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\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\)
\(A=\left|x-1990\right|+\left|1991-x\right|\ge\left|x-1990+1991-x\right|=1\)
\(A_{min}=1\) khi \(1990\le x\le1991\)
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 \(A_{min}=2\Leftrightarrow1\le x\le3\)
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a) \(\left|x-2000\right|+\left|x-2002\right|=\left|x-2000\right|+\left|2002-x\right|\)
\(\ge\left|x-2000+2002-x\right|=2\) (1)
Dấu "=" \(\Leftrightarrow\left(x-2000\right)\left(2002-x\right)\ge0\)
\(\Leftrightarrow2000\le x\le2002\)
+ \(\left|x-2001\right|\ge0\forall x\). "=" \(\Leftrightarrow x=2001\) (2)
Từ (1) và (2) suy ra \(A\ge2\)
Dấu "=" \(\Leftrightarrow x=2001\)
b) \(B=\left|x-8\right|+\left|x-9\right|+\left|x-10\right|+\left|x+11\right|\)
+ \(\left|x-10\right|+\left|x+11\right|=\left|x+11\right|+\left|10-x\right|\)
\(\ge\left|x+11+10-x\right|=21\) (3)
Dấu "=" \(\Leftrightarrow\left(x+11\right)\left(10-x\right)\ge0\Leftrightarrow-11\le x\le10\)
+ \(\left|x-8\right|+\left|x-9\right|\ge\left|x-8+9-x\right|=1\) (4)
"=" \(\Leftrightarrow\left(x-8\right)\left(9-x\right)\ge0\Leftrightarrow8\le x\le9\)
Từ (3) và (4) suy ra \(B\ge22\)
"=" \(\Leftrightarrow8\le x\le9\)