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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 \(\left(I\right)+\left(II\right)\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{15}\left(đpcm\right)\)
a: AC/AB=4/7 nên \(\dfrac{CA}{CB}=\dfrac{4}{7+4}=\dfrac{4}{11}\)
=>AB/CB=7/11
hay BC/AB=11/7
b: AC/BC=5/4
nên BC/AC=4/5
=>BA/AC=1/5
AB/BC=1/4
c: BC/AB=11/5
nên AB/BC=5/11
=>AC/BC=6/11
=>AB/AC=5/6
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ac}{a+c}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{a+c}{ac}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c\\a=b\end{matrix}\right.\) \(\Rightarrow a=b=c\)
Thay vào M ta được:
\(M=\dfrac{ab+bc+ac}{a^2+b^2+c^2}=\dfrac{a.a+a.a+a.a}{a^2+a^2+a^2}=\dfrac{3a^2}{3a^2}=1\)
theo đề bài ta có:
\(\Rightarrow\dfrac{abc}{ab+bc}=\dfrac{abc}{ab+ac}=\dfrac{abc}{bc+ab}\)
\(\Rightarrow ac+bc=ab+ac=bc+ab\)
\(\Rightarrow M=\dfrac{ab+bc+ca}{a^2+b^2+c^2}=\dfrac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
Bài 2:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b: \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7\cdot b^2k^2+5\cdot bk\cdot dk}{7\cdot b^2k^2-5\cdot bk\cdot dk}\)
\(=\dfrac{7b^2k^2+5bdk^2}{7b^2k^2-5bdk^2}=\dfrac{7b^2+5bd}{7b^2-5bd}\)(đpcm)
Ta có:
\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)
\(c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)
Từ (1) và (2), suy ra: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\)
Vậy \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)(đpcm)
~ Học tốt!~
Do \(a,b,c\ne0\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ac}{a+c}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{a+c}{ac}\)
\(\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{b}{bc}+\dfrac{c}{bc}=\dfrac{a}{ac}+\dfrac{c}{ac}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c\\b=a\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow M=\dfrac{a.a+a.a+a.a}{a^2+a^2+a^2}=\dfrac{3a^2}{3a^2}=1\)
