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Ta có:
\(a+b+c-abc=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+c\left(a+b\right)\right)-abc\)
\(=\left(a+b\right)ab+\left(a+b\right)^2c+abc+c^2\left(a+b\right)-abc\)
\(=\left(a+b\right)\left(ab+c^2+c\left(a+b\right)\right)\)
\(=\left(a+b\right)\left(ab+ac+c^2+bc\right)\)
\(=\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Đồng thời:
\(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự:
\(b^2+1=\left(a+b\right)\left(b+c\right)\)
\(c^2+1=\left(a+c\right)\left(b+c\right)\)
Từ đó:
\(P=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}=1\)
Đặt \(\dfrac{a}{2017}=\dfrac{b}{2018}=\dfrac{c}{2019}=k\Rightarrow a=2017k;b=2018k;c=2019k\)
M = 4(2017k - 2018k)(2018k - 2019k) - (2019k - 2017k)2
= 4(-k)(-k) - (2k)2
= 4k2 - 4k2
= 0
4. (3/4-81)(3^2/5-81)(3^3/6-81)....(3^6/9-81).....(3^2011/2014-81)
mà 3^6/9-81=0 => (3/4-81)(3^2/5-81)....(3^2011/2014-81)=0