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Bài 2:
a+b+c+d=0
nên b+c=-(a+d)
\(a^3+b^3+c^3+d^3\)
\(=\left(a+d\right)^3-3ad\left(a+d\right)+\left(b+c\right)^3-3bc\left(b+c\right)\)
\(=-\left(b+c\right)^3+3ad\left(b+c\right)+\left(b+c\right)^3-3bc\left(b+c\right)\)
\(=3ad\left(b+c\right)-3bc\left(b+c\right)\)
\(=\left(b+c\right)\left(3ad-3bc\right)\)
\(=3\left(b+c\right)\left(ad-bc\right)\)
b: \(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ac+c^2\right)+\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)=0\)
=>(a-c)^2+(a-b)^2+(b-c)^2=0
=>a=b=c
c: \(\Leftrightarrow a^2+b^2+c^2-ab-ac-bc=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ac+c^2\right)+\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)=0\)
=>(a-b)^2+(a-c)^2+(b-c)^2=0
=>a=b=c
p giúp mk câu b đk k? Mk đọc mãi cũng không hiểu lắm câu a thì làm đk r
Với a + b + c = 0 ta có:
\(B=\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}\)
\(\Leftrightarrow B=\dfrac{ab}{\left(a+b\right)^2-2ab-c^2}+\dfrac{bc}{\left(b+c\right)^2-2bc-a^2}+\dfrac{ca}{\left(c+a\right)^2-2ca-b^2}\)
\(\Leftrightarrow B=\dfrac{ab}{\left(a+b+c\right)\left(a+b-c\right)-2ab}+\dfrac{bc}{\left(b+c-a\right)\left(b+c+a\right)-2bc}+\dfrac{ac}{\left(a+c+b\right)\left(c+a-b\right)-2ca}\)
\(\Leftrightarrow B=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ac}{-2ac}\)
\(\Leftrightarrow B=\dfrac{-1}{2}+\dfrac{-1}{2}+\dfrac{-1}{2}\)
\(\Leftrightarrow B=\dfrac{-3}{2}\)