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Ta có:
a+b-c/c = b+c-a/a = c+a-b/b
=>a+b-c/c + 2 = b+c-a/a +2 = c+a-b/b +2
=>a+b-c/c + 2c/c =b+c-a/a +2a/a = c+a-b/b +2/b
=>a+b+c/c = a+b+c/a =a+b+c/b
* Nếu a+b+c=0 thì a= 0-b-c= -(b+c)
b= 0-a-c= -(a+c)
c= 0-b-a= -(b+a)
Thay a= -(b+c) ; b=-(a+c);c=-(b+a) vào B ta được
B=(1+b/a)(1+a/c)(1+c/b)=(a/a + b/a )(c/c +a/c)(b/b+c/b)=(a+b)/a * (a+c)/c * (c+b)/b
=(-c)/a * (-b)/c * (-a)/b =-1
* Nếu a+b+c\(\ne\)0 thì a=b=c
Khi đó
B=(1+b/a)(1+a/c)(1+c/b)=(1+1)(1+1)(1+1)=2*2*2=8
Vậy B=-1 hoặc B=8
nhớ k nha bạn
\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Do a+b+c=0=>b+c=-a ; a+c=-b ; a+b=-c
=>M=\(\frac{\left(-a\right)\left(-b\right)\left(-c\right)}{abc}=-1\)
a+b +c = 0 => a + b = -c ; a+ c = -b ; b+ c = - a thay vào ta có :
\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{a+c}{a}=-\frac{c}{b}\cdot-\frac{a}{c}\cdot-\frac{b}{a}=-\frac{abc}{abc}=-1\)
Nhơ đúng nha
\(\left(a+b\right)^2\ge4ab\Rightarrow\frac{a^2+b^2}{ab\left(a+b\right)}\ge\frac{4ab}{ab\left(a+b\right)}\)bài1
a) ta có \(\left(a-b\right)^2\ge0\) với mọi a,b\(\in\)N*
=> \(a^2-2ab+b^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow\frac{a^2}{ab}+\frac{b^2}{ab}\ge2\Rightarrow\frac{a}{b}+\frac{b}{a}\ge2\)
b) tương tự ta có \(a^2+b^2\ge2ab\)
\(\left(a+b\right)^2\ge4ab\Rightarrow\frac{\left(a+b\right)^2}{ab\left(a+b\right)}\ge\frac{4ab}{ab\left(a+b\right)}\)(do a,b\(\in\)N*)
\(\Rightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\Rightarrow\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)
bài 2 chịu
\(\text{Vì }\left[a,b\right],\left[b,c\right],\left[c,a\right]\text{ là BCNN}\)
\(\Rightarrow\left[a,b\right]=a.b;\left[b,c\right]=b.c;\left[c,a\right]=c.a\)
\(\Rightarrow\frac{1}{\left[a+b\right]}+\frac{1}{\left[b+c\right]}+\frac{1}{\left[c+a\right]}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(\text{Giả sử }a< b< c\)
\(\Rightarrow a\le2;b\le3;c\le5\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\le\frac{1}{2.3}+\frac{1}{3.5}+\frac{1}{5.2}=\frac{1}{3}\)
\(\text{hay }\frac{1}{\left[a+b\right]}+\frac{1}{\left[b+c\right]}+\frac{1}{c+a}\le\frac{1}{3}\left(đpcm\right)\)
Bài 1: D
Bài 2:
Ta có: \(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}\pm1=\frac{c}{d}\pm1\)
\(\Rightarrow\frac{a\pm b}{b}=\frac{c\pm d}{d}\)(đpcm)
\(\frac{b+c-a}{a}=\frac{a+c-b}{b}=\frac{a+b-c}{c}=\frac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{b+c-a}{a}=1\\\frac{a+c-b}{b}=1\\\frac{a+b-c}{c}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\a+b=2c\end{matrix}\right.\)
\(\Rightarrow Q=\frac{\left(a+b\right)}{b}.\frac{\left(b+c\right)}{c}.\frac{\left(a+c\right)}{c}=\frac{2c.2a.2b}{abc}=8\)