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5 tháng 11 2018

Theo T/C dãy tỉ số bằng nhau 

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\frac{a+b}{c}=2\Rightarrow a+b=2c\)

Tương tự ta có 

\(b+c=2a\)

\(c+a=2b\)

Xét \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{c+a}{a}\right)\)

\(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2a\cdot2b\cdot2c}{abc}=8\)

AH
Akai Haruma
Giáo viên
3 tháng 5 2018

Lời giải:
Ta có:

\(\frac{b-c}{(a-b)(a-c)}+\frac{c-a}{(b-a)(b-c)}+\frac{a-b}{(c-a)(c-b)}=2013\)

\(\Leftrightarrow \frac{-(b-c)^2}{(a-b)(b-c)(c-a)}+\frac{-(c-a)^2}{(a-b)(b-c)(c-a)}+\frac{-(a-b)^2}{(a-b)(b-c)(c-a)}=2013\)

\(\Leftrightarrow \frac{-[(a-b)^2+(b-c)^2+(c-a)^2]}{(a-b)(b-c)(c-a)}=2013\)

\(\Rightarrow \frac{2(a^2+b^2+c^2-ab-bc-ac)}{(a-b)(b-c)(c-a)}=-2013(*)\)

Lại có:

\(P=\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)

\(=\frac{(b-c)(c-a)+(c-a)(a-b)+(a-b)(b-c)}{(a-b)(b-c)(c-a)}\)

\(=\frac{bc-ba-c^2+ca+ca-bc-a^2+ab+ab-ac-b^2+bc}{(a-b)(b-c)(c-a)}\)

\(=\frac{ab+bc+ac-(a^2+b^2+c^2)}{(a-b)(b-c)(c-a)}=-\frac{1}{2}.\frac{2(a^2+b^2+c^2-ab-bc-ac)}{(a-b)(b-c)(c-a)}\)

\(=\frac{-1}{2}.-2013=\frac{2013}{2}\) (theo $(*)$)

5 tháng 11 2018

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}+\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\Rightarrow a=b=c}\)

\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)+\left(1+\dfrac{c}{a}\right)=2.2.2=8\)

5 tháng 11 2018

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)

<=>\(\dfrac{a+b}{c}-1=\dfrac{b+c}{a}-1=\dfrac{c+a}{b}-1\)

=\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)

=\(\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}\)=\(\dfrac{a+b+c}{a+b+c}\)=1

=>\(\left\{{}\begin{matrix}a+b-c=c\\b+c-a=a\\c+a-b=b\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)

P=\(\dfrac{b+a}{b}\).\(\dfrac{c+b}{c}\).\(\dfrac{a+c}{a}\)=\(\dfrac{2c}{b}\).\(\dfrac{2a}{c}.\dfrac{2b}{a}\)=8

27 tháng 12 2018

áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=1\)

\(\Rightarrow\dfrac{a+b-c}{c}=1\Leftrightarrow a+b-c=c\Leftrightarrow a+b=2c\)

\(\Rightarrow\dfrac{b+c-a}{a}=1\Leftrightarrow b+c-a=a\Leftrightarrow b+c=2a\)

ta có

\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\dfrac{a+b}{a}\times\dfrac{c+a}{c}\times\dfrac{b+c}{b}=\dfrac{2c}{a}\times\dfrac{2b}{c}\times\dfrac{2a}{b}=8\)

\(\Rightarrow M=8\)

4 tháng 12 2021

Áp dụng t/c dtsbn ta có:

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)

\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\\ \dfrac{c+a-b}{b}=1\Rightarrow c+a-b=b\Rightarrow c+a=2b\)

\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\\ =\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}\\ =\dfrac{2c.2b.2a}{abc}\\ =\dfrac{8abc}{abc}\\ =8\)

5 tháng 12 2021

Cảm ơn bn.

19 tháng 3 2017

ta có:\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)do đó:

+)\(\dfrac{a+b-c}{c}=1\)

=> a+b-c=c

=> a+b=2c

=> a+b+c =3c (1)

cm tương tự ta đươc (bạn cần làm chi tiết hơn)

+)3a=a+b+c (2)

+) 3b=a+b+c(3)

từ (1);(2) và (3)=> 3a=3b=3c

=> a=b=c

=>B=\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{c}{c}\right)\left(1+\dfrac{b}{b}\right)=2.2.2=8\)

vậy ...

8 tháng 8 2018

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)

ADTCDTSBN ta có:

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+c+a}{c+a+b}\)

\(=\dfrac{2a+2b+2c}{a+b+c}=2\)

Do đó: \(\dfrac{a+b}{c}=2=>a+b=2c\)

\(\dfrac{b+c}{a}=2=>b+c=2a\)

\(\dfrac{c+a}{b}=2=>c+a=2b\)

Có: P=\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

P=\(\dfrac{b+a}{b}.\dfrac{c+b}{c}=\dfrac{a+c}{a}\)

P=\(\dfrac{b+a}{c}.\dfrac{c+b}{a}=\dfrac{a+c}{b}\)

P=\(\dfrac{2c}{c}.\dfrac{2a}{a}=\dfrac{2b}{b}\)

P=2.2.2=8

28 tháng 1 2018

Lời giải \(B=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)

Ta có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)

\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{b+c-a}{a}+2=\dfrac{c+a-b}{b}+2\)

\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\)

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

Khi \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\Leftrightarrow B=\dfrac{-abc}{abc}=-1\)

Khi \(a=b=c\Leftrightarrow B=\dfrac{8abc}{abc}=8\)

21 tháng 10 2017

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a+b-c}{3c}=\dfrac{b+c-a}{3a}=\dfrac{c+a-b}{3b}=\dfrac{a+b-c+b+c-a+c+a-b}{3a+3b+3c}=\dfrac{a+b+c+\left(a-a\right)+\left(b-b\right)+\left(c-c\right)}{3a+3b+3c}=\dfrac{a+b+c}{3\left(a+b+c\right)}=\dfrac{1}{3}\)

Khi đó:

\(\left\{{}\begin{matrix}\dfrac{a+b-c}{3c}=\dfrac{1}{3}\\\dfrac{b+c-a}{3a}=\dfrac{1}{3}\\\dfrac{c+a-b}{3b}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b-3c=3c\\3b+3c-3a=3a\\3c+3a-3b=3b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6c\\3b+3c=6a\\3c+3a=6b\end{matrix}\right.\)Thay vào \(P\)

\(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\left(\dfrac{a+b}{a}\right)\left(\dfrac{c+a}{c}\right)\left(\dfrac{b+c}{b}\right)\)

\(27P=3\left(\dfrac{a+b}{a}\right).3\left(\dfrac{c+a}{c}\right).3\left(\dfrac{b+c}{b}\right)\)

\(27P=\left(\dfrac{3a+3b}{a}\right)\left(\dfrac{3c+3a}{c}\right)\left(\dfrac{3b+3c}{b}\right)\)

\(27P=\)\(\dfrac{6c}{a}.\dfrac{6b}{c}.\dfrac{6a}{b}=\dfrac{216abc}{abc}=216\Leftrightarrow P=\dfrac{216}{27}=8\)

22 tháng 10 2017

thank