Cho 3 số thực a, b, c khác nhau. Rút gọn \(\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{a+c}{c-a}.\frac{b+c}{b-c}+\frac{a+c}{c-a}.\frac{b+a}{a-b}\)
K
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AH
17 tháng 9 2018
Có : a/b+c = b/a+c = c/a+b => b+c/a = a+c/b = a+b/c
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
b+c/a = a+c/b = a+b/c = b+c+a+c+a+b/a+b+c = 2
=> P = 2+ 2 + 2 =6
k mk nha
PN
11 tháng 2 2016
Vì \(a+b+c=0\) \(\Rightarrow\) \(c=-a-b\)
Gọi \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\) , ta có:
\(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)=1+\frac{c}{a-b}.\frac{\left(b^2-bc+ac-a^2\right)}{ab}=1+\frac{c}{a-b}.\frac{\left(a-b\right)\left(c-a-b\right)}{ab}=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)
Tương tự, \(M.\frac{a}{b-c}=1+\frac{2a^3}{abc};\) \(M.\frac{b}{c-a}=1+\frac{2b^3}{abc}\)
Mặt khác, ta cũng có: từ \(a+b+c=0\), suy ra \(a^3+b^3+c^3=3abc\)
Vậy, \(B=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+6=9\) (vì \(a,b,c\ne0\) nên \(abc\ne0\) )
KN
8 tháng 9 2020
\(\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b}\)\(=\frac{\left(a+b\right)\left(b+c\right)\left(c-a\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)+\left(c+a\right)\left(a+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(b^2+ab+bc+ca\right)\left(c-a\right)+\left(c^2+ab+bc+ca\right)\left(a-b\right)+\left(a^2+ab+bc+ca\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(b^2c+bc^2+c^2a-ab^2-a^2b-ca^2\right)+\left(c^2a+a^2b+ca^2-bc^2-ab^2-b^2c\right)+\left(a^2b+ab^2+b^2c-ca^2-bc^2-c^2a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(a^2b-ca^2\right)+\left(b^2c-bc^2\right)-\left(ab^2-c^2a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b+c\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(b-c\right)\left(a^2+bc-ab-ac\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)
AH
Akai Haruma
Giáo viên
2 tháng 6
Lời giải:
Ta có:
$\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{a+b}{a-b}.\frac{c+a}{c-a}+\frac{b+c}{b-c}.\frac{c+a}{c-a}$
$=\frac{(a+b)(b+c)(c-a)+(a+b)(c+a)(b-c)+(b+c)(c+a)(a-b)}{(a-b)(b-c)(c-a)}$
$=\frac{[b^2+(ab+bc+ac)](c-a)+[a^2+(ab+bc+ac)](b-c)+[c^2+(ab+bc+ac)](a-b)}{(a-b)(b-c)(c-a)}$
$=\frac{b^2(c-a)+a^2(b-c)+c^2(a-b)+(ab+bc+ac)(c-a+b-c+a-b)}{(a-b)(b-c)(c-a)}$
$=\frac{b^2(c-a)+a^2(b-c)+c^2(a-b)}{(a-b)(b-c)(c-a)}$
$=\frac{(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)}{-[(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)]}=-1$
Ta có đpcm.
19 tháng 10 2015
\(\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{a+c}{c-a}.\frac{b+c}{b-c}+\frac{a+c}{c-a}.\frac{b+a}{a-b}=\frac{a+b}{a-b}.\left(\frac{b+c}{b-c}+\frac{a+c}{c-a}\right)+\frac{a+c}{c-a}.\frac{b+c}{b-c}=\frac{a+b}{a-b}.\frac{2c\left(b-a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{a+c}{c-a}.\frac{b+c}{b-c}\)
\(=\frac{2c\left(a+b\right)}{\left(b-c\right)\left(a-c\right)}+\frac{\left(a+c\right)\left(b+c\right)}{\left(c-a\right)\left(b-c\right)}=\frac{2ac+2bc-ab-ac-bc-c^2}{\left(b-c\right)\left(a-c\right)}=\frac{\left(b-c\right)\left(c-a\right)}{\left(b-c\right)\left(a-c\right)}=-1\)
tick nha công mk đánh máy
S
7 tháng 12 2018
Ta có:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\Leftrightarrow\)
\(\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=\frac{b+c+a+c+a+b}{a+b+c}=2\)
\(\Rightarrow P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=3.2=6\)
N
7 tháng 12 2018
bài này có 2 trường hợp nhé =))
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\Rightarrow1+\frac{a}{b+c}=1+\frac{b}{a+c}=1+\frac{c}{a+b}\)
\(\Rightarrow\frac{a+b+c}{b+c}=\frac{a+b+c}{a+c}=\frac{a+b+c}{a+b}\)
\(TH1:a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}b+c=-a\\a+c=-b\\a+b=-c\end{cases}\Rightarrow P=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-3}\)
\(TH2:a+b+c\ne0\)
\(\Rightarrow\hept{\begin{cases}b+c=a+c\Rightarrow a=b\\a+c=a+b\Rightarrow c=b\\a+b=b+c\Rightarrow a=c\end{cases}\Rightarrow a=b=c}\)
\(\Rightarrow P=\frac{a+a}{a}+\frac{b+b}{b}+\frac{c+c}{c}=2.3=6\)
Vậy P=-3 hay P=6