Question

[Hiếu Phạm]

cho a,b,c thuộc R thỏa a,b,c khác nhau đôi một
và a khác b+c,b khác c+a,c khác a+b
CMR:đặt x=b+c-a ; y=c+a-b; z=a+b-c thì
\(\frac{4}{xyz}=\frac{1}{x\left(a-b\right)\left(a-c\right)}+\frac{1}{y\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)

Answer 1

[๖ۣۜҨž ♫ ℱ¡ɗℰ£¡ɑ๖²⁴ʱ]

Ta có: \(\frac{1}{x\left(a-b\right)\left(a-c\right)}+\frac{1}{y\left(b-a\right)\left(b-c\right)}+\frac{1}{z\left(c-a\right)\left(c-b\right)}\)

\(=\frac{1}{x\left(a-b\right)\left(a-c\right)}-\frac{1}{y\left(a-b\right)\left(b-c\right)}+\frac{1}{z\left(a-c\right)\left(b-c\right)}\)

\(=\frac{yz\left(b-c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{xz\left(a-c\right)}{yxz\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{xy\left(a-b\right)}{zxy\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)

\(=\frac{yz\left(b-c\right)-xz\left(a-c\right)+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)\(=\frac{yz\left(b-c\right)-xz\left[\left(b-c\right)+\left(a-b\right)\right]+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{yz\left(b-c\right)-xz\left(b-c\right)-xz\left(a-b\right)+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(b-c\right)z\left(y-x\right)-\left(a-b\right)x\left(z-y\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(b-c\right)z\left(c+a-b-b-c+a\right)-\left(a-b\right)x\left(a+b-c-c-a+b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(b-c\right)z\left(2a-2b\right)-\left(a-b\right)x\left(2b-2c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(b-c\right)2z\left(a-b\right)-\left(a-b\right)2x\left(b-c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(a-b\right)\left(b-c\right)\left(2z-2x\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

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

\(=\frac{2\left(2a-2c\right)}{xyz\left(a-c\right)}=\frac{2.2\left(a-c\right)}{xyz\left(a-c\right)}=\frac{4}{xyz}\Rightarrowđpcm\)