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

Ta có: \(B=\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)

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

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

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

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

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

\(=\frac{b}{-2b}+\frac{c}{-2c}+\frac{a}{-2a}=\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)=-\frac{3}{2}\)

cái này tự nghĩ thui,k có cách giải nhá bạn

x=0 nhá :D

2x^2- 50x=0

2x( x- 25)= 0

2x= 0 hoặc  x-25 = 0

x= 0 hoặc x= 25

ta có :3(x-2)-x=0

=>3x-6-x=0

=>3x-x=0+6

=>2x=6

=>x=3

k cho  minh nhé

bạn ơi giải thích cho mình ở chỗ =>2x=6

3x^2 -6xy +3y^2

=3x^2-3xy-3xy+3y^2

=3x(x-y)-3y(x-y)

=(x-y)(3x-3y)

=(x-y)3(x-y)

=3(x-y)^2

x=căn bậc 2 -27/8

làm giúp tui vs mn oi

\(xy^2-5xz=x\left(y^2-5z\right)\)