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\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}\)
\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{a+c-b}{b}+2=\dfrac{b+c-a}{a}+2\)
\(\Rightarrow\dfrac{a+b-c}{c}+\dfrac{2c}{c}=\dfrac{a+c-b}{b}+\dfrac{2b}{b}+\dfrac{b+c-a}{a}+\dfrac{2a}{a}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+c+b}{b}=\dfrac{b+c+a}{a}\)
\(\Rightarrow a=b=c\) Thay \(M\) vào ta được:
\(M=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a.a.a}=\dfrac{2a.2a.2a}{a^3}=\dfrac{8a^3}{a^3}=8\)
Vậy .............
Chúc bạn học tốt!
Lời giải:
TH1: $a+b+c=0$
Khi đó: \(a+b=-c; b+c=-a; c+a=-b\)
\(\Rightarrow M=\frac{(-c)(-a)(-b)}{abc}=\frac{-abc}{abc}=-1\)
TH2: \(a+b+c\neq 0\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=\frac{a+b-c+a-b+c-a+b+c}{c+b+a}=\frac{a+b+c}{a+b+c}=1\)
\(\Rightarrow \left\{\begin{matrix} a+b-c=c\\ a-b+c=b\\ -a+b+c=a\end{matrix}\right.\Rightarrow a+b=2c; a+c=2b; b+c=2a\)
\(\Rightarrow M=\frac{2c.2a.2b}{abc}=\frac{8abc}{abc}=8\)
Ta có:
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}\)
\(=\dfrac{a+b-c+a-b+c-a+b+c}{a+b+c}\)
\(=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+b-c}{c}=1\\\dfrac{a-b+c}{b}=1\\\dfrac{-a+b+c}{a}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a-b+c=b\\-a+b+c=a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)(1)
Thay (1) vào M ta được
\(M=\dfrac{2c.2a.2b}{abc}=\dfrac{8abc}{abc}=8\)
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}\)
\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{a-b+c}{b}+2=\dfrac{-a+b+c}{a}+2\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{b}=\dfrac{a+b+c}{a}\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
\(\circledast\) Với \(a+b+c=0\) thì \(\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
\(m=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{-abc}{abc}=-1\)
\(\circledast\) Với \(a=b=c\) thì \(m=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a.a.a}=\dfrac{8a^3}{a^3}=8\)
b)\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}\) và \(\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
\(\Rightarrow1+\dfrac{a+b}{c}=1+\dfrac{b+c}{a}\)và \(1+\dfrac{b+c}{a}=1 +\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{c}{c}+\dfrac{a+b}{c}=\dfrac{a}{a}+\dfrac{b+c}{a}\)và \(\dfrac{a}{a}+\dfrac{b+c}{a}=\dfrac{b}{b}+\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}\)và \(\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}-\dfrac{a+b+c}{a}=0\) \(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{c}-\dfrac{1}{a}\right)=0\)
và \(\dfrac{a+b+c}{a}-\dfrac{a+b+c}{b}=0\)
\(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=0\)
+) Vì a,b,c đôi một khác 0
\(\Rightarrow a+b+c=0\)
\(\rightarrow a+b=\left(-c\right)\)
\(\rightarrow a+c=\left(-b\right)\)
\(\rightarrow b+c=\left(-a\right)\)
+) Ta có:
\(M=\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{a}\right)\cdot\left(\dfrac{c+a}{c}\right)\)
\(=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}\)
\(=\left(-1\right)\)
TH1:a+b+c=0
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
\(\Rightarrow H=\dfrac{b+a}{b}.\dfrac{c+b}{c}.\dfrac{a+c}{a}=\dfrac{\left(-c\right)\left(-b\right)\left(-a\right)}{b.c.a}=-1\)
TH2:\(a+b+c\ne0\)
Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
\(\Rightarrow H=\dfrac{b+a}{b}.\dfrac{c+b}{c}.\dfrac{a+c}{a}=\dfrac{\left(2c\right)\left(2b\right)\left(2a\right)}{b.c.a}=8\)
Vậy H=-1 hoặc H=8
c)
Ta có \(a< b< c< d< m< n\)
\(\Rightarrow\left\{{}\begin{matrix}a< b\\c< d\\m< n\end{matrix}\right.\)
\(\Rightarrow a+c+m\le b+d+n\)
\(\dfrac{a+c+m}{a+b+c+d+m+n}< \dfrac{1}{2}\)
\(\Leftrightarrow2a+2c+2m< a+b+c+d+m+n\)
\(\Leftrightarrow a+c+m< b+d+n\) ( thỏa mãn đề bài )
\(\Rightarrow\) đpcm
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 \(a=b=c\Leftrightarrow B=\dfrac{8abc}{abc}=8\)
sao biết trước đề vậy
cái này là đề cương bạn nha, nhỡ cho cái câu này là chết mình đó