chứng minh rằng nếu a+b+c=0 thì :
\(A=\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)
K
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26 tháng 11 2019
Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)=\left(x,y,z\right)\)
Khi đó :
\(Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\)
Ta có :
\(x+y=\frac{a-b}{c}+\frac{b-c}{a}=\frac{a^2-ab+bc-c^2}{ac}=\frac{b\left(c-a\right)-\left(c-a\right)\left(c+a\right)}{ca}\)
\(=\frac{b\left(c-a\right)-\left(c-a\right)\left(-b\right)}{ac}=\frac{2b\left(c-a\right)}{ca}\) ( do \(a+b+c=0\))
\(\Rightarrow\frac{x+y}{z}=\frac{2b\left(c-a\right)}{ca}.\frac{b}{c-a}=\frac{2b^2}{ca}=\frac{2b^3}{abc}\)
Hoàn toàn tương tự
\(\frac{y+z}{x}=\frac{2c^3}{abc};\frac{x+z}{y}=\frac{2a^3}{abc}\)
Do đó :
\(Q=3+\frac{x+y}{z}+\frac{y+z}{x}+\frac{x+z}{y}=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=3\)
\(=3+\frac{2\left[\left(-c\right)^3-3ab\left(-c\right)^3+c^3\right]}{abc}=3+\frac{2.3abc}{abc}=3+6=9\)
Ta có đpcm
19 tháng 8 2017
\(VT=\frac{b-a+a-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-b+b-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-c+c-b}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{-1}{a-c}+\frac{1}{a-b}+\frac{-1}{b-a}+\frac{1}{b-c}+\frac{-1}{c-b}+\frac{1}{c-a}\)
\(=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}=VP\)
12 tháng 8 2019
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Leftrightarrow\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}\)
\(=\frac{b}{a-c}+\frac{c}{b-a}\)
\(=\frac{b^2-ab+ac-c^2}{\left(c-a\right)\left(a-b\right)}\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ab+ac-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\) ( 1 )
Tương tự,ta có:
\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-ba+ba-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\) ( 2 )
\(\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+cb-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\) ( 3 )
Cộng vế theo vế của ( 1 );( 2 );( 3 ) suy ra đpcm
LH
22 tháng 4 2019
*Đặt P = (a-b)/c + (b-c)/a + (c-a)/b, ta có:
P = (a-b)/c + (b-c)/a + (c-a)/b
=> abc.P = ab(a-b) + bc(b-c) + ca(c-a)
= ab(a-b) + bc(b-a + a-c) + ca(c-a)
= ab(a-b) - bc(a-b) - bc(c-a) + ca(c-a)
= b(a-b)(a-c) + c(c-a)(a-b)
= (a-b)(a-c)(b-c)
=> P = (a-b)(a-c)(b-c)/abc
*Đặt Q = c/(a-b) + a/(b-c) + b/(c-a), ta có:
Vì a+b+c = 0 => a+b = -c ; b+c = -a ; c+a = -b
Q = c/(a-b) + a/(b-c) + b/(c-a)
=> (a-b)(b-c)(c-a).Q = c(b-c)(c-a) + a(a-b)(c-a) + b(a-b)(b-c)
= c(b-c)(c-a) + (-b-c)(a-b)(c-a) + b(a-b)(b-c)
= c(b-c)(c-a) – c(a-b)(c-a) – b(a-b)(c-a) + b(a-b)(b-c)
= c(c-a)(2b-a-c) + b(a-b)(a+b-2c)
= 3bc(c-a) – 3bc(a-b)
= 3bc(b+c-2a)
= 3bc(-a-2a)
= -9abc
=> Q = -9abc/(a-b)(b-c)(c-a) = 9abc /(a-b)(b-c)(a-c)
Vậy P.Q = 9 (đpcm)
18 tháng 7 2016
Ta có ; \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}-\frac{1}{a-c}=\frac{1}{a-b}+\frac{1}{c-a}\)
\(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{\left(b-a\right)-\left(b-c\right)}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}-\frac{1}{b-a}=\frac{1}{b-c}+\frac{1}{a-b}\)
\(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{\left(c-b\right)-\left(c-a\right)}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}-\frac{1}{c-b}=\frac{1}{c-a}+\frac{1}{b-c}\)
Cộng các vế lại với nhau được điều phải chứng minh.
LM
18 tháng 7 2016
A , B , C khác nhau thì bạn làm sao có thể cho : A-C = B đc ?
12 tháng 9 2017
Gọi \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
Ta có : \(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)=+\frac{c}{a-b}\left(\frac{b^2-bc+ac-a^2}{ab}\right)\)
\(=1+\frac{c}{a-b}.\frac{\left(a-b\right)\left(c-a-b\right)}{ab}=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)
Tương tự : \(M.\frac{a}{b-c}=1+\frac{2a^3}{abc};M.\frac{b}{c-a}=+\frac{2b^3}{abc}\)
\(\Rightarrow A=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=9\)(vì \(a^3+b^3+c^3=3abc\))
30 tháng 8 2017
Ta có:
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}\)
\(=-\left(\frac{b-c}{\left(a-b\right)\left(c-a\right)}+\frac{c-a}{\left(b-c\right)\left(a-b\right)}+\frac{a-b}{\left(c-a\right)\left(b-c\right)}\right)\)
\(=2.\frac{-a^2-b^2-c^2+ab+bc+ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=2.\frac{\left(a-b\right)\left(b-c\right)+\left(b-c\right)\left(c-a\right)+\left(c-a\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\)
LC
1 tháng 5 2016
Đặt \(\frac{a-b}{c}=x,\frac{b-c}{a}=y,\frac{c-a}{b}=z\)
=>\(\frac{c}{a-b}=\frac{1}{x},\frac{a}{b-c}=\frac{1}{y},\frac{b}{c-a}=\frac{1}{z}\)
=>\(A=\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
=>\(A=x.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+y.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+z.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
=>\(A=1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1\)
=>\(A=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Ta thấy: \(\frac{y+z}{x}=\frac{\frac{b-c}{a}+\frac{c-a}{b}}{\frac{a-b}{c}}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right):\frac{a-b}{c}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right).\frac{c}{a-b}\)
\(=\left[\frac{\left(b-c\right).b}{a.b}+\frac{\left(c-a\right).a}{a.b}\right].\frac{c}{a-b}=\left(\frac{b^2-bc}{ab}+\frac{ac-a^2}{ab}\right).\frac{c}{a-b}\)
\(=\left(\frac{b^2-bc+ac-a^2}{ab}\right).\frac{c}{a-b}=\left[\frac{\left(ac-bc\right)-\left(a^2-b^2\right)}{ab}\right].\frac{c}{a-b}\)
\(=\left[\frac{c.\left(a-b\right)-\left(a+b\right).\left(a-b\right)}{ab}\right].\frac{c}{a-b}=\left[\frac{\left(c-a-b\right).\left(a-b\right)}{ab}\right].\frac{c}{a-b}\)
\(=\frac{c-a-b}{ab}.\left(a-b\right).\frac{c}{a-b}=\frac{c-a-b}{ab}.c=\left(c-a-b\right).\frac{c}{ab}=\left(2c-a-b-c\right).\frac{c}{ab}\)
Vì a+b+c=0=>2a-(a+b+c)=2c=>2c-a-b-c=2c
=>\(\frac{y+z}{x}=\left(2c-a-b-c\right).\frac{c}{ab}=2c.\frac{c}{ab}=\frac{2c^2}{ab}=\frac{2c^3}{abc}\)
=>\(\frac{y+z}{x}=\frac{2c^3}{abc}\)
Chứng minh tương tự, ta có:
\(\frac{z+x}{y}=\frac{2a^3}{abc},\frac{x+y}{z}=\frac{2b^3}{abc}\)
=>\(A=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}=3+\frac{2c^3}{abc}+\frac{2a^3}{abc}+\frac{2b^3}{abc}\)
=>\(A=3+\frac{2c^3+2a^3+2b^3}{abc}=3+\frac{2.\left(a^3+b^3+c^3\right)}{abc}\)
Lại có:
Áp dụng bất đẳng thức, ta có: \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3=>a^3+b^3=\left(a+b\right)^3-3a^2b-3ab^2\)
=>\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3abc-3a^2b-3ab^2\)
\(=\left[\left(a+b\right)^3+c^3\right]-3ab.\left(a+b+c\right)\)
\(=\left(a+b+c\right).\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab.\left(a+b+c\right)\)
\(=\left(a+b+c\right).\left[\left(a+b\right)^2-\left(a+b\right).c+c^2-3ab\right]\)
Vì a+b+c=0
=>\(\left(a+b+c\right).\left[\left(a+b\right)^2-\left(a+b\right).c+c^2-3ab\right]=0\)
=>\(a^3+b^3+c^3-3abc=0=>a^3+b^3+c^3=3abc\)
Thay vào A, ta có:
\(A=3+\frac{2.\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+2.3=3+6=9\)
=>ĐPCM
HT
2 tháng 5 2016
Từ chỗ lại có bạn làm hơi dài mình sẽ làm cách khác ngắn hơn
Xét \(a^3+b^3+c^3=\left(a+b\right)^3-3a^2b-3ab^2+c^3\)
\(=\text{[}\left(a+b\right)^3+c^3\text{]}-3ab\left(a+b\right)\) (I)
Mà \(\text{ }a+b+c=0\Rightarrow a+b=-c\) thay vào (I) , ta được
\(a^3+b^3+c^3=\text{[}\left(-c\right)^3+c^3\text{]}-3ab\left(-c\right)\)
\(=3abc\)
Sau đó thay vào rồi tính