Cho a+b+c=0. Tinh Q=\((\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b})(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a})\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
KK
2 tháng 1 2019
Giải: Ta có :
\(\frac{a+b+c-2011d}{d}=\frac{b+c+d-2011a}{a}=\frac{c+d+a-2011b}{b}=\frac{d+a+b-2011c}{c}\)
=> \(\frac{a+b+c}{d}-2011=\frac{b+c+d}{a}-2011=\frac{c+d+a}{b}-2011=\frac{d+a+b}{c}-2011\)
=> \(\frac{a+b+c}{d}=\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{d+a+b}{c}\)
=> \(\frac{a+b+c}{d}+1=\frac{b+c+d}{a}+1=\frac{c+d+a}{b}+1=\frac{d+a+b}{c}+1\)
=> \(\frac{a+b+c+d}{d}=\frac{b+c+d+a}{a}=\frac{c+d+a+b}{b}=\frac{d+a+b+c}{c}\)
TH1: a + b + c + d = 0
=> a + b = -(c + d)
b + c = -(a + d)
khi đó, ta có : S = \(\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{a+d}+\frac{c+d}{-\left(c+d\right)}+\frac{d+a}{-\left(a+d\right)}\)
= \(-1+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
= -4
TH2 : a + b + c + d \(\ne\)0
=> a = b = c = d
khi đó, ta có : S = \(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{b+a}+\frac{d+a}{b+c}\)
= 1 + 1 + 1 + 1
= 4
QH
2
25 tháng 12 2016
quy đồng lên ta có bc/abc+ac/abc+ab/abc=0
bc+ac+ab/abc=0
suy ra bc+ac+ab=0
quy đồng M ta có (b+c)bc/abc+(c+a)ac/abc+(a+b)ab/abc
=(b^2c+bc^2+ac^2+a^2c+a^2b+ab^2)/abc
=(b^2c+ab^2+abc+bc^2+ac^2+abc+a^2c+a^2b+abc-3abc)/abc
=(b(bc+ab+ac)+c(bc+ac+ab)+a(ac+ab+bc)-3abc)/abc
=-3abc/abc=-3
NK
23 tháng 12 2015
\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\frac{1}{b-c}=0\Rightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(b-c\right)}=0\)(1)
\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\frac{1}{c-a}=0\Rightarrow\frac{a}{\left(b-c\right)\left(c-a\right)}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)\left(c-a\right)}=0\)(2)
\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\frac{1}{a-b}=0\Rightarrow\frac{a}{\left(b-c\right)\left(a-b\right)}+\frac{b}{\left(c-a\right)\left(a-b\right)}+\frac{c}{\left(a-b\right)^2}=0\)(3)
Cộng (1);(2);(3) ta có: \(M+\frac{b\left(a-b\right)+c\left(c-a\right)+a\left(a-b\right)+c\left(b-c\right)+a\left(c-a\right)+b\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\Rightarrow\)\(M+\frac{ab-b^2+c^2-ac+a^2-ab+bc-c^2+ac-a^2+b^2-bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(\Rightarrow M+0=0\Rightarrow M=0\)
tick cho mk nka ^^
9 tháng 11 2018
ta có (a+b-c/c)+2=(a-b+c/b)+2=(-a+b+c/a)+2
=>a+b-c+2c/c=a-b+c+2b/b=-a+b+c+2a/a
=>a+b+c/c=a+b+c/b=a+b+c/a (1)
Trường hợp 1
Nếu a+b+c=0 => a+b=-c
=> b+c=-a
=> a+c=-b
M= (-c)(-a)(-a)/abc = -1
Trường hợp 2
Từ (1) =>(a+b+c). 1/c =(a+b+c). 1/b =(a+b+c). 1/a
=>1/a=1/b=1/c
Từ (1) =>3(a+b+c)/a+b+c=3
hay (a+b/c)+1=(a+c/b)+1=(b+c/a)=2
N
9 tháng 11 2018
Nguyễn Trọng Tâm Đạt làm sai một TH nhé =)
trường hợp 2
\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\)
\(2+\frac{a+b-c}{c}=2+\frac{a-b+c}{b}=2+\frac{-a+b+c}{a}\)
\(\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}\)
\(\Rightarrow a=b=c\)
thay a=b=c vào M ta có
\(M=\frac{\left(b+b\right).\left(b+c\right).\left(c+a\right)}{a.b.c}=\frac{2a.2a.2a}{aaa}=\frac{8.a^3}{a^3}=8\)
Ta có: a + b + c = 0 => a + b = -c; b + c = -a; a + c = -b
a + b + c = 0 <=> a + b = -c
<=> (a + b)3 = (-c)3
<=> a3 + 3a2b + 3ab2 + b3 = -c3
<=> a3 + b3 + c3 = -3ab(a + b)
<=> a3 + b3 + c3 = 3abc (vì a + b = -c)
Khi đó: Q = \(\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)
Q = \(1+\frac{a\left(a-b\right)}{c\left(b-c\right)}+\frac{b\left(a-b\right)}{c\left(c-a\right)}+\frac{c\left(b-c\right)}{a\left(a-b\right)}+1+\frac{b\left(b-c\right)}{a\left(c-a\right)}+\frac{c\left(c-a\right)}{b\left(a-b\right)}+\frac{a\left(c-a\right)}{b\left(b-c\right)}+1\)
Q = \(3+\left(\frac{a\left(a-b\right)}{c\left(b-c\right)}+\frac{a\left(c-a\right)}{b\left(b-c\right)}\right)+\left(\frac{b\left(a-b\right)}{c\left(c-a\right)}+\frac{b\left(b-c\right)}{a\left(c-a\right)}\right)+\left(\frac{c\left(b-c\right)}{a\left(a-b\right)}+\frac{c\left(c-a\right)}{b\left(a-b\right)}\right)\)
Q = \(3+\frac{ab\left(a-b\right)+ac\left(c-a\right)}{bc\left(b-c\right)}+\frac{ab\left(a-b\right)+bc\left(b-c\right)}{ac\left(c-a\right)}+\frac{bc\left(b-c\right)+ca\left(c-a\right)}{ab\left(a-b\right)}\)
Q = \(3+\frac{a\left(ab-b^2+c^2-ac\right)}{bc\left(b-c\right)}+\frac{b\left(a^2-ab+bc-c^2\right)}{ac\left(c-a\right)}+\frac{c\left(b^2-bc+ac-a^2\right)}{ab\left(a-b\right)}\)
Q = \(3+\frac{a\left[a\left(b-c\right)-\left(b-c\right)\left(b+c\right)\right]}{bc\left(b-c\right)}+\frac{b\left[b\left(c-a\right)-\left(c-a\right)\left(c+a\right)\right]}{ac\left(c-a\right)}+\frac{c\left[c\left(a-b\right)-\left(a-b\right)\left(a+b\right)\right]}{ab\left(a-b\right)}\)
Q = \(3+\frac{a\left[a-\left(b+c\right)\right]}{bc}+\frac{b\left(b-\left(c+a\right)\right)}{ac}+\frac{c\left[c-\left(a+b\right)\right]}{ab}\)
Q = \(3+\frac{a\left(a+a\right)}{bc}+\frac{b\left(b+b\right)}{ac}+\frac{c\left(c+c\right)}{ab}\)
Q = \(3+\frac{2a^2}{bc}+\frac{2b^2}{ac}+\frac{2c^2}{ab}\)
Q = \(3+\frac{2a^3+2b^3+2c^3}{abc}\)
Q = \(3+\frac{2\left(a^3+b^3+c^3\right)}{abc}\)
Q = \(3+\frac{2.3abc}{abc}=3+6=9\)
Bài làm:
Đặt \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
\(\Leftrightarrow abc.M=ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(\Leftrightarrow abc.M=ab\left(a-b\right)+b^2c-bc^2+c^2a-ca^2\)
\(\Leftrightarrow abc.M=ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow abc.M=\left(a-b\right)\left(ab+c^2-ac-bc\right)\)
\(\Leftrightarrow abc.M=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
\(\Rightarrow M=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{abc}\)
Đặt \(N=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
\(\Rightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right).N=c\left(b-c\right)\left(c-a\right)+a\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)\)
Mà \(a+b+c=0\Rightarrow\hept{\begin{cases}a=-b-c\\b=-c-a\\c=-a-b\end{cases}}\)
Thay vào ta được:
\(N=\frac{c\left(b-c\right)\left(c-a\right)-\left(b+c\right)\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{c\left(c-a\right)\left(b-c-a+b\right)+b\left(a-b\right)\left(b-c-c+a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{c\left(c-a\right)\left(2b-c-a\right)+b\left(a-b\right)\left(a+b-2c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{c\left(c-a\right)\left(2b+b\right)+b\left(a-b\right)\left(-c-2c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{3bc\left(c-a\right)-3bc\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{3bc\left(b+c-2a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{9abc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
Mà \(Q=M.N=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{abc}.\frac{9abc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=9\)
Vậy Q = 9