Cho A=\(\frac{4bc-a^2}{bc+2a^2}\),B=\(\frac{4ca-b^2}{ac+2b^2}\),C=\(\frac{4ab-c^2}{ab+2c^2}\).CMR nếu a+b+c=0 thì A.B.C=1

Cho \(a+b+c=0\), đặt \(A=\frac{4bc-a^2}{bc+2a^2}\);\(B=\frac{4ca-b^2}{ca+2b^2}\);\(C=\frac{4ab-c^2}{ab+2c^2}\).Chứng minh rằng: \(A.B.C=1\)

cho a+b+c=0 .

Chứng minh a, \(\frac{4bc-a^2}{bc+2a^2}.\frac{4ab-c^2}{ab+2c^2}.\frac{4ac-b^2}{ac+2b^2}\)=1

b, \(\frac{4bc-a^2}{bc+2a^2}+\frac{4ab-c^2}{ab+2c^2}+\frac{4ac-b^2}{ac+2b^2}\)=3

Cho \(a+b+c=0\) , Đặt \(A=\frac{4bc-a^2}{bc+2a^2},B=\frac{4ca-b^2}{ca+2b^2},C=\frac{4ab-c^2}{ab+2c^2}\)

Chứng minh rằng : \(A.B.C=1\)

Giúp mk vs thanks mn 

Cho phân thức A=\(\frac{4bc-a^2}{bc+2a^2}\);B=\(\frac{4ca-b^2}{ca+2b^2}\);C=\(\frac{4ab-c^2}{ab+2c^2}\)

Cmr nếu a+b+c=0 a khác b khác c thì A.B.C=1

Bạn nào giải nhanh đúng mình tick cho nha ^ ^.

\(A=\dfrac{4bc-a^2}{bc+2a^2}\\ B=\dfrac{4ca-b^2}{ca+2b^2}\\ C=\dfrac{4ab-c^2}{ab+2c^2}\\ \)

CMR: nếu a+b+c=0 thì A.B.C=1

cho A=(4bc-a²)/(bc+2a²); B=(4ca-b²)/(ca+2a²); C=(4ab-c²)/(ab+2c²)

Chứng minh rằng nếu a+b+c=0 thì a.b.c=1

Cho a, b, c > 0 thỏa mãn a.b.c=1. Chứng minh rằng: \(\frac{bc}{a^2b+a^2c}+\frac{ac}{b^2a+b^2c}+\frac{ab}{c^2a+c^2b}\ge\frac{3}{2}\)

Cho a + b + c = 0. Hãy tính M = (4bc - a^2) / (bc + 2a^2) . (4ca - b^2) / (ca + 2b^2) . (4ab - c^2) / (ab + 2c^2)