Cho ab+bc+ca=11.Tìm GTNN của P=\(\frac{5a+5b+2c}{\sqrt{12\left(a^2+11\right)}+\sqrt{12\left(b^2+11\right)}+\sqrt{c^2+11}}\)
Cho các số thực dương a, b, c thỏa mãn ab+bc+ca=11. Tìm GTNN của P=\(\frac{5a+5b+2c}{\sqrt{12\left(a^2+11\right)}+\sqrt{12\left(b^2+11\right)}+\sqrt{c^2+11}}\)
\(P=\frac{5a+5b+2c}{\sqrt{12\left(a^2+ab+bc+ca\right)}+\sqrt{12\left(b^2+ab+bc+ca\right)}+\sqrt{c^2+ab+bc+ca}}\)
\(=\frac{5a+5b+2c}{\sqrt{12\left(a+b\right)\left(a+c\right)}+\sqrt{12\left(a+b\right)\left(b+c\right)}+\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(=\frac{5a+5b+2c}{\sqrt{\left(6a+6b\right)\left(2a+2c\right)}+\sqrt{\left(6a+6b\right)\left(2b+2c\right)}+\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(\Rightarrow P\ge\frac{2\left(5a+5b+2c\right)}{6a+6b+2a+2c+6a+6b+2b+2c+a+c+b+c}\)
\(\Rightarrow P\ge\frac{2\left(5a+5b+2c\right)}{3\left(5a+5b+2c\right)}=\frac{2}{3}\)
\(P_{min}=\frac{2}{3}\) khi \(\left\{{}\begin{matrix}a=b=1\\c=5\end{matrix}\right.\)
Cho a,b,c là các số thực dương thỏa mãn ab + bc + ca = 28
Tìm min \(A=\frac{5a+5b+2c}{\sqrt{12\left(a^2+28\right)}+\sqrt{12\left(b^2+28\right)}+\sqrt{c^2+28}}\)
Cho các số thực dương a,b,c thỏa mãn ab+bc+ca=28
Tìm GTLN của \(P=\dfrac{5a+5b+2c}{\sqrt{12\left(a^2+28\right)}+\sqrt{12\left(b^2+28\right)}+\sqrt{12\left(c^2+28\right)}}\)
thay 28 vao pt nhan tu roi am-gm cho cai do luon
Ps: tim Min
Bài 1 Rút gọn biểu thức
a) \(5\sqrt{4a^2}-5a^2\)
b) \(7\sqrt{\left(a-1\right)^2}+5a\)(a>1)
c)\(\sqrt{\left(2-a\right)^2}-5\sqrt{a+2}\) (a>0)
Bài 2 tính
a) \(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-5\right)^2}\)
b)\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
c)\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}\)
Bài 1:
a) \(=5.|2a|-5a^2\)
b) \(=7\left(a-1\right)+5a=12a-7\)
c) \(|a-2|-5\sqrt{a+2}\)
Bài 2:
a) \(=3-\sqrt{2}+5-\sqrt{2}=8-2\sqrt{2}\)
b) \(=3+\sqrt{2}-\left(3-\sqrt{2}\right)\)
\(=2\sqrt{2}\)
c) \(=6-\sqrt{5}-\left(6+\sqrt{5}\right)\)
\(=-2\sqrt{5}\)
a) \(5\sqrt{4a^2}-5a^2\)
\(=5.|2a|-5a^2\)
b) \(7\sqrt{\left(a-1\right)^2}+5a\)
\(=7\left(a-1\right)+5a\)
\(=12a-7\)
c) \(\sqrt{\left(2-a\right)^2}-5\sqrt{a+2}\)
\(=|a-2|-5\sqrt{a+2}\)
bài 2:
a)\(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-5\right)^2}\)
\(=3-\sqrt{2}+5-\sqrt{2}\)
\(=8-2\sqrt{2}\)
b) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=3+\sqrt{2}-\left(3-\sqrt{2}\right)\)
\(=2\sqrt{2}\)
c)\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}\)
\(=6-\sqrt{5}-\left(6+\sqrt{5}\right)\)
\(=-2\sqrt{5}\)
Bài 1 :
a, \(5\sqrt{4a^2}-5a^2=5\sqrt{2^2a^2}-5a^2=5.2a-5a^2=10a-5a\)
b, \(7\sqrt{\left(a-1\right)^2}+5a=7a-7+5a=12a-7\)
c.\(\sqrt{\left(2-a\right)^2}-5\sqrt{a+2}=2-a-5\sqrt{a+2}\)
Cho a,b,c >0 thỏa mãn \(15\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=10\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+2007\) .Tìm GTNN của \(P=\frac{1}{\sqrt{5a^2+2ab+2b^2}}+\frac{1}{\sqrt{5b^2+2bc+2c^2}}+\frac{1}{\sqrt{5c^2+2ac+2a^2}}\)
Mẫu thức như vầy thì tìm max còn được chứ tìm min sao nổi bạn?
\(15\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+30\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=40\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2007\)
\(\Leftrightarrow15\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=40\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2007\)
\(\Leftrightarrow15\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le\frac{40}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2007\)
\(\Leftrightarrow\frac{5}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le2007\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{\frac{6021}{5}}\)
Ta có:
\(5a^2+2ab+2b^2=4a^2+2ab+b^2+a^2+b^2\ge4a^2+2ab+b^2+2ab=\left(2a+b\right)^2\)
\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}=\frac{1}{a+a+b}+\frac{1}{b+b+c}+\frac{1}{c+c+a}\)
\(\Rightarrow P\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow P\le\frac{1}{3}\sqrt{\frac{6021}{5}}\)
Dấu "=" xảy ra khi \(a=b=c=3\sqrt{\frac{5}{6021}}\)
Cho 3 số thực a,b,c thỏa mãn: \(3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)-2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=404\)
Tìm MaxP \(=\dfrac{1}{\sqrt{5a^2+2ab+2b^2}}+\dfrac{1}{\sqrt{5b^2+2bc+2c^2}}+\dfrac{1}{\sqrt{5c^2+2ca+2a^2}}\)
\(404=3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)-2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\ge\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-\dfrac{2}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\le1212\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le2\sqrt{303}\)
Ta có:
\(5a^2+2ab+2b^2=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow P\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{2}{c}+\dfrac{1}{a}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{2\sqrt{303}}{3}\)
Rút gọn
\(C=\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{12+2\sqrt{\left(\sqrt{13}+1\right)^2}}-\sqrt{\left(\sqrt{11}+1\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{14+2\sqrt{13}}-\left(\sqrt{11}+1\right)\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{13}+\sqrt{11}\right)\)
C \(\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{13}+\sqrt{11}\right)\) = \(13-11\) = \(2\)
\(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}-\sqrt{12+2\sqrt{11}}}\right)\left(\sqrt{11}+\sqrt{3}\right)\)
\(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{12+2\sqrt{\left(\sqrt{13+1}\right)^2}}-\sqrt{\left(\sqrt{11+1}\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{12+2\sqrt{13+2}}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)\(=\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{11}+\sqrt{13}\right)=13-11=2\)
sao dấu= thứ 2 lại ra như vậy
cho ba số thực dương a,b,c. cmr : \(\sqrt[3]{5a^2b+3}+\sqrt[3]{5b^2c+3}+\sqrt[3]{5c^2a+3}\le\frac{21}{12}\left(a+b+c\right)+\frac{1}{4}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
help me!