Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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: \(\left(\overline{abc}\right)^{123}=111^{123}.a^{40}.b^{41}.c^{42}\)
a) \(\left(\frac{1}{3}\right)^n=\frac{1}{27}\)
\(\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow n=3\)
b) \(\left(\frac{3}{5}\right)^n=\frac{81}{625}\)
\(\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^4\)
\(\Rightarrow n=4\)
c) \(3^n\cdot2^n=36\)
\(\left(3\cdot2\right)^n=36\)
\(6^n=6^2\)
\(\Rightarrow n=2\)
d) \(\frac{2^n}{3^n}=\frac{8}{27}\)
\(\left(\frac{2}{3}\right)^n=\left(\frac{2}{3}\right)^3\)
\(\Rightarrow n=3\)
Ta có \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Leftrightarrow\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
Từ \(\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}\Rightarrow\frac{1}{a}=\frac{1}{c}\)
Tương tự suy ra \(\frac{1}{c}=\frac{1}{b};\frac{1}{b}=\frac{1}{a}\)
\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
Ta có \(ab^2+bc^2+ca^2=a^3+b^3+c^3\)(đccm)
\(\text{Một cách khác}\)
\(\text{Ta có:}\)
\(\frac{ab}{a+b}=\frac{bc}{b+c}\)
\(\Leftrightarrow ab\left(b+c\right)=bc\left(a+b\right)\)
\(\Leftrightarrow ab^2+abc=abc+b^2c\)
\(\Leftrightarrow a=c\left(1\right)\)
\(\frac{bc}{b+c}=\frac{ca}{a+c}\)
\(\Rightarrow bc\left(a+c\right)=ca\left(b+c\right)\)
\(\Rightarrow abc+bc^2=abc+c^2a\)
\(\Rightarrow b=a\left(2\right)\)
\(Từ\)\(\text{(1) và (2)}\)\(\Rightarrow a=b=c\)
\(\text{Ta có :}\)\(ab^2+bc^2+ca^2=a^3+b^3+c^3\)