Bài 1:

có: a + b - 2c = 0

⇒ a = 2c − b thay vào a² + b² + ab - 3c² = 0 ta có:

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Vậy 

giúp mik vớiii

Đề sai: a=b=-c

a^3+b^3+c^3=3abc

=> (a+b)^3-3ab(a+b)-3abc+c^3=0

=>[(a+b)^3+c^3]-3ab(a+b+c)=0

=>(a+b+c).[(a+b)^2+(a+b).c+c^2]-3ab(a+b+c)=0

=>(a+b+c).(a^2+b^2+c^2-ab+ac+bc)=0

TH1: a+b+c=0

TH2: a^2+b^2+c^2-ab+ac+bc=0

=> 2a^2+2b^2+2c^2-2ab+2ac+2bc=0

=> (a-b)^2+(a+c)^2+(b+c)^2=0

=>a=b=-c

Vậy: a+b+c=0 hoặc a=b=-c thì a^3+b^3+c^3=3abc

\(\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{a+d}\ge\frac{a-d}{a+b}\)

\(\Leftrightarrow\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{a+d}-\frac{a-d}{a+b}\ge0\)

\(\Leftrightarrow\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{a+d}+\frac{d-a}{a+b}\ge0\)

\(\Leftrightarrow\left(\frac{a-b}{b+c}+1\right)+\left(\frac{b-c}{c+d}+1\right)+\left(\frac{c-d}{a+d}+1\right)+\left(\frac{d-a}{a+b}+1\right)\ge4\)

\(\Leftrightarrow\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{a+d}+\frac{d+b}{a+b}\ge4\)

\(\Leftrightarrow\left(a+c\right)\left(\frac{1}{b+c}+\frac{1}{a+d}\right)+\left(b+d\right)\left(\frac{1}{c+d}+\frac{1}{a+b}\right)\ge4\)(1)

Áp dụng BĐT AM-GM ta có:

\(\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{a+d}+\frac{d+b}{a+b}\ge\)\(\left(a+c\right)\frac{2}{\sqrt{\left(b+c\right)\left(a+d\right)}}+\left(b+d\right)\frac{2}{\sqrt{\left(c+d\right)\left(a+b\right)}}\ge\frac{4\left(a+c\right)}{a+b+c+d}+\frac{4\left(b+d\right)}{a+b+c+d}=\frac{4\left(a+b+c+d\right)}{a+b+c+d}=4 \left(2\right)\)Từ (1) và (2) \(\Rightarrow\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{a+d}\ge\frac{a-d}{a+b}\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{1}{b+c}=\frac{1}{a+d}\\\frac{1}{c+d}=\frac{1}{a+b}\end{cases}}\Leftrightarrow\hept{\begin{cases}b+c=a+d\\c+d=a+b\end{cases}}\Leftrightarrow a=b=c=d\)

vì sao                                          

 (a+c)(2/căn bậc 2 của(b+c)(a+d))+(b+d)(2/căn bậc 2 của (c+d)(a+b))
>=(4(a+c)/a+b+c+d) +4(b+d)/a+b+c+d

(căn bậc 2 máy mink ko viết đc)

ta có (a+b)(b+c)(c+a)+abc

=(a+b)(bc+ab+c^2+ca)+abc

=(a+b)(bc+ab+ca+c^2)+abc

=(a+b).c^2+abc

=ac^2+bc^2+abc

=c(ac+bc+ab)=c.0=0 (đpcm)

Ta có : a+b/b+c = c+d/d+a 
=> (a+b)/(c+d)= (b+c)/(d+a) 
=> (a+b)/(c+d)+1=(b+c)/(d+a)+1 
hay: (a+b+c+d)/(c+d)=(b+c+d+a)/(d+a) 
- Nếu a+b+c+d khác 0 thì : c+d=d+a => c=a (1)
- Nếu a+b+c+d = 0 (2) 

Từ (1) và (2) 

\(\RightarrowĐPCM\)

Ta có : \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)

\(\Rightarrow\)\(\frac{a+b}{c+d}=\frac{b+c}{d+a}\)

\(\Rightarrow\)\(\frac{a+b}{c+d}+1=\frac{b+c}{d+a}+1\)

Hoặc \(\frac{a+b+c+d}{c+d}=\frac{b+c+d+a}{d+a}\)

Nếu a + b + c + d khác 0 thì c + d = d + a => c = a ( hoặc a = c )

Nếu a + b + c + d = 0 ( đpcm )

Ta có:\(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)

\(\implies\)\(\frac{a+b}{c+d}=\frac{b+c}{d+a}\)

\(\implies\) \(\frac{a+b}{c+d}+1=\frac{b+c}{d+a}+1\)

\(\implies\) \(\frac{a+b+c+d}{c+d}=\frac{a+b+c+d}{d+a}\)

\(\implies\) \(\frac{a+b+c+d}{c+d}-\frac{a+b+c+d}{d+a}=0\)

\(\implies\) \(\left(a+b+c+d\right)\left(\frac{1}{c+d}-\frac{1}{d+a}\right)=0\)

\(\implies\)\(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}-\frac{1}{d+a}=0\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}=\frac{1}{d+a}\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\c+d=d+a\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\c=a\end{cases}}\)