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Võ Quang Nhân
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Nguyễn Thị Minh Nhã
22 tháng 5 2022 lúc 19:42

P≤a2+2aab+2b2+b2+22bc+2c2+c2+22ca+2a2

P≤(a+2b)2+(b+2c)2+(c+2a)2

P≤(1+2)(a+b+c)=1+2

Dấu "=" xảy ra khi (a;b;c)=(0;0;1) và các hoán vị

Hi Mn
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Lizy
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Nguyễn Việt Lâm
14 tháng 1 lúc 13:07

Min:

\(\left(a+b+c\right)^3=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\ge a^3+b^3+c^3\)

\(\Rightarrow a+b+c\ge\sqrt[3]{a^3+b^3+c^3}=\sqrt[3]{3}\)

\(\Rightarrow P=\dfrac{a}{7-3bc}+\dfrac{b}{7-3ca}+\dfrac{c}{7-3ab}\ge\dfrac{a}{7}+\dfrac{b}{7}+\dfrac{c}{7}=\dfrac{a+b+c}{7}\ge\dfrac{\sqrt[3]{3}}{7}\)

Dấu "=" xảy ra tại \(\left(a;b;c\right)=\left(0;0;\sqrt[3]{3}\right)\) và các hoán vị

Max:

\(\left(a^3+1+1\right)+\left(b^3+1+1\right)+\left(c^3+1+1\right)\ge3a+3b+3c\)

\(\Rightarrow a+b+c\le\dfrac{a^3+b^3+c^3+6}{3}=3\)

 

Khi đó:

\(7P=\dfrac{7a}{7-3bc}+\dfrac{7b}{7-3ca}+\dfrac{7c}{7-3ab}=\dfrac{a\left(7-3bc\right)+3abc}{7-3bc}+\dfrac{b\left(7-3ca\right)+3abc}{7-3ca}+\dfrac{c\left(7-3ab\right)+3abc}{7-3ab}\)

\(=a+b+c+\dfrac{3abc}{7-3bc}+\dfrac{3abc}{7-3ca}+\dfrac{3abc}{7-3ab}\)

Ta có:

\(7-3ab\ge\dfrac{7}{9}\left(a+b+c\right)^2-3ab=\dfrac{1}{9}\left[\dfrac{13}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(a^2+b^2\right)+7c^2+14bc+14ca\right]\)

Do \(\dfrac{13}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(a^2+b^2\right)\ge\dfrac{1}{2}\left(a^2+b^2\right)\ge ab\)

\(\Rightarrow7-3ab\ge\dfrac{1}{9}\left(ab+7c^2+14bc+14ca\right)\)

\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{27abc}{ab+7c\left(c+2a+2b\right)}\le\dfrac{27abc}{36^2}\left(\dfrac{1^2}{ab}+\dfrac{35^2}{7c\left(c+2a+2b\right)}\right)\)

\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{c}{48}+\dfrac{175}{48}.\dfrac{ab}{c+2a+2b}=\dfrac{c}{48}+\dfrac{175}{48}.\dfrac{ab}{\left(a+b+c\right)+\left(a+b\right)}\)

\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{c}{48}+\dfrac{175}{48}.\dfrac{ab}{5^2}\left(\dfrac{3^2}{a+b+c}+\dfrac{2^2}{a+b}\right)\)

\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{c}{48}+\dfrac{21}{16}.\dfrac{ab}{a+b+c}+\dfrac{7}{12}.\dfrac{ab}{a+b}\le\dfrac{c}{48}+\dfrac{21}{16}.\dfrac{ab}{a+b+c}+\dfrac{7}{48}.\dfrac{\left(a+b\right)^2}{a+b}\)

\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{7a+7b+c}{48}+\dfrac{21}{16}.\dfrac{ab}{a+b+c}\)

Tương tự:

\(\dfrac{3abc}{7-3bc}\le\dfrac{a+7b+7c}{48}+\dfrac{21}{16}.\dfrac{bc}{a+b+c}\)

\(\dfrac{3abc}{7-3ca}\le\dfrac{7a+b+7c}{48}+\dfrac{21}{16}.\dfrac{ca}{a+b+c}\)

\(\Rightarrow7P\le\dfrac{21}{16}\left(a+b+c\right)+\dfrac{21}{16}\left(\dfrac{ab+bc+ca}{a+b+c}\right)\le\dfrac{21}{16}\left(a+b+c\right)+\dfrac{21}{48}.\dfrac{\left(a+b+c\right)^2}{a+b+c}\)

\(\Rightarrow7P\le\dfrac{7}{4}\left(a+b+c\right)\)

\(\Rightarrow P\le\dfrac{a+b+c}{4}\le\dfrac{3}{4}\)

Vậy \(P_{max}=\dfrac{3}{4}\) khi \(a=b=c=1\)

 

Lê Minh Trí
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Ruynn
25 tháng 9 2021 lúc 8:12

Kham khảo bài lm này nhé:

Nguyễn Hoàng Minh
25 tháng 9 2021 lúc 8:13

\(2a^2+2b^2=5ab\\ \Leftrightarrow2a^2-5ab+2b^2=0\\ \Leftrightarrow2a^2-4ab-ab+2b^2=0\\ \Leftrightarrow2a\left(a-2b\right)+b\left(a-2b\right)=0\\ \Leftrightarrow\left(2a+b\right)\left(a-2b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=-\dfrac{b}{2}\\a=2b\end{matrix}\right.\)

Với \(a=-\dfrac{b}{2}\Leftrightarrow Q=\dfrac{-\dfrac{b}{2}+b}{-\dfrac{b}{2}-b}=\dfrac{b}{2}:\dfrac{-3b}{2}=\dfrac{b}{-3b}=-\dfrac{1}{3}\)

Với \(a=2b\Leftrightarrow Q=\dfrac{3b}{b}=3\)

Lấp La Lấp Lánh
25 tháng 9 2021 lúc 8:13

\(2a^2+2b^2=5ab\)

\(\Leftrightarrow\left(2a^2-4ab\right)+\left(2b^2-ab\right)=0\)

\(\Leftrightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\b=2a\end{matrix}\right.\)

TH1: a=2b

\(Q=\dfrac{a+b}{a-b}=\dfrac{2b+b}{2b-b}=\dfrac{3b}{b}=3\)

TH2: b=2a

\(Q=\dfrac{a+b}{a-b}=\dfrac{a+2a}{a-2a}=\dfrac{3a}{-a}=-3\)

Nguyễn Hồng Phúc
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ILoveMath
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Phúc Phúc
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Nguyễn Hoàng Minh
26 tháng 11 2021 lúc 14:09

\(M=a^2-a\left|a\right|-\dfrac{b}{2}\cdot2\left|b\right|-b^2\\ M=a^2+a^2-b^2-b^2\\ M=2\left(a^2-b^2\right)\\ D\)

Rhider
26 tháng 11 2021 lúc 14:05

D . \(2.\left(a^2-b^2\right)\)

Hi Mn
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Mai Thị Thanh
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Mai Thị Thanh
21 tháng 8 2021 lúc 20:51

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