\(BĐT\Leftrightarrow a^2+5b^2-3a-b-3ab+5\ge0\)
\(\Leftrightarrow2a^2+10b^2-6a-2b-6ab+10\ge0\)
\(\Leftrightarrow\left(a^2-6ab+9b^2\right)+\left(a^2-2a+9\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-36b\right)^2+\left(a-3\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng với mọi a; b)
Vậy bất đẳng thức được chứng minh.Đẳng thức xảy ra <=> (a; b) = (3; 1)
Cho a,b,c thực dương .CMR
\(\sqrt{\frac{\left(a+b\right)^3}{ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(b+c\right)^3}{bc\left(4b+4c+a\right)}}+\sqrt{\frac{\left(c+a\right)^3}{ca\left(4c+4c+b\right)}}\ge2\sqrt{2}\)
Tính \(\frac{a}{b\left(a+b\right)}+\frac{a}{\left(a+b\right)\left(2a+b\right)}+\frac{a}{\left(2a+b\right)\left(3a+b\right)}+...\) với mọi a và b là số tự nhiên dương
Cho a,b,c>0 thỏa mãn a+b+c = 3. CMR:
\(25\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+351\ge88\left(a^2+b^2+c^2\right)\)
CHo a,b,c > 0 thỏa mãn: abc=1 .CMR:
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(a+c\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\) (1)
Cho a,b,c là các số thực dương. CMR:
\(\frac{1}{2a+b+\sqrt{8bc}}-\frac{8}{\sqrt{2b^2+2\left(a+c\right)^2+3}}\ge\frac{-3}{2}\)
\(A=\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)\(B=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{4}\right)x.......x\left(1-\frac{1}{2015}\right)x\left(1-\frac{1}{2016}\right)\)
\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2x2\frac{1}{3}\right):\frac{7}{4}\)
a.,\(\dfrac{4}{5}+5\dfrac{1}{2}\text{x }\left(4,5-2\right)=\dfrac{7}{10}\) b,125%x\(\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
c,\(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\) d, \(\dfrac{3}{17}+\dfrac{11}{4}+\dfrac{5}{8}+\dfrac{14}{17}+\dfrac{3}{8}\)
Tìm GTNN
a)\(A=3.\left|1-2x\right|-5\)
b)\(B=\left(2x^2+1\right)^4-3\)
c)\(C=_{\left|x-\frac{1}{2}\right|+\left(y+2\right)^2+11}\)
1, Tính giá trị biểu thức :
\(a,A=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}.4\frac{1}{2}-2.2\frac{1}{3}\right):\frac{7}{4}\)
\(b,B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).............\left(1-\frac{1}{2017}\right).\left(1-\frac{1}{2018}\right)\)
Mọi người giúp mình giả toán nha !
