Question

Cho 3 số a,b,c khác 0 thỏa mãn a+b+c=0.

Tính \(P=\frac{a}{c}\cdot\frac{a^2-b^2-c^2}{b^2-c^2-a^2}\cdot\frac{c^2+a^2-b^2}{c^2-a^2-b^2}\)

HELP Tớ với ạ.Thanks trước:v

Answer 1

a+b+c=0=> a²-b²-c²=2bc,b²-c²-a²=2ac,c²+a²-b²=-2ac,c²-a²-b²=2ab

=>\(P=\frac{a}{c}.\frac{2bc}{2ac}.\frac{-2ac}{2ab}=-1\)

Answer 2

a+b+c=0 <=> a+b=-c; b+c=-a;c+a=-b

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

\(=\frac{b\left[c-\left(a+b\right)\right]}{c\left[a-\left(b+c\right)\right]}=\frac{b\left[c-\left(-c\right)\right]}{c\left[a-\left(-a\right)\right]}=\frac{b.2c}{c.2a}=\frac{b}{a}\)

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

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

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