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Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{ab+ac}{2}=\dfrac{bc+ba}{3}=\dfrac{ca+cb}{4}\)
\(=\dfrac{ab+ac+bc+ba-ca-cb}{2+3-4}=\dfrac{2ab}{1}\) \(\left(1\right)\)
\(=\dfrac{bc+cb+bc+ba-ab-ac}{3+4-2}=\dfrac{2bc}{5}\left(2\right)\)
\(=\dfrac{ab+ac+ca+cb-bc-ba}{2+4-3}=\dfrac{2ac}{3}\)\(\left(3\right)\)
Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow\dfrac{2ab}{1}=\dfrac{2bc}{5}=\dfrac{2ac}{3}\)
\(\dfrac{2ab}{1}=\dfrac{2bc}{5}\Leftrightarrow\dfrac{a}{1}=\dfrac{c}{15}\) \(\Leftrightarrow\dfrac{a}{3}=\dfrac{c}{15}\left(I\right)\)
\(\dfrac{2bc}{5}=\dfrac{2ac}{3}\Leftrightarrow\dfrac{b}{5}=\dfrac{a}{3}\left(II\right)\)
Từ \(\left(I\right)+\left(II\right)\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{15}\left(đpcm\right)\)
\(\dfrac{ab+1}{3}=\dfrac{bc+2}{8}=\dfrac{ca-1}{2}=\dfrac{ab+bc+ca+1+2-1}{3+8+2}=\dfrac{11+2}{13}=1\)
\(\Rightarrow\left\{{}\begin{matrix}ab+1=3\\bc+2=8\\ca-1=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ab=2\\bc=6\\ca=3\end{matrix}\right.\)
\(\Rightarrow\left(abc\right)^2=36\)
\(\Rightarrow abc=6\) (vì a,b,c là số thực dương)
Mà \(ab=2\Rightarrow c=3\)
Tiếp \(bc=6\Rightarrow a=1;b=2\)
Vậy \(\left(a,b,c\right)=\left(1;2;3\right)\)
Do \(a,b,c\ne0\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ac}{a+c}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{a+c}{ac}\)
\(\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{b}{bc}+\dfrac{c}{bc}=\dfrac{a}{ac}+\dfrac{c}{ac}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c\\b=a\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow M=\dfrac{a.a+a.a+a.a}{a^2+a^2+a^2}=\dfrac{3a^2}{3a^2}=1\)
Ta có: \(0\le a\le b\le c\le1\Leftrightarrow\left\{{}\begin{matrix}1-a\ge0\\1-b\ge0\end{matrix}\right.\)
\(\Rightarrow\left(1-a\right)\left(1-b\right)\ge0\Leftrightarrow1\left(1-b\right)-a\left(1-b\right)\ge0\)
\(\Rightarrow1-b-a+ab\ge0\Leftrightarrow1+ab\ge a+b\)
Tiếp tục chứng minh ta có: \(\left\{{}\begin{matrix}1\ge c\\0\le a\le b\Leftrightarrow ab\ge0\end{matrix}\right.\)
cộng theo vế: \(1+ab+1+ab\ge a+b+c+0\)
\(\Rightarrow2\left(1+ab\right)\ge a+b+c\)
Ta có: \(\dfrac{c}{ab+1}=\dfrac{2c}{2\left(ab+1\right)}\le\dfrac{2c}{a+b+c}\) (1)
chứng minh tương tự suy ra đpcm
a) \(ab=\dfrac{3}{5};bc=\dfrac{4}{5};ca=\dfrac{3}{4}\)
\(\Leftrightarrow ab.bc.ca=\dfrac{3}{5}.\dfrac{4}{5}.\dfrac{3}{4}\)
\(\Leftrightarrow a^2.b^2.c^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left(abc\right)^2=\left(\dfrac{3}{5}\right)^2=\left(-\dfrac{3}{5}\right)^2\)
+ Khi \(\left(abc\right)^2=\left(\dfrac{3}{5}\right)^2\Leftrightarrow abc=\dfrac{3}{5}\)
Vậy \(\left\{{}\begin{matrix}a=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4}\\b=\dfrac{3}{5}:\dfrac{3}{4}=\dfrac{4}{5}\\c=\dfrac{3}{5}:\dfrac{3}{5}=1\end{matrix}\right.\)
+ Khi \(\left(abc\right)^2=\left(-\dfrac{3}{5}\right)^2\Leftrightarrow abc=-\dfrac{3}{5}\)
Vậy \(\left\{{}\begin{matrix}a=\left(-\dfrac{3}{5}\right):\dfrac{4}{5}=-\dfrac{3}{4}\\b=\left(-\dfrac{3}{5}\right):\dfrac{3}{4}=-\dfrac{4}{5}\\c=\left(-\dfrac{3}{5}\right):\dfrac{3}{5}=-1\end{matrix}\right.\)
b) \(a\left(a+b+c\right)=-12;b\left(a+b+c\right)=18;c\left(a+b+c\right)=30\)
\(\Leftrightarrow a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=\left(-12\right)+18+30\)
\(\Leftrightarrow\left(a+b+c\right)\left(a+b+c\right)=36\)
\(\Leftrightarrow\left(a+b+c\right)^2=6^2=\left(-6\right)^2\)
+ Khi \(\left(a+b+c\right)^2=6^2\Leftrightarrow a+b+c=6\)
Vậy \(\left\{{}\begin{matrix}a=\left(-12\right):6=-2\\b=18:6=3\\c=30:6=5\end{matrix}\right.\)
+ Khi \(\left(a+b+c\right)^2=\left(-6\right)^2\Leftrightarrow a+b+c=-6\)
Vậy \(\left\{{}\begin{matrix}a=\left(-12\right):\left(-6\right)=2\\b=18:\left(-6\right)=-3\\c=30:\left(-6\right)=-5\end{matrix}\right.\)
c) \(ab=c;bc=4a;ac=9b\)
Kiểm tra lại đề bài xem có thiếu điều kiện không.
Cứ theo khẳng định của Nguyễn Thị Ngọc Linh thì đề c) không thiếu gì. Xin giải tiếp.
c) \(ab=c;bc=4a;ac=9b\)
\(\Leftrightarrow ab.bc.ac=c.4a.9b\)
\(\Leftrightarrow\left(abc\right)\left(abc\right)=36\left(abc\right)\)
\(\Leftrightarrow abc=36\)
+ Vì \(ab=c\Leftrightarrow cc=36\Leftrightarrow c^2=6^2=\left(-6\right)^2\)
+ Vì \(bc=4a\Leftrightarrow a.4a=36\Leftrightarrow4a^2=36\Leftrightarrow a^2=9=3^2=\left(-3\right)^2\)
+ Vì \(ac=9b\Leftrightarrow b.9b=36\Leftrightarrow9b^2=36\Leftrightarrow b^2=4=2^2=\left(-2\right)^2\)
Vậy \(\left\{{}\begin{matrix}a_1=3;a_2=-3\\b_1=2;b_2=-2\\c_1=6;c_2=-6\end{matrix}\right.\)
Đặt \(\dfrac{ab+ac}{4}=\dfrac{bc+ab}{6}=\dfrac{ca+cb}{8}=k\)
=>ab+ac=4k; bc+ab=6k; ac+bc=8k
=>ac-bc=-2k; ac+bc=8k; ab+ac=4k
=>ac=3k; bc=5k; ab=k
=>c/b=3; c/a=5
=>c=3b=5a
=>a/3=b/5=c/15