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选修\(4-5\):不等式选讲
已知函数\(f\left( x \right)=\left| x-2 \right|\).
\((\)Ⅰ\()\)解不等式\(f\left( x \right)+f\left( x+1 \right)\geqslant 5\);
\((\)Ⅱ\()\)若\(\left| a \right| > 1\),且\(f\left( ab \right) > \left| a \right|\cdot f\left( \dfrac{b}{a} \right)\),证明:\(\left| b \right| > 2\).
把\(2{{a}^{2}}-3ab-2{{b}^{2}}\)分解因式,结果正确的是( )
已知,分别求,,的值,然后归纳猜想一般性结论,并证明你的结论.
已知函数\(f(x)=\left| x-1 \right|\).
\((\)Ⅰ\()\)解不等式\(f(x-1)+f(x+3)\geqslant 6\);
\((\)Ⅱ\()\)若\(\left| a \right| < 1,\left| b \right| < 1\),且\(a\ne 0\),求证:\(f(ab) > \left| a \right|f(\dfrac{b}{a})\).
已知函数\(f\left( x \right)=\left| 2x+1\left| - \right|x-2 \right|\),不等式\(f\left( x \right)\leqslant 2\)的解集为\(M\).
\((1)\)求\(M\);
\((2)\)记集合\(M\)的最大元素为\(m\),若正数\(a,b,c\)满足\(abc=m\),求证:\(\sqrt{a}+\sqrt{b}+\sqrt{c}\leqslant \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\).
\((1)\)设\(a\),\(b\)是两个不相等的正数,若\( \dfrac{1}{a} + \dfrac{1}{b} =1\),用综合法证明:\(a+b > 4\)
\((2)\)已知\(a > b > c\),且\(a+b+c=0\),用分析法证明:\( \dfrac{ \sqrt{{b}^{2}-ac}}{a} < \sqrt{3} \).
\((II)\)证明:设 的三边,求证: \((\)可直接应用第\((I)\)小题结论\()\)
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