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a) \(A=\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-3\right|\ge\left|\left(x-1\right)+\left(3-x\right)\right|=2\)
Vậy\(A_{min}=2\Leftrightarrow\left(x-1\right)\left(3-x\right)\ge0\)
\(TH1:\hept{\begin{cases}x-1\ge0\\3-x\ge0\end{cases}}\Leftrightarrow1\le x\le3\)
\(TH1:\hept{\begin{cases}x-1\le0\\3-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge3\end{cases}}\left(L\right)\)
Vậy \(A_{min}=2\Leftrightarrow1\le x\le3\)
A = \(\sqrt{\left(x-3\right)-2\sqrt{x-3}+1+2}\)
= \(\sqrt{\left(\sqrt{x-3}-1\right)^2+2}\)\(\ge\)\(\sqrt{0+2}\)=\(\sqrt{2}\)
''='' <=> x = 4
=> Min A = \(\sqrt{2}\)và x = 4
B = |x-2011| + |x-1|
TH1: x \(\le\)1
=> B = 2012 - 2x \(\ge\)2010 ''='' <=> x = 1
TH2: 1\(\le\)x\(\le\)2011
=> B = x - 1 + 2011 - x = 2010 với mọi x t/m đkiện
TH3: x \(\ge\)2011
=> B = 2x - 2012 \(\ge\)2010 ''='' <=> x = 2011
Vậy Min B = 2010 <=> 1\(\le\)x\(\le\)2011
a) Đề có lẽ là:
đk: \(x\ge0\)
\(\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}+2\right)x=x\sqrt{x}-\sqrt{x}+3\)
\(\Leftrightarrow x+2\sqrt{x}+1+x\sqrt{x}+2x-x\sqrt{x}+\sqrt{x}-3=0\)
\(\Leftrightarrow3x+3\sqrt{x}-2=0\)
\(\Leftrightarrow3\left(x+\sqrt{x}+\frac{1}{4}\right)-\frac{11}{4}=0\)
\(\Leftrightarrow\left(\sqrt{x}+\frac{1}{2}\right)^2-\frac{11}{12}=0\)
\(\Leftrightarrow\left(\sqrt{x}+\frac{3+\sqrt{33}}{6}\right)\left(\sqrt{x}+\frac{3-\sqrt{33}}{6}\right)=0\)
Vì \(\sqrt{x}\ge0\left(\forall x\right)\)
=> \(\sqrt{x}=\frac{3-\sqrt{33}}{6}\Rightarrow x=\frac{7-\sqrt{33}}{6}\)
b) đk: \(x\ge1\)
Ta có: \(\sqrt{4\left(x^2-1\right)}-2\sqrt{15}=0\)
\(\Leftrightarrow\sqrt{x^2-1}=\sqrt{15}\)
\(\Leftrightarrow x^2-1=15\)
\(\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
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\(A=\sqrt{x-2\sqrt{x-3}}\\ A=\sqrt{x-3-2\sqrt{x-3}+1+2}\\ A=\sqrt{\left(\sqrt{x-3}-1\right)^2+2}\)
vì:
\(\sqrt{x-3}\ge0\\ \Rightarrow\sqrt{x-3}+1\ge1\\ do\:đó\:\left(\sqrt{x-3}+1\right)^2+2\ge2\Rightarrow A\ge\sqrt{2}\)
đẳng thức xảy ra khi x=3
vậy \(MIN_A=\sqrt{2}\) tại x=3
\(B=\sqrt{\left(x-2007\right)^2}+\sqrt{\left(x-1\right)^2}\\ B=\left|x-2007\right|+\left|x-1\right|\\ B=\left|2007-x\right|+\left|x-1\right|\ge\left|2007-x+x-1\right|=2006\)
đẳng thức xảy ra khi :
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}2007-x\ge0\\x-1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2007-x< 0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le2007\\x\ge1\end{matrix}\right.\left(nhận\right)\\\left\{{}\begin{matrix}x>2007\\x< 1\end{matrix}\right.\left(loại\right)\end{matrix}\right.\)
vậy GTNN của B= 2006 tại \(1\le x\le2007\)