\(\frac{\sqrt{a^2+5}+\sqrt{b^2+5}+\sqrt{6\left(c^2+5\right)}}{7a+7b+8c}\)ab+bc+ca=3 Tìm gtln của biểu thức trên( Sử dụng bđt cauchy)
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9 tháng 8 2020
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
16 tháng 10 2019
Tự tìm ĐKXĐ nhé
\(P=\frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}+2}-\frac{5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\sqrt{x}-3}\)
\(=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
c, \(P=\frac{\sqrt{x}+4}{\sqrt{x}+2}=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
Để \(P\in Z\Rightarrow1+\frac{2}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ\left(2\right)=\left\{1;2;-1;-2\right\}\)
\(\Rightarrow\sqrt{x}=\left\{-1;0\right\}\)
\(\Rightarrow x=\left\{0\right\}\)
Kết hợp với ĐKXĐ =>...
21 tháng 5 2018
\(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+3\ge7\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le3\)Áp dụng BĐT AM-GM ta có :
\(A=\frac{1}{\sqrt{a^3+b^3+1}}+\frac{1}{\sqrt{b^3c^3+1+1}}+\frac{4\sqrt{3}}{c^6+1+2a^3+8}\)
\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{4\sqrt{3}}{2c^3+2a^3+8}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+4}\)
\(=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+1+1+1+1}\)
\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{6\sqrt{ac}}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{1}{\sqrt{3ac}}\)\(=\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}\right)\)
\(\le\frac{1}{\sqrt{3}}\sqrt{3\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}=\sqrt{\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}\le\sqrt{3}\) (Bunhiacopxki)
Dấu "=" xảy ra\(\Leftrightarrow a=b=c=1\)
PS : Thánh cx đc phết ha; chế đc bài này tui mới khâm phục :)))
28 tháng 5 2018
nó ko chém đâu anh nó chép trong toán tuổi thơ đấy,thk này khốn nạn lắm
MN
4 tháng 4 2020
Bài 1 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b) Để \(A< -1\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)
\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}< 1\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)
\(\Leftrightarrow x< \frac{1}{4}\)
Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)
20 tháng 5 2019
Ta có:\(7\left(\frac{1}{a^2}+...\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+2015\)
Mà \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le2015\)=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{6045}\)
\(P=\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+...\)
Mà \(\left(2+1\right)\left(2a^2+b^2\right)\ge\left(2a+b\right)^2\)(bất dẳng thức buniacoxki)
=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
Lại có \(\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
=> \(P\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\le\frac{\sqrt{6045}}{3}\)
Vậy \(MaxP=\frac{\sqrt{6045}}{3}\)khi \(a=b=c=\frac{\sqrt{6045}}{2015}\)
H9
HT.Phong (9A5)
CTVHS
23 tháng 8 2023
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
H
23 tháng 8 2023
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
KN
6 tháng 8 2020
\(P=\frac{3a+3b+2c}{\sqrt{6\left(a^2+5\right)}+\sqrt{6\left(b^2+5\right)}+\sqrt{c^2+5}}\)
\(=\frac{3a+3b+2c}{\sqrt{6\left(a^2+ab+bc+ca\right)}+\sqrt{6\left(b^2+ab+bc+ca\right)}+\sqrt{c^2+ab+bc+ca}}\)(Do ab + bc + ca = 5)
\(=\frac{3a+3b+2c}{\sqrt{6\left(a+b\right)\left(a+c\right)}+\sqrt{6\left(b+a\right)\left(b+c\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Áp dụng BĐT AM - GM, ta được:
\(\sqrt{6\left(a+b\right)\left(a+c\right)}=2\sqrt{\frac{6}{4}\left(a+b\right)\left(a+c\right)}\)\(\le\frac{6}{4}\left(a+b\right)+\left(a+c\right)=\frac{5}{2}a+\frac{6}{4}b+c\)
\(\sqrt{6\left(b+a\right)\left(b+c\right)}=2\sqrt{\frac{6}{4}\left(b+a\right)\left(b+c\right)}\)\(\le\frac{6}{4}\left(a+b\right)+\left(b+c\right)=\frac{6}{4}a+\frac{5}{2}b+c\)
\(\sqrt{\left(c+a\right)\left(c+b\right)}\le\frac{\left(c+a\right)+\left(c+b\right)}{2}=c+\frac{a}{2}+\frac{b}{2}\)
Cộng theo vế của 3 BĐT trên, ta được: \(\sqrt{6\left(a+b\right)\left(a+c\right)}+\sqrt{6\left(b+a\right)\left(b+c\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}\)\(\le\frac{9}{2}a+\frac{9}{2}b+3c\)
\(\Rightarrow\frac{3a+3b+2c}{\sqrt{6\left(a+b\right)\left(a+c\right)}+\sqrt{6\left(b+a\right)\left(b+c\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}}\)\(\ge\frac{3a+3b+2c}{\frac{9}{2}a+\frac{9}{2}b+3c}=\frac{2}{3}\)
Đẳng thức xảy ra khi \(a=b=1;c=2\)
4 tháng 7 2017
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\). Thiếp lập 2 BĐT còn lại:
\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{b}{a+b}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(A\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\)
Xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
6 tháng 5 2020
Áp dụng Bunhia cho bộ số (1;1;1) vfa (a;b;c) ta có 3(a2+b2+c2) >= (a+b+c)2
=> 3(2a2+b2) >=(2a+b2); 3(2b2+c2) >= (2b+c)2; 3(2c2+a2) >= (2c+a)2
=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
Ta có \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\Rightarrow\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x+y+z}\)
=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+b}\le\frac{1}{9}\left[\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\right]\)
=> \(P\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(I\right)\)
Ta có \(10\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2015\)
\(=3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2015\left(II\right)\)
Áp dụng Bunhia cho bộ số (1;1;1) và \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\)
Ta được \(3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)\(\Rightarrow\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
=> \(10\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge10\cdot\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\left(III\right)\)
Từ (I)(II)(III) => \(3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2015\ge10\cdot\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le3\cdot2015\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\sqrt{3\cdot2015}\left(IV\right)\)
Từ (I)(IV) => \(P\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}\cdot\sqrt{3\cdot2015}=\sqrt{\frac{2015}{3}}\)
Vậy GTNN của P=\(\sqrt{\frac{2015}{3}}\)khi a=b=c và \(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2015\)
=> \(a=b=c=\sqrt{\frac{3}{2015}}\)
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