Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Có abc=1 nên
1/(1+a+ab)=abc/(abc+a+ab)
=abc/[a(1+b+bc)]
=bc/(1+b+bc)
1/(1+c+ac)=abc/(abc+c.abc+ac)
=abc/[ca(1+b+bc)]=b/(1+b+bc)
=>1/(1+a+ab) + 1/(1+b+bc)+ 1/(1+c+ac)
=bc/(1+b+bc)+1/(1+b+bc)+b/(1+b+bc)
=(1+b+bc)/(1+b+bc)
=1
=>1/(1+a+ab) + 1/(1+b+bc)+ 1/(1+c+ac)=1
ràu xong
Ta có \(\frac{1}{ab+a+1}+\frac{b}{bc+b+1}+\frac{1}{abc+bc+b}\)mình chỉnh sửa đề 1 chút , chắc bạn viết sai
\(=\frac{1}{ab+a+1}+\frac{b}{bc+b+1}+\frac{1}{1+bc+b}\)(vì abc=1)
\(=\frac{1}{ab+a+1}+\frac{a.b}{a.\left(bc+b+1\right)}+\frac{a}{a.\left(1+bc+b\right)}\)
\(=\frac{1}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{a}{a+abc+ab}\)
\(=\frac{1}{ab+a+1}+\frac{ab}{1+ab+a}+\frac{a}{a+1+ab}\)
\(=\frac{1+ab+a}{ab+a+1}\)
\(=1\)
\(S=\frac{abc}{abc+a+ab}+\frac{1}{1+b+bc}+\frac{bc}{bc+bc^2+c^2ab}=\frac{bc}{bc+1+b}+\frac{1}{1+b+bc}+\frac{b}{b+bc+1}\)
\(=\frac{1+b+bc}{1+bc+b}=1\rightarrow S=1\)
\(S=\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ca}\)
\(\Rightarrow S=\frac{abc}{abc+a+ab}+\frac{1}{1+b+bc}+\frac{abc}{abc+c.abc+ca}\)
\(S=\frac{abc}{a.\left(bc+b+1\right)}+\frac{1}{1+b+bc}+\frac{abc}{ac.\left(bc+b+1\right)}\)
\(S=\frac{bc}{bc+b+1}+\frac{1}{1+b+bc}+\frac{b}{bc+b+1}\)
\(S=\frac{bc+b+1}{bc+b+1}\)
\(S=1\)
Điều kiện \(c\ge0\);\(a;b>0\)
Ta có: \(a>b\)
\(\Rightarrow ac\ge bc\)
\(\Rightarrow ac+ab\ge bc+ab\)
\(a.\left(b+c\right)\ge b.\left(c+a\right)\)
\(\Rightarrow\frac{a+c}{b+c}\ge\frac{a}{b}\)
Tham khảo nhé~
Ta có : \(\frac{a^2+b^2}{2}=ab\Rightarrow a^2+b^2=2ab\)
\(\Rightarrow a^2-ab+b^2=0\Rightarrow\left(a-b\right)^2=0\Rightarrow a=b\)
Tương tự : \(\frac{b^2+c^2}{2}=bc\Rightarrow b=c\)
\(\frac{a^2+c^2}{2}=ac\Rightarrow a=c\)
Áp dụng t/c bắc cầu ta dc : \(a=b=c\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a\times3=9a\)
=>a2+b2=2ab
=>a2-2ab+b2=0
=>(a-b)2=0=>a=b
tương tự=>b=c
=>a=b=c
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a.3=9a\)