Ta có a+b+c=0

<=> a+b=-c <=>a²+b²-c²=-2ab

   b+c=-a <=> b²+c²-a²=-2bc

  c+a=-b <=> c²+a²-b²=-2ca

Thay vào biểu thức ta có

\(B=\frac{ab}{-2ab}-\frac{bc}{2bc}-\frac{ca}{2ca}=\frac{-3}{2}\)

1. a + b + c = 0 \(\Rightarrow\)a + b = -c \(\Rightarrow\)( a + b )² = ( -c )² \(\Rightarrow\)a² + b² - c² = -2ab

Tương tự : b² + c² - a2 = -2bc ; c² + a² - b² = -2ac

Ta có : \(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}\)

\(=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}=\frac{-1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(=\frac{-1}{2}\left(\frac{a+b+c}{abc}\right)=0\)

2. tương tự

3,4 . có ở dưới, câu hỏi của Quyết Tâm chiến thắng

1 bai thoi cung dc

\(B=\Sigma\frac{ab}{a^2+b^2-c^2}\)

\(B=\frac{ab}{a^2+\left(b-c\right)\left(b+c\right)}+\frac{bc}{b^2+\left(c-a\right)\left(c+a\right)}+\frac{ac}{c^2+\left(a-b\right)\left(a+b\right)}\)

\(B=\frac{ab}{a^2-a\left(b-c\right)}+\frac{bc}{b^2-b\left(c-a\right)}+\frac{ac}{c^2-c\left(a-b\right)}\)

\(B=\frac{ab}{a\left(a-b+c\right)}+\frac{bc}{b\left(b-c+a\right)}+\frac{ac}{c\left(c-a+b\right)}\)

\(B=\frac{b}{a+b+c-2b}+\frac{c}{a+b+c-2c}+\frac{a}{a+b+c-2a}\)

\(B=\frac{-b}{2b}+\frac{-c}{2c}+\frac{-a}{2a}\)

\(B=\frac{-1}{2}+\frac{-1}{2}+\frac{-1}{2}\)

\(B=\frac{-3}{2}\)

\(a+b=c\Rightarrow\left(a+b\right)^2=c^2\Rightarrow a^2+2ab+b^2=c^2\Rightarrow a^2+b^2-c^2=-2ab\)

Tượng tự: \(b^2+c^2-a^2=2bc,c^2+a^2-b^2=2ac\)

Khi đó: \(B=\frac{-1}{2ab}+\frac{1}{2bc}+\frac{1}{2ac}=\frac{-c+a+b}{2abc}=0\)

Chúc bạn học tốt.

a+b+c=0 =>a+b=-c =>(a+b)²=(-c)²=>a²+b²+2ab=c²=>a²+b²-c²=-2ab

tương tự , b²+c²-a²=-2bc ; c²+a²-b²=-2ca 

Thay vào P=1/-2ab + 1/-2bc + 1/-2ca = 0

đề ko có d nha bạn : 

=> sửa lại : cho a+b+c =0 . CM: ...........

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a , Ta có : \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)

=> M = \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

\(=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1\)

\(a+b+c=0\) nha

a có bạn làm rồi mình làm ý b thôi nak

\(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)

\(N=\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}+\frac{1}{a^2+b^2-c^2}\)

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

\(\frac{1}{\left(b+c\right)^2-a^2-2bc}+\frac{1}{\left(a+c\right)^2-b^2-2ac}+\frac{1}{\left(a+b\right)^2-c^2-2ab}\)

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

\(=\frac{a+b+c}{-2abc}=0\)