A=((a-b)/c+(b-c)/a+(c-a)/b)(c/(a-b)+a/(b-c)+b/(c-a)) va a+b+c=0
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ND
25 tháng 3 2018
Với a + b + c = 0 ta có:
\(B=\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}\)
\(\Leftrightarrow B=\dfrac{ab}{\left(a+b\right)^2-2ab-c^2}+\dfrac{bc}{\left(b+c\right)^2-2bc-a^2}+\dfrac{ca}{\left(c+a\right)^2-2ca-b^2}\)
\(\Leftrightarrow B=\dfrac{ab}{\left(a+b+c\right)\left(a+b-c\right)-2ab}+\dfrac{bc}{\left(b+c-a\right)\left(b+c+a\right)-2bc}+\dfrac{ac}{\left(a+c+b\right)\left(c+a-b\right)-2ca}\)
\(\Leftrightarrow B=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ac}{-2ac}\)
\(\Leftrightarrow B=\dfrac{-1}{2}+\dfrac{-1}{2}+\dfrac{-1}{2}\)
\(\Leftrightarrow B=\dfrac{-3}{2}\)
TH
0
LT
0
LT
0
NN
1
XO
1 tháng 1 2021
Ta có \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
=> \(\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)
=> \(\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
Nếu a + b + c = 0
=> a + b = -c
b + c = -a
a + c = -b
Khi đó P = \(\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\frac{a+b}{a}.\frac{b+c}{b}.\frac{a+c}{c}=\frac{-c}{a}.\frac{-a}{b}.\frac{-b}{c}=\frac{-abc}{abc}=-1\)
Nếu a + b + c \(\ne\)0
=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)
=> a = b = c
Khi đó P \(\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2.2.2=8\)
Vậy khi a + b + c = 0 thì P = -1
khi a + b + c \(\ne\)0 thì P = 8
S
1
HP
27 tháng 4 2016
\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(\Rightarrow S=\left(\frac{a+b+c}{b+c}\right)+\left(\frac{a+b+c}{c+a}\right)+\left(\frac{a+b+c}{a+b}\right)-3\)
\(\Rightarrow S=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3=2016.\frac{1}{90}-3=\frac{97}{5}\)
Vậy....................
em khong biet
Đặt \(p=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
\(=>abc.P=\text{ab(a-b) + bc(b-c) + ca(c-a)}\)
\(=>abc.P=a^2b-ab^2+b^2c-bc^2+ca\left(c-a\right)\)
\(=>abc.P=b\left(a^2-c^2\right)-b^2\left(a-c\right)+ca\left(c-a\right)\)
\(=>abc.P=b\left(a-c\right)\left(a+c\right)-b^2\left(a-c\right)-ca\left(a-c\right)\)
\(=>abc.P=\left(a-c\right)\left(ab+bc-b^2-ca\right)\)
\(=>abc.P=\left(a-c\right)\left[a\left(b-c\right)-b\left(b-c\right)\right]\)
\(=>abc.P=\left(a-c\right)\left(b-c\right)\left(a-b\right)\)
\(=>P=\frac{\left(a-c\right)\left(b-c\right)\left(a-b\right)}{abc}\)
Đặt \(Q=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
\(=>\left(a-b\right)\left(b-c\right)\left(c-a\right).Q=\text{ c(b-c)(c-a) + a(a-b)(c-a) + b(a-b)(b-c) }\)
\(=\text{= c(b-c)(c-a) + (-b-c)(a-b)(c-a) + b(a-b)(b-c) }\)
\(\text{= c(b-c)(c-a) – c(a-b)(c-a) – b(a-b)(c-a) + b(a-b)(b-c) }\)
\(=\text{ c(c-a)(2b-a-c) + b(a-b)(a+b-2c) }\)
\(=\text{3bc(c-a) – 3bc(a-b) }\text{= 3bc(b+c-2a) }\text{= 3bc(-a-2a) = -9abc }\)
\(=>Q=\frac{9abc}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
Vậy A=P.Q=9