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Xét \(a+b+c=0\) thì \(\hept{\begin{cases}a+2b=c\\b+2c=a\\c+2a=b\end{cases}}\)\(\Rightarrow P=\frac{\left(2a+b\right)\left(2b+c\right)\left(2c+a\right)}{abc}=1\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(a+b+c=\frac{a+2b-c}{c}=\frac{b+2c-a}{a}+\frac{c+2a-b}{b}=\frac{a+2b-c+b+2c-a+c+2a-b}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=2\)
\(\Rightarrow\hept{\begin{cases}a+2b=3c\\b+2c=3a\\c+2a=3b\end{cases}}\)\(\Rightarrow P=\frac{3a.3b.3c}{abc}=27\)
Có a+2b-c/c=b+2c-a/a=c+2a-b/b
suy ra a+2b-c/c=b+2c-a/a=c+2a-b/b=a+2b-c+b+2c-a+c+2a-b/a+b+c=2a+2b+2c/a+b+c=2
suy ra a+2b-c=2c suy ra a+2b=3c
b+2c-a=2a suy ra b+2c=3a
c+2a-b=2b suy ra c+2a=3b
Có P=(2+a/b)(2+b/c)(2+c/a)=(2b+a/b)(2c+b/c)(2a+c/a)=(3c/b)(3a/c)(3b/a)=27abc/abc=27
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-a\right)\left(b-c\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=2013\)
<=>\(\frac{\left(b-a\right)-\left(c-a\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(c-b\right)-\left(a-b\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(a-c\right)-\left(b-c\right)}{\left(c-a\right)\left(c-b\right)}=2013\)
<=>\(\frac{1}{c-a}+\frac{1}{a-b}+\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{b-c}+\frac{1}{c-a}=2013\)
<=>\(2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2013\)
<=>\(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}=\frac{2013}{2}=1006,5\)
Đặt \(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2022}=k\Rightarrow\hept{\begin{cases}a=2020k\\b=2021k\\c=2022k\end{cases}}\)
Khi đó M = 4(a - b)(b - c) - (c - a)2
= 4(2020k - 2021k)(2021k - 2022k) - (2022k - 2020k)2
= 4(-k)(-k) - (2k)2
= 4k2 - 4k2 = 0
Vậy M = 0
Đặt \(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2022}=k\)( \(k\ne0\))
\(\Rightarrow a=2020k\); \(b=2021k\); \(c=2022k\)
Thay a, b, c vào biểu thức M ta có:
\(M=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
\(=4\left(2020k-2021k\right)\left(2021k-2022k\right)-\left(2022k-2020k\right)^2\)
\(=4.\left(-k\right).\left(-k\right)-\left(2k\right)^2=4k^2-4k^2=0\)
Vậy \(M=0\)
Ta có : \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}=\frac{a-c}{\left(a-b\right)\left(a-c\right)}-\frac{a-b}{\left(a-b\right)\left(a-c\right)}\)
\(=\frac{1}{a-b}-\frac{1}{a-c}=\frac{1}{a-b}+\frac{1}{c-a}\left(1\right)\)
Tương tự ta cũng chứng minh được :
\(\hept{\begin{cases}\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}+\frac{1}{a-b}\left(2\right)\\\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}+\frac{1}{b-c}\left(3\right)\end{cases}}\)
Từ (1), (2), (3), suy ra : \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\)
\(=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\left(đpcm\right)\)
Ta có :
\(VT=\frac{1}{2}\left[\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}\right]\)
\(=\frac{1}{2}\left[\frac{\left(b-c\right)^2}{\left(a-b\right)\left(a-c\right)}+\frac{\left(a-c\right)^2}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}+\frac{\left(a-b\right)^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)
\(=\frac{1}{2}\left[\frac{\left(b-c\right)^2+\left(a-c\right)^2+\left(a-b\right)^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)
\(=\frac{1}{2}\left[\frac{b^2-2bc+c^2+a^2-2ac+c^2+a^2-2ab+b^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)
\(=\frac{1}{2}\left[\frac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\right]\)
\(=\frac{a^2+b^2+c^2-ab-bc-ac}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)(1)
Lại có :
\(VP=\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)
\(=\frac{\left(b-c\right)\left(a-c\right)+\left(a-b\right)\left(a-c\right)-\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{ab-bc-ac+c^2+a^2-ac-ab+bc-ab+ac+b^2-bc}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2+b^2+c^2-ab-ac-bc}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)(2)
Từ (1) và (2) \(\RightarrowĐPCM\)
Đặt \(\hept{\begin{cases}a-b=x\\b-c=y\\c-a=z\end{cases}}\)
\(A=\frac{2}{x}+\frac{2}{y}+\frac{2}{z}+\frac{x^2y^2z^2}{xyz}\)
\(A=\frac{\left(2y+2x\right).z+2xy}{xyz}+\frac{x^2+y^2+x^2}{xyz}\)
\(A=\frac{2yz+2xz+2xy}{xyz}+\frac{x^2+y^2+z^2}{xyz}\)
\(A=\frac{2yz+2xz+2xy+x^2+y^2+z^2}{xyz}=\frac{\left(x+y+z\right)^2}{xyz}\)
Có đúng k nhỉ k chắc