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Bài 1 :
a) Ta thấy : \(\left(x^2-9\right)^2\ge0\)
\(\left|y-2\right|\ge0\)
\(\Leftrightarrow A=\left(x^2-9\right)^2+\left|y-2\right|-1\ge-1\)
Dấu " = " xảy ra :
\(\Leftrightarrow\hept{\begin{cases}x^2-9=0\\y-2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\in\left\{3;-3\right\}\\y=2\end{cases}}\)
Vậy \(Min_A=-1\Leftrightarrow\left(x;y\right)\in\left\{\left(3;2\right);\left(-3;2\right)\right\}\)
b) Ta thấy : \(B=x^2+4x-100\)
\(=\left(x+4\right)^2-104\ge-104\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy \(Min_B=-104\Leftrightarrow x=-4\)
c) Ta thấy : \(C=\frac{4-x}{x-3}\)
\(=\frac{3-x+1}{x-3}\)
\(=-1+\frac{1}{x-3}\)
Để C min \(\Leftrightarrow\frac{1}{x-3}\)min
\(\Leftrightarrow x-3\)max
\(\Leftrightarrow x\)max
Vậy để C min \(\Leftrightarrow\)\(x\)max
p/s : riêng câu c mình không tìm được C min :( Mong bạn nào giỏi tìm hộ mình
Bài 2 :
a) Ta thấy : \(x^2\ge0\)
\(\left|y+1\right|\ge0\)
\(\Leftrightarrow3x^2+5\left|y+1\right|-5\ge-5\)
\(\Leftrightarrow C=-3x^2-5\left|y+1\right|+5\le-5\)
Dấu " = " xảy ra :
\(\Leftrightarrow\hept{\begin{cases}x=0\\y+1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)
Vậy \(Max_A=-5\Leftrightarrow\left(x;y\right)=\left(0;-1\right)\)
b) Để B max
\(\Leftrightarrow\left(x+3\right)^2+2\)min
Ta thấy : \(\left(x+3\right)^2\ge0\)
\(\Leftrightarrow\left(x+3\right)^2+2\ge2\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy \(Max_B=\frac{1}{2}\Leftrightarrow x=-3\)
c) Ta thấy : \(\left(x+1\right)^2\ge0\)
\(\Leftrightarrow x^2+2x+1\ge0\)
\(\Leftrightarrow-x^2-2x-1\le0\)
\(\Leftrightarrow C=-x^2-2x+7\le8\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy \(Max_C=8\Leftrightarrow x=-1\)
b)\(B=\frac{x^2-3x+7}{x-3}=\frac{x\left(x-3\right)+7}{x-3}=x+\frac{7}{x-3}\)
\(\Rightarrow B\in Z\Leftrightarrow x+\frac{7}{x-3}\in Z\Leftrightarrow x\in Z,\frac{7}{x-3}\in Z\Leftrightarrow7⋮x-3\Leftrightarrow x-3\inƯ\left\{7\right\}\)
\(\Rightarrow x-3\in\left\{-1;-7;1;7\right\}\)
\(\Rightarrow x\in\left\{2;-4;4;10\right\}\)
c)\(C=\frac{x^2+1}{x-1}=\frac{x^2-1+2}{x-1}=\frac{\left(x-1\right)\left(x+1\right)+2}{x-1}=\left(x+1\right)+\frac{2}{x-1}\)
\(\Rightarrow C\in Z\Leftrightarrow\left(x+1\right)+\frac{2}{x-1}\in Z\Leftrightarrow x-1\in Z;\frac{2}{x-1}\in Z\)
\(\Leftrightarrow x\in Z;2⋮x-1\Rightarrow x-1\inƯ\left(2\right)\)
\(\Rightarrow x-1\in\left\{-1;-2;1;2\right\}\)
\(\Rightarrow x\in\left\{0;-1;2;3\right\}\)
\(3\left|2x+5\right|-4=1\)
\(\Rightarrow\hept{\begin{cases}3\left(2x+5\right)-4=1\\3\left(5-2x\right)-4=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}6x+15-4=1\\15-6x-4=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}6x+11=1\\11-6x=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{-10}{6}\\x=\frac{10}{6}\end{cases}}\)
a) \(A=\left|x+2\right|+\left|x-3\right|\)
\(A=\left|x+2\right|+\left|3-x\right|\ge\left|x+2+3-x\right|=5\)
\(\Rightarrow A\ge5\)
Dấu bằng xảy ra
\(\Leftrightarrow\left(x+2\right)\left(3-x\right)\ge0\)
\(\Leftrightarrow-2\le x\le3\)
Vậy .............................
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