共50条信息
若正实数\(x\),\(y\)满足\(2x+y=2\),则\(\dfrac{4{{x}^{2}}}{y+1}+\dfrac{{{y}^{2}}}{2x+2}\)的最小值是_____.
\((2)\)若\(p\),\(q\),\(r\)为正实数,且\( \dfrac{1}{3p}\)\(+\)\( \dfrac{1}{2q}\)\(+\)\( \dfrac{1}{r}\)\(=4\),求\(3p+2q+r\)的最小值.
设\(a_{1}\),\(a_{2}\),\(…\),\(a_{n} > 0\),\(\dfrac{n}{\dfrac{1}{1+{{a}_{1}}}+\dfrac{1}{1+{{a}_{2}}}+\cdots +\dfrac{1}{1+{{a}_{n}}}}-\dfrac{n}{\dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{2}}}+\cdots +\dfrac{1}{{{a}_{n}}}}\geqslant 1\).
已知定义在\(R\)上的函数\(f\)\((\)\(x\)\()=|\)\(x\)\(+1|+|\)\(x\)\(-2|\)的最小值为\(a\).
\((1)\)求\(a\)的值;
\((2)\)若\(p\),\(q\),\(r\)是正实数,且满足\(p\)\(+\)\(q\)\(+\)\(r\)\(=\)\(a\),求证:\(p\)\({\,\!}^{2}+\)\(q\)\({\,\!}^{2}+\)\(r\)\({\,\!}^{2}\geqslant 3\).
已知函数\(f(x)=4-|x|-|x-3|\)
\((\)Ⅰ\()\)求不等式\(f(x+ \dfrac{3}{2} )\geqslant 0\)的解集;
\((\)Ⅱ\()\)若\(p\),\(q\),\(r\)为正实数,且\( \dfrac{1}{3p}+ \dfrac{1}{2q}+ \dfrac{1}{x}=4 \),求\(3p+2q+r\)的最小值.
选修\(4-5\):已知函数\(f(x)=|x-a|+|x+b|(a > 0,b > 0)\).
\((1)\)若\(a=1\),\(b=2\),解不等式\(f(x)\leqslant 5\);\((2)\)若\(f(x)\)的最小值为\(3\),求\( \dfrac{{a}^{2}}{b}+ \dfrac{{b}^{2}}{a} \)的最小值.
著名数学家华罗庚说过:“数形结合百般好,格列分家万事休”。的确,有很多代数问题可以转化为几何问题加以解决。如图:三角形是边长为\(1\)的等边\(\triangle ABC\),点\(P\),\(Q\),\(R\)分别在边\(AB\),\(BC\),\(CA\)上运动,显然,\(\triangle PQR\)的面积不大于\(\triangle ABC\)的面积。记\(AP=x\),\(BQ=y\),\(CR=z\).
由此可知:若\(0\leqslant x\),\(y\),\(z\leqslant 1\),则\((1-x)y+(1-y)z+(1-z)x\leqslant \)________.
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