\left(m+1\right)^2+3\\ =m^2+m^2+2m+1+3\\ =2m^2+2m+4\\ =2\left(m^2+m+2\right)\)
\(=2\left(m^2+m+\dfrac{1}{4}+\dfrac{7}{4}\right)\)
\(=2\left[\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]\)
\(=2\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{2}\ge\dfrac{7}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow m=-\dfrac{1}{2}\)
Vậy ...
Câu 1:
=>\(\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}-1< 0\)
=>\(\dfrac{\sqrt{x}-1-x+\sqrt{x}-1}{x-\sqrt{x}+1}< 0\)
=>\(-x+2\sqrt{x}-2< 0\)
=>\(x-2\sqrt{x}+2>0\)
=>(căn x-1)^2+1>0(luôn đúng)
Vậy: x>0
Câu 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2x=2m+4\\x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\y=5-x=5-m-2=3-m\end{matrix}\right.\)
\(A=xy+x-1=\left(m+2\right)\left(3-m\right)+m+2-1\)
\(=3m-m^2+6-2m+m+1\)
\(=-m^2+2m+7\)
\(=-\left(m^2-2m-7\right)\)
\(=-\left(m^2-2m+1-8\right)\)
\(=-\left(m-1\right)^2+8< =8\)
Dấu = xảy ra khi m=1
b: Thay x=2/3 và y=0 vào (d), tađược:
2/3(2m-3)-3=0
=>4/3m-2-3=0
=>4/3m-5=0
=>m*4/3=5
=>m=5:4/3=5*3/4=15/4
Ta có: \(\left\{{}\begin{matrix}3x+y=2m+9\\x+y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x+5-x=2m+9\\y=5-x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x=2m+4\\y=5-x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\y=5-m-2\end{matrix}\right.\)
Gọi A=xy+x-1, ta có: \(A=\left(m+2\right)\left(5-m-2\right)+m+2-1\)
\(A=\left(m+2\right)\left(3-m\right)+m+1\)
\(A=-m^2+m+6+m+1\)
\(A=-m^2+2m+7=-\left(m-1\right)^2+8\)
\(A_{max}=7\Leftrightarrow m=1\) Khi đó x=3, y=2