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\(P=\frac{a^3b^2c^2}{ab+a^2bc+abc}+\frac{ab^2c}{bc+b+abc}+\frac{abc^2}{ac+c+1}\)
\(=\frac{ }{ab\left(1+ac+c\right)}+\frac{ }{b\left(c+1+ac\right)}+\frac{ }{ac+c+1}\)
ta có: \(\frac{2013a^2bc}{ab+2013a+2013}\)= \(\frac{2013.ab.ac}{ab+ab.ac+abc}\)= \(\frac{2013.ab.ac}{ab.\left(ac+c+1\right)}\)= \(\frac{2013ac}{ac+c+1}\)
\(\frac{ab^2c}{bc+b+2013}\)= \(\frac{abc.b}{bc+b+abc}\)= \(\frac{2013b}{b\left(ac+c+1\right)}\)= \(\frac{2013}{ac+c+1}\)
\(\frac{abc^2}{ac+c+1}\)= \(\frac{abc.c}{ac+c+1}\)= \(\frac{2013c}{ac+c+1}\)
Cộng cả 3 phân thức cùng mẫu thức ta có phân thức cuối cùng là:
P=\(\frac{2013.\left(ac+c+1\right)}{ac+c+1}\)=2013
\(\frac{P}{abc}=\frac{P}{2013}=\frac{2013a}{ab+2013a+2013}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)
\(=\frac{2013ac}{abc+2013ac+2013c}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)
\(=\frac{2013ac}{2013\left(ac+c+1\right)}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}\)
\(=\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}=\frac{ac+c+1}{ac+c+1}=1\)
\(\Rightarrow P=2013\)
Thay abc=2013 vào P
P= \(\dfrac{abc.a^2bc}{ab+abc.a+abc}\)+\(\dfrac{ab^2c}{bc+b+abc}+\dfrac{abc^2}{ac+c+1}\)
P=\(\dfrac{a^3b^2c^2}{ab\left(1+ac+c\right)}+\dfrac{ab^2c}{b\left(c+1+ac\right)}+\dfrac{abc^2}{ac+c+1}\)
P=\(\dfrac{a^2bc^2}{ac+c+1}+\dfrac{abc}{c+ac+1}+\dfrac{abc^2}{ac+1+c}\)
P=\(\dfrac{a^2bc^2+abc+abc^2}{ac+c+1}\)
P=abc (*)
Thay abc=2013 vào (*)
P=2013
\(=\frac{2013ac}{abc+2013ac+2013c}+\frac{abc}{abc^2+abc+2013ac}+\frac{2013c}{2013ac+2013c+2013}\)
\(=\frac{2013ac}{2013+2013ac+2013c}+\frac{2013}{2013c+2013+2013ac}+\frac{2013c}{2013ac+2013c+2013}\)
\(=\frac{2013ac+2013c+2013}{2013ac+2013c+2013}=1\left(đpcm\right)\)