bn đang nằm mơ hay sao jậy

ừ :v t đang mơ đấy..nên đừng phá để t mơ tiếp

We have : 

\(A=\frac{-2a}{2ab+2a+1}-\frac{b}{bc+b+1}+\frac{c}{-2ac-c-1}\)

\(=\frac{-2a}{2ab+2a+2abc}-\frac{b}{bc+b+1}+\frac{bc}{-2abc-bc-b}\)(\(abc=\frac{1}{2}\))

\(=\frac{-2a}{2a\left(bc+b+1\right)}-\frac{b}{bc+b+1}+\frac{bc}{-\frac{2.1}{2}-bc-b}\)(\(abc=\frac{1}{2}\))

\(=\frac{-1}{bc+b+1}-\frac{b}{bc+b+1}-\frac{bc}{bc+b+1}\)

\(=\frac{-bc-b-1}{bc+b+1}=-1\)

The value of A is - 1 because \(abc=\frac{1}{2}\)

các bạn giúp mình nha càng nhanh càng tốt

Chờ mình nhé 

1+2=2+1

=> em thích anh = em yêu anh !

Mk hứa sẽ tk lại

anh ứ thích em

theo cô -si ta có 

1/2*[ab/(c+a) + ab/(c+b)] >= 1/2*2căn(ab*ab/[(c+a)(c+b)] = ab/căn[(c+a)(c+b)]

cô si bạn ơi

Theo mình thì đề thiếu

a+b+c = 1 bạn

\(b^4+c^4\ge\)\(b^3c+bc^3\) (bn tu cm nhé)

\(\Rightarrow\frac{a}{b^4+c^4+a}\le\frac{a}{bc\left(b^2+c^2\right)+a}=\frac{abc}{b^2c^2\left(b^2+c^2\right)+abc}=\frac{1}{b^2c^2\left(b^2+c^2\right)+1}=\)

\(\frac{a^2b^2c^2}{b^2c^2\left(b^2+c^2\right)+a^2b^2c^2}=\frac{a^2b^2c^2}{b^2c^2\left(a^2+b^2+c^2\right)}=\frac{a^2}{a^2+b^2+c^2}\)

ttu \(T\le\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\) dau = xay ra khi va chi khi a=b=c=1

\(\Sigma\frac{a}{c^4+b^4+a}\le\Sigma\frac{a^2}{abc\left(c^2+b^2\right)+a^2}=1\)

\(\sqrt{c+ab}\) =\(\sqrt{c\left(a+b+c\right)+ab}=\sqrt{c^2+ac+cb+ab}=\sqrt{\left(c+a\right)\left(c+b\right)}\)

\(\frac{ab}{\sqrt{c+ab}}\le\frac{ab}{2}\left(\frac{1}{c+a}+\frac{1}{b+c}\right)\)

ttu \(\frac{bc}{\sqrt{a+bc}}\le\frac{1}{2}\left(\frac{1}{a+b}+\frac{1}{a+c}\right);\frac{ac}{\sqrt{b+ca}}\le\frac{1}{2}\left(\frac{1}{b+a}+\frac{1}{a+c}\right)\)

\(\Rightarrow P\le\frac{bc+ac}{2\left(a+b\right)}+\frac{ac+ab}{2\left(a+b\right)}+\frac{bc+ab}{2\left(c+b\right)}=\frac{1}{2}\left(a+b+c\right)=\frac{1}{2}\)

dau = xay ra khi a=b=c=1/3

trả lời 

=1/2

chúc bn

học tốt