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Câu 2:
Theo đề, ta có: \(\dfrac{10a+b}{a+b}=\dfrac{10b+c}{b+c}\)
=>10ab+10ac+b^2+bc=10ab+10b^2+ac+bc
=>9ac-9b^2=0
=>ac-b^2=0
=>ac=b^2
=>a/b=b/c
\(\Leftrightarrow\dfrac{10a+b}{10b+c}=\dfrac{b}{c}\)
=>10ac+bc=10b^2+cb
=>10ac=10b^2
=>ac=b^2
=>a/b=b/c=k
=>a=bk; b=ck
=>a=ck*k=k^2*c
\(\dfrac{a}{c}=\dfrac{k^2c}{c}=k^2\)
\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{b^2k^2+b^2}{c^2k^2+c^2}=\dfrac{b^2}{c^2}=\dfrac{c^2k^2}{c^2}=k^2\)
=>ĐPCM
\(\Leftrightarrow\dfrac{10a+b}{a+b}=\dfrac{10b+c}{b+c}\)
=>10ac+bc=10b^2+cb
=>10ac=10b^2
=>ac=b^2
Ta có:
\(\dfrac{\overline{ab}}{b}=\dfrac{\overline{bc}}{c}=\dfrac{\overline{ca}}{a}\)
\(\Rightarrow\dfrac{10a}{b}+\dfrac{b}{b}=\dfrac{10b}{c}+\dfrac{c}{c}=\dfrac{10c}{a}+\dfrac{a}{a}\)
\(\Rightarrow\dfrac{10a}{b}+1=\dfrac{10b}{c}+1=\dfrac{10c}{a}+1\)
\(\Rightarrow\dfrac{10a}{b}=\dfrac{10b}{c}=\dfrac{10c}{a}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{10a}{b}=\dfrac{10b}{c}=\dfrac{10c}{a}=\dfrac{10a+10b+10c}{b+c+a}=\dfrac{10\left(a+b+c\right)}{a+b+c}=10\)
\(\Rightarrow\left\{{}\begin{matrix}10a=10b\\10b=10c\\10c=10a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c\)
\(\Rightarrow\left(\overline{abc}\right)^{123}=\left(\overline{aaa}\right)^{123}\)(1)
\(\Rightarrow c=111^{123}.a^{40}.a^{41}.a^{42}=111^{123}.a^{123}=\left(111.a\right)^{123}=\left(\overline{aaa}\right)^{123}\)(2)
Từ (1) và (2) suy ra: \(\left(\overline{abc}\right)^{123}=111^{123}.a^{40}.b^{41}.c^{42}\)
Ta có:
\(\frac{\overline{ab}}{a+b}=\frac{\overline{bc}}{b+c}\Rightarrow\frac{10a+b}{a+b}=\frac{10b+c}{b+c}\Rightarrow\left(10a+b\right)\left(b+c\right)=\left(10b+c\right)\left(a+b\right)\)
\(\Rightarrow10ab+b^2+10ac+bc=10ab+ac+10b^2+bc\Rightarrow9b^2=9ac\Rightarrow b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\left(đpcm\right)\)
=>\(\dfrac{10a+b}{10b+c}=\dfrac{b}{c}\)
=>10ac+bc=10b^2+bc
=>ac=b^2
=>a/b=b/c=k
=>a=bk; b=ck
=>a=ck^2; b=ck
\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{c^2k^4+c^2k^2}{c^2k^2+c^2}=k^2\)
\(\dfrac{a}{c}=\dfrac{ck^2}{c}=k^2\)
=>\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)