Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(P=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-c\right)\left(b-a\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
Đặt \(x=\frac{b}{c-a},y=\frac{c}{a-b},z=\frac{a}{b-c}\) , suy ra : \(P=-xy-yz-xz\)
Lại có : \(\left(x-1\right)\left(y-1\right)\left(z-1\right)=\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
\(\Rightarrow xy+yz+xz=-1\Rightarrow P=1\)
\(Q=\frac{\left[\left(x+\frac{1}{x}\right)^2\right]^3-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\)
\(=3x+\frac{3}{x}=3\left(x+\frac{1}{x}\right)\)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{c-b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{a-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{b-a}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{c-b+b-a+a-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
MTC: \(abc\left(a-b\right)\left(b-c\right)\left(a-c\right)\)nên
\(A=\frac{bc\left(b-c\right)\left(a-2\right)\left(a-1014\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{ac\left(a-c\right)\left(b-2\right)\left(b-1004\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)\left(c-2\right)\left(c-1004\right)}{abc\left(a-c\right)\left(a-b\right)\left(b-c\right)}\)
\(=\frac{2008b^2c+2008a^2c+2008a^2b-2008bc^2-2008a^2c-2008ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left[\left(c^2a-c^2b\right)+\left(a^2b-a^2c\right)+\left(b^2a-b^2c\right)\right]}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008}{abc}\) ( với \(abc\ne0\))
Theo ra ta có : A= 1/ (a−b)(a−c) + 1/ (b−a)(b−c)+1/ (c−a)(c−b)
ĐKXĐ : a khác b khác c
\(\Leftrightarrow\)A= 1/ (a−b)(a−c) - 1/(a−b) (b−c)+1/ (a−c)(b−c)
\(\Leftrightarrow\)A= ( b-c)-(a-c)+(a-b) /(a−b)(a−c)(b−c)
\(\Leftrightarrow\)A= 0
chúc bn học tốt
