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Cho a,b>0, \(a^2+b^2\le16\). Tìm max của \(M=a\sqrt{9b\left(a+8b\right)}+b\sqrt{9a\left(b+8a\right)}\)
22 tháng 5 2020 lúc 21:19
Cho \(a,b>0\); \(a^2+b^2\le16\).
Tìm max của \(N=a\sqrt{9b\left(a+8b\right)}+b\sqrt{9a\left(b+8a\right)}\)
Cho a, b > 0. CMR :
\(\dfrac{a+b}{\sqrt{a\left(8a+b\right)}+\sqrt{b\left(8b+a\right)}}\ge\dfrac{1}{3}\)
cho a,b \(\ge\) 0 , \(a^2+b^2=2\)tìm GTLN của
M=\(A\sqrt{9B\left(4a+5b\right)}\)+\(b\sqrt{9a\left(4b+5a\right)}\)
Cho a;b\(\ge\)0 và \(a^2+b^2=2\)
Tìm giá trị lớn nhất của:
\(M=a\sqrt{9b\left(4a+5b\right)}+b\sqrt{9a\left(4b+5a\right)}\)
cho 4 số thực a,b,c,d tm a+b+c+d=4
cmr \(\frac{\left(a+\sqrt{b}\right)^2}{\sqrt{a^2-ab+b^2}}+\frac{\left(b+\sqrt{c}\right)^2}{\sqrt{b^2-bc+c^2}}+\frac{\left(c+\sqrt{d}\right)^2}{\sqrt{c^2-cd+d^2}}+\frac{\left(d+\sqrt{a}\right)^2}{\sqrt{d^2-ad+a^2}}\le16\)
Cho a,b,c > 0 thỏa a+b+c=abc. Tìm GTLN của BT :
\(\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}+\dfrac{b}{\sqrt{ac\left(1+b^2\right)}}+\dfrac{c}{\sqrt{ab\left(1+c^2\right)}}\)
B1:Giải bpt sau:left(sqrt{13}-sqrt{2x^2-2x+5}-sqrt{2x^2-4x+4}right).left(x^6-x^3+x^2-x+1right)ge0B2:Cho a;b;c0 thỏa mãn a+b+cfrac{1}{a}+frac{1}{b}+frac{1}{c}.CMR 3left(a+b+cright)gesqrt{8a^2+1}+sqrt{8b^2+1}+sqrt{8c^2+1}B3:giải pt nghiệm nguyên sau : 6left(y^2-1right)+3left(x^2+y^2z^2right)+2left(z^2-9xright)0
Đọc tiếp
B1:Giải bpt sau:\(\left(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\right).\left(x^6-x^3+x^2-x+1\right)\ge0\)
B2:Cho a;b;c>0 thỏa mãn \(a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\).CMR \(3\left(a+b+c\right)\ge\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\)
B3:giải pt nghiệm nguyên sau : \(6\left(y^2-1\right)+3\left(x^2+y^2z^2\right)+2\left(z^2-9x\right)=0\)
Pfrac{a}{sqrt{left(b+1right)left(b^2-b+1right)}}+frac{b}{sqrt{left(c+1right)left(c^2-c+1right)}}+frac{c}{sqrt{left(a+1right)left(a^2-a+1right)}}gefrac{2a}{b^2+2}+frac{2b}{c^2+2}+frac{2c}{a^2+2}left(a+b+cright)-left(frac{ab^2}{b^2+2}+frac{bc^2}{c^2+2}+frac{ca^2}{a^2+2}right)6-left(frac{2ab^2}{b^2+4+b^2}+frac{2bc^2}{c^2+4+c^2}+frac{2ca^2}{a^2+4+a^2}right)ge6-left(frac{2ab}{b+4}+frac{2bc}{c+4}+frac{2ca}{a+4}right)6-left(2a+2b+2c-frac{8a}{b+4}-frac{8b}{c+4}-frac{8c}{a+4}right)frac{8a}{b+4}+frac{8b}{...
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\(P=\frac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}+\frac{b}{\sqrt{\left(c+1\right)\left(c^2-c+1\right)}}+\frac{c}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\)
\(\ge\frac{2a}{b^2+2}+\frac{2b}{c^2+2}+\frac{2c}{a^2+2}=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+2}+\frac{bc^2}{c^2+2}+\frac{ca^2}{a^2+2}\right)\)
\(=6-\left(\frac{2ab^2}{b^2+4+b^2}+\frac{2bc^2}{c^2+4+c^2}+\frac{2ca^2}{a^2+4+a^2}\right)\ge6-\left(\frac{2ab}{b+4}+\frac{2bc}{c+4}+\frac{2ca}{a+4}\right)\)
\(=6-\left(2a+2b+2c-\frac{8a}{b+4}-\frac{8b}{c+4}-\frac{8c}{a+4}\right)\)
\(=\frac{8a}{b+4}+\frac{8b}{c+4}+\frac{8c}{a+4}-6=\frac{8a^2}{ab+4a}+\frac{8b^2}{bc+4b}+\frac{8c^2}{ca+4c}-6\)
\(\ge\frac{8\left(a+b+c\right)^2}{\left(ab+bc+ca\right)+4\left(a+b+c\right)}-6\ge\frac{288}{\frac{\left(a+b+c\right)^2}{3}+24}-6=2\)