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

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

de ma abc=3 dua ti thoi kho day

Nâng cao và phát triển toán 8 tập 1 bài 153

\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\frac{1}{b-c}=0\Rightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(b-c\right)}=0\)(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 ^^

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

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

\(M=\frac{\left(b+b\right).\left(b+c\right).\left(c+a\right)}{a.b.c}=\frac{2a.2a.2a}{aaa}=\frac{8.a^3}{a^3}=8\)

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