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Ta có: \(M=\frac{2010a}{ab+2010a+2010}+\frac{b}{bc+b+2010}+\frac{c}{ac+c+1}\)
Thế: abc = 2010 ta được:
\(M=\frac{a^2bc}{ab+a^2bc+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)
\(\Leftrightarrow\frac{a^2bc}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)
\(\Leftrightarrow\frac{a^2bc}{ab\left(1+ac+c\right)}+\frac{ab}{ab\left(c+1+ac\right)}+\frac{abc}{ab\left(ac+c+1\right)}\)
\(\Leftrightarrow\frac{a^2bc+ab+abc}{ab\left(1+ac+c\right)}=\frac{ab\left(ac+1+c\right)}{ab\left(1+ac+c\right)}=1\)
Vậy \(M=1\)
Áp dụng giả thiết từ đề bài :
\(M=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(\Leftrightarrow M=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}\)
\(\Leftrightarrow M=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+b}\)
\(\Leftrightarrow M=\frac{1+b+bc}{b+1+bc}=1\)
Vậy M = 1
M=a/ab+a+1 +b/bc+b+1 +c/ca+c+1
=ac/abc+ca+c +abc/abc^2+abc+ac +c/ca+c+1
=ac/1+ca+c +1/c+1+ac +c/ca+c+1
=ac+1+c/1+ca+c
=1
Tham khảo: Câu hỏi của Nguyễn Thị Nhàn - Toán lớp 8 - Học toán với OnlineMath
Học tốt=)
tth : mẫu nó khác bạn nhé
- mẫu nó là 2bc 2ac 2ab
mẫu mk ko có nhân 2
\(P=\frac{a^3b^2c^2}{ab+a^2bc+abc}+\frac{ab^2c}{bc+b+abc}+\frac{abc^2}{ac+c+1}\)
\(=\frac{ }{ab\left(1+ac+c\right)}+\frac{ }{b\left(c+1+ac\right)}+\frac{ }{ac+c+1}\)
Dễ dàng chứng minh được:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)với \(x,y>0\)(1)
Dấu bằng xảy ra \(\Leftrightarrow x=y>0\)
Ta có:
\(\frac{a}{bc\left(a+1\right)}=\frac{a}{abc+bc}=\frac{a}{ab+bc+ca+bc}=\frac{a}{\left(ab+bc\right)+\left(bc+ca\right)}\)
Áp dụng (1), ta được:
\(\frac{1}{ab+bc}+\frac{1}{bc+ca}\ge\frac{4}{\left(ab+bc\right)+\left(bc+ca\right)}\)
\(\Leftrightarrow\frac{1}{4\left(ab+bc\right)}+\frac{1}{4\left(bc+ca\right)}\ge\frac{1}{ab+bc+bc+ca}\)
\(\Leftrightarrow\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ca}\right)\ge\frac{a}{ab+bc+bc+ca}\)
\(\Leftrightarrow\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ca}\right)\ge\frac{a}{bc\left(a+1\right)}\left(2\right)\)
Dấu bằng xảy ra \(\Leftrightarrow b=c>0\)
Chúng minh tương tự, ta được:
\(\frac{b}{4}\left(\frac{1}{ab+ca}+\frac{1}{bc+ca}\right)\ge\frac{b}{ca\left(b+1\right)}\left(3\right)\)
Dấu bằng xảu ra \(\Leftrightarrow a=c>0\).
\(\frac{c}{4}\left(\frac{1}{ac+ab}+\frac{1}{ab+bc}\right)\ge\frac{c}{ab\left(c+1\right)}\left(4\right)\)
Từ (2), (3) và (4), ta được:
\(\frac{a}{bc\left(a+1\right)}+\frac{b}{ca\left(b+1\right)}+\frac{c}{ab\left(c+1\right)}\le\)\(\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ac}\right)+\frac{b}{4}\left(\frac{1}{ac+bc}+\frac{1}{ac+ab}\right)\)\(+\frac{c}{4}\left(\frac{1}{ab+bc}+\frac{1}{ab+ac}\right)\)
\(\Leftrightarrow P\le\frac{1}{4}.\left(\frac{a}{ab+bc}+\frac{c}{ab+bc}\right)+\frac{1}{4}\left(\frac{a}{bc+ac}+\frac{b}{bc+ac}\right)\)\(+\frac{1}{4}\left(\frac{b}{ab+ac}+\frac{c}{ab+ac}\right)\)
\(\Leftrightarrow P\le\frac{a+c}{4\left(ab+bc\right)}+\frac{a+b}{4\left(bc+ac\right)}+\frac{b+c}{4\left(ab+ac\right)}\)
\(\Leftrightarrow P\le\frac{a+c}{4b\left(a+c\right)}+\frac{a+b}{4c\left(a+b\right)}+\frac{b+c}{4a\left(b+c\right)}\)
\(\Leftrightarrow P\le\frac{1}{4b}+\frac{1}{4c}+\frac{1}{4a}\)
\(\Leftrightarrow P\le\frac{1}{4}\left(\frac{ab+bc+ca}{abc}\right)\)
\(\Leftrightarrow P\le\frac{1}{4}.\frac{abc}{abc}=\frac{1}{4}.1=\frac{1}{4}\)( vì \(ab+bc+ca=abc\))
Dấu bằng xảy ra
\(\Leftrightarrow\hept{\begin{cases}a=b=c>0\\ab+bc+ca=abc\end{cases}}\Leftrightarrow a=b=c=3\)
Vậy \(minP=\frac{1}{4}\Leftrightarrow a=b=c=3\)
\(M=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{a.b}{a.\left(bc+b+1\right)}+\frac{c}{ac+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{c}{ac+c+1}\)
Vì abc=1
\(=>M=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{c}{ac+c+abc}=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{c}{c\left(a+ab+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}=\frac{ab+a+1}{ab+a+1}=1\)
Vậy M=1
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{c}{ac+c+abc}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{1+ab+a}+\frac{1}{a+1+ab}=\frac{ab+a+1}{ab+a+1}=1\)