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Áp dụng tính chất dãy tỉ số bằng nhau ta có : \(\frac{2b+c-a}{a}=\frac{2c-b+a}{b}=\frac{2a+b-c}{c}=\frac{\left(2b+c-a\right)+\left(2c-b+a\right)+\left(2a+b-c\right)}{a+b+c}\)\(=\frac{2a+2c+2a}{a+b+c}=2\)
vậy : \(\frac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\Rightarrow2b+c-3a=0\Rightarrow3a-2c=c\Rightarrow3a-c=2b\)
\(\frac{2c-b+a}{b}=2\Rightarrow2c-b+a=2b\Rightarrow2c+a-3b=0\Rightarrow3b-2c=a\Rightarrow3b-a=2c\)
\(\frac{2a+b-c}{c}=2\Rightarrow2a+b-c=2c\Rightarrow2a+b-3c=0\Rightarrow3c-2a=b\Rightarrow3c-b=2a\)
Vậy \(P=\frac{\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)}{\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)}=\frac{c.a.b}{2b.2c.2a}=\frac{1}{8}\)
a) Áp dụng TC của dãy tỉ số bằng nhau ta có :
\(\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\frac{a-2c+3e}{b-2d+3f}\left(đpcm\right)\)
a, Ta có
\(\frac{c}{d}=\frac{2c}{2d};\frac{e}{f}=\frac{3e}{3f}\)
\(\Rightarrow\frac{a}{b}=\frac{2c}{2d}=\frac{3e}{3f}=\frac{a-2c+3e}{b-2d+3f}\)( t/c dãy tỉ số bằng nhau )
b, \(\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\frac{a+c+e}{b+d+f}\)( t/c dãy tỉ số bằng nhau )
\(\Rightarrow\frac{a}{b}=\frac{a+c+e}{b+d+f}\)
\(\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{a+c+e}{b+d+f}\right)^3\)
Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
\(\Rightarrow2ab=c.\left(a+b\right)\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b.\left(a-c\right)=a.\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
Áp dụng t/c dãy tỉ số = nhau
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c+b+c-a+c+a-b}{c+a+b}=\frac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\frac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\)
Tương tự \(b+c=2a;;c+a=2b\)
\(\Rightarrow D=\left(\frac{a+b}{a}\right)\left(\frac{b+c}{b}\right)\left(\frac{c+a}{c}\right)=\left(\frac{2c}{a}\right)\left(\frac{2a}{b}\right)\left(\frac{2b}{c}\right)=8\)
Theo đề ta có :
\(\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{a+c-b}{b}+2\)
\(\Rightarrow\frac{a+b-c+2c}{c}=\frac{b+c-a+2a}{a}=\frac{a+c-b+2b}{b}\)
\(\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
\(\Rightarrow\left(a+b+c\right).\frac{1}{c}=\left(a+b+c\right)\frac{1}{c}=\left(a+b+c\right)\frac{1}{b}\)
(vì \(a\ne b\ne c\ne0\) \(\frac{\Rightarrow1}{a}\ne\frac{1}{b}\ne\frac{1}{c}\ne0\) \(\Rightarrow a+b+c=0\))
* a+b+c=0
=>a+b=-c ; b+c=-a ; a+c =-b
\(D=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\)
\(=\frac{a+b}{a}.\frac{b+c}{b}.\frac{a+c}{c}=\frac{-c.-a.-b}{a.b.c}=\frac{-1.\left(a.b.c\right)}{a.b.c}=-1\)
Vậy : D=-1
\(A=\left(\frac{a+b}{b}\right).\left(\frac{b+c}{c}\right).\left(\frac{a+c}{a}\right)\)
Vì \(a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}}\)
\(\Rightarrow A=\frac{-c}{b}.\left(\frac{-a}{c}\right).\left(\frac{-b}{a}\right)\)
\(\Rightarrow A=-1\)
ta có: \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=\frac{1}{c}\times2=\frac{1}{a}+\frac{1}{b}\)
\(=\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
\(=\frac{2}{c}=\frac{b+a}{ab}\)
= \(c\left(b+a\right)=ab\times2\)
= cb +ca = ab+ab
= ab - cb = ac-ab
\(=b\left(a-c\right)=a\left(c-b\right)\)
= \(\frac{a}{b}=\frac{a-c}{c-b}\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}\)
\(\frac{1}{c}=\frac{a+b}{2ab}\)
\(2ab=c\left(a+b\right)\)
\(ab+ab=ac+bc\)
\(ab-bc=ac-ab\)
\(b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
Ta có
\(\frac{2b+c-a}{a}=\frac{2c-b+a}{b}=\frac{2a+b-c}{c}=\frac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=\frac{2a+2b+2c}{a+b+c}\)
\(=2\)
Từ \(\frac{2b+c-a}{a}=2\Rightarrow2a=2b+c-a\Rightarrow3a-2b=c\)và \(3a-c=2b\)
Tương tự có \(3b-2c=a;3b-a=2c\) và \(3c-2a=b;3c-b=2a\)
Thay vào biểu thức M ta có
\(M=\frac{a\cdot b\cdot c}{2\cdot b\cdot2\cdot a\cdot2\cdot c}=\frac{1}{8}\)