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 ^ ^.
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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
a/ \(\frac{4bc-a^2}{bc+2a^2}.\frac{4ab-c^2}{ab+2c^2}.\frac{4ac-b^2}{ac+2b^2}\)
\(=\frac{4bc-\left(b+c\right)^2}{bc+2\left(b+c\right)^2}.\frac{4\left(-b-c\right)b-c^2}{\left(-b-c\right)b+2c^2}.\frac{4\left(-b-c\right)c-b^2}{\left(-b-c\right)c+2b^2}\)
\(=\frac{-\left(b-c\right)^2}{\left(c+2b\right)\left(b+2c\right)}.\frac{-\left(c+2b\right)^2}{-\left(b-c\right)\left(b+2c\right)}.\frac{-\left(b+2c\right)^2}{\left(b-c\right)\left(c+2b\right)}=1\)
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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.CMR:
\(\frac{ab}{a^2+3b^2+4ab+5bc+3ac}+\frac{bc}{2a^2+b^2+3c^2+3ab+4bc+5ac}+\frac{ac}{3a^2+2b^2+c^2+5ab+3bc+4ac}\le\frac{1}{6}\)
Cho A=\(\frac{4bc-a^2}{bc+2a^2}\),B=\(\frac{4ca-b^2}{ac+2b^2}\),C=\(\frac{4ab-c^2}{ab+2c^2}\)
Chứng minh : 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 \(a+b+c=0\) , 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 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)
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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 \(\ge\)1 . Chứng minh
\(\frac{1}{2a-1}+\frac{1}{2b-1}+\frac{1}{2c-1}+\frac{4ab}{1+ab}+\frac{4bc}{1+bc}+\frac{4ac}{1+ac}\ge9\)
Không có điều kiện gì nữa à? Chẳng hạn như a + b +c = 3;..