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\(s=\frac{bc}{bc\left(1+a+ab\right)}+\frac{1}{1+b+bc}+\frac{b}{b\left(1+c+ac\right)}=>\) \(s=\frac{bc}{bc+abc+ab^2c}+\frac{1}{1+b+bc}+\frac{b}{b+bc+abc}\)=>
\(s=\frac{bc}{1+b+bc}+\frac{1}{1+b+bc}+\frac{b}{1+b+bc}\)=>
\(s=\frac{1+b+bc}{1+b+bc}=1\)Vậy với a.b.c=1 S=1
Từ đề bài:A=\(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}=\dfrac{abc}{a^2}+\dfrac{abc}{b^2}+\dfrac{abc}{c^2}=abc\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=8\cdot\dfrac{3}{4}=6\)
\(A=\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\)
\(=\dfrac{abc}{a^2}+\dfrac{abc}{b^2}+\dfrac{abc}{c^2}\\ =abc\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\\ =8\cdot\dfrac{3}{4}\\ =6\)
Vì: \(0\le a\le b\le c\le1\) nên:
\(\left(a-1\right).\left(b-1\right)\ge0\Leftrightarrow ab-a-b+1\ge0\Leftrightarrow ab+1\ge a+b\)
\(\Leftrightarrow\dfrac{1}{ab+1}\le\dfrac{1}{a+b}\Leftrightarrow\dfrac{c}{ab+1}\le\dfrac{c}{a+b}\) (1)
\(\left(a-1\right).\left(c-1\right)\ge0\Leftrightarrow ac-a-c+1\ge0\Leftrightarrow ac+1\ge a+c\)
\(\Leftrightarrow\dfrac{1}{ac+1}\le\dfrac{1}{a+c}\Leftrightarrow\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\) (2)
\(\left(b-1\right).\left(c-1\right)\ge0\Leftrightarrow bc-b-c+1\ge0\Leftrightarrow bc+1\ge b+c\)
\(\Leftrightarrow\dfrac{1}{bc+1}\le\dfrac{1}{b+c}\Leftrightarrow\dfrac{a}{bc+1}\le\dfrac{a}{b+c}\) (3)
Cộng vế với vế của (1)(2) và (3) ta được:
\(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
\(\Leftrightarrow\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{2a+2b+2c}{a+b+c}\)
\(\Leftrightarrow\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{2.\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ac+1}\le2\left(đpcm\right)\)
Ta có :
\(A=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(A=\dfrac{a}{ab+a+1}+\dfrac{ab}{abc+ab+a}+\dfrac{abc}{aabc+abc+ab}\)
\(A=\dfrac{a}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{1}{a+1+ab}\)
\(A=\dfrac{a+ab+1}{ab+a+1}\)
\(\Rightarrow A=1\left(đpcm\right)\)
Lời giải:
Ta có:
\(S=\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ac}\)
\(S=\frac{c}{1.c+ac+abc}+\frac{ac}{ac+b.ac+bc.ac}+\frac{1}{1+c+ac}\)
Thay \(abc=1\) ta có:
\(S=\frac{c}{c+ac+1}+\frac{ac}{ac+1+c}+\frac{1}{1+c+ac}\)
\(S=\frac{a+ac+1}{c+ac+1}=1\)
\(a,A=\dfrac{-3\left(2n-3\right)-8}{2n-3}=-3-\dfrac{8}{2n-3}\in Z\\ \Leftrightarrow2n-3\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Leftrightarrow n\in\left\{1;2\right\}\left(n\in Z\right)\)
\(b,\dfrac{ab}{a+2b}=\dfrac{3}{2}\Leftrightarrow\dfrac{a+2b}{ab}=\dfrac{2}{3}\Leftrightarrow\dfrac{1}{b}+\dfrac{2}{a}=\dfrac{2}{3}\\ \dfrac{bc}{b+2c}=\dfrac{4}{3}\Leftrightarrow\dfrac{b+2c}{bc}=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{c}+\dfrac{2}{b}=\dfrac{3}{4}\\ \dfrac{ca}{c+2a}=3\Leftrightarrow\dfrac{c+2a}{ca}=\dfrac{1}{3}\Leftrightarrow\dfrac{1}{a}+\dfrac{2}{c}=\dfrac{1}{3}\)
Cộng vế theo vế \(\Leftrightarrow\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}=\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{7}{4}\)
\(\Leftrightarrow3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{7}{4}\\ \Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{7}{12}\\ \Leftrightarrow\dfrac{ab+bc+ca}{abc}=\dfrac{7}{12}\\ \Leftrightarrow T=\dfrac{12}{7}\)
Ta có: \(0\le a\le b\le c\le1\Leftrightarrow\left\{{}\begin{matrix}1-a\ge0\\1-b\ge0\end{matrix}\right.\Leftrightarrow\left(1-a\right)\left(1-b\right)\ge0\)
\(\Rightarrow1-b-a+ab\ge0\Leftrightarrow1+ab\ge a+b\)(1)
Tiếp tục chứng minh ta được: \(0\le a\le b\le c\le1\Leftrightarrow\left\{{}\begin{matrix}1\ge c\\ab\ge0\end{matrix}\right.\)(2)
Cộng theo vế pt(1) với pt(2) ta được:
\(1+ab+1+ab\ge a+b+c+0\)
\(\Rightarrow2\left(ab+1\right)\ge a+b+c\)
Nên: \(\dfrac{c}{ab+1}=\dfrac{2c}{2\left(ab+1\right)}\le\dfrac{2c}{a+b+c}\)
Chứng minh tương tự suy ra đpcm
Câu hỏi của Phạm Quốc Anh - Toán lớp 7 - Học toán với OnlineMath
CTHH có mà (=.=") https://hoc24.vn/hoi-dap/question/384421.html