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Thêm đk \(a,b,c\ne0\)
Ta có: \(\frac{ab}{a+b}=\frac{1}{3}\Rightarrow\frac{a+b}{ab}=3\)
\(\frac{bc}{b+c}=\frac{1}{4}\Rightarrow\frac{bc}{b+c}=4\)
\(\frac{ca}{c+a}=\frac{1}{5}\Rightarrow\frac{c+a}{ca}=5\)
\(\Rightarrow\frac{a+b}{ab}+\frac{b+c}{bc}+\frac{c+a}{ca}=12\)
\(\Leftrightarrow\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}=12\)
\(\Leftrightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=12\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
Ta có abbcca=\(\frac{3}{5}.\frac{4}{5}.\frac{3}{7}\)
=>a2b2c2=\(\frac{36}{175}\)
=>abc=\(\sqrt{\frac{36}{175}}=\frac{6\sqrt{7}}{35}\)
=>a=\(\frac{6\sqrt{7}}{35}:\frac{4}{5}=\frac{3\sqrt{7}}{14}\)=>b=\(\frac{6\sqrt{7}}{35}:\frac{3}{7}=\frac{2\sqrt{7}}{5}\)=>c
ta có A= \(\frac{8^{18}+1}{8^{19} +1}\)=> 8A=\(\frac{8^{19}+8}{8^{19}+1}\)= \(\frac{\left(8^{19}+1\right)+7}{8^{19}+1}\)=\(\frac{8^{19}+1}{8^{19} +1}\)+\(\frac{7}{8^{19}+1}\) =1+\(\frac{7}{8^{19}+1}\) =\(\frac{7}{8^{19}+1}\)
B= \(\frac{8^{23}+1}{8^{24}+1}\)=> 8B=\(\frac{8^{24}+8}{8^{24}+1}\)= \(\frac{\left(8^{24}+1\right)+7}{8^{24}+1}\)=\(\frac{8^{24}+1}{8^{24}+1}\)+\(\frac{7}{8^{24}+1}\) =1+\(\frac{7}{8^{24} +1}\)=\(\frac{7}{8^{24}+1}\)
vì \(8^{19}\)<\(8^{24}\)=> \(8^{19}\)+1 >\(8^{24}\)+1 => \(\frac{7}{8^{19}+1}\)<\(\frac{7}{8^{24}+1}\)=> A<B
a) ta có \(8A=\frac{8^{19}+8}{8^{19}+1}=1+\frac{7}{8^{19}+1}\\ 8B=\frac{8^{24}+8}{8^{24}+1}=1+\frac{7}{8^{24}+1}\)
Vì \(8^{24}+1>8^{19}+1\\\frac{7}{8^{24}+1}< \frac{7}{8^{19}+1} \)
vậy 8A>8B nên A>B
1/ a = 2100 = (24)25 = 1625
b = 375 = (33)25 = 2725
c = 550 = (52)25 = 2525
Do: 16 < 25 < 27 => 1625 < 2525 < 2725 => 2100 < 550 < 375 => a < c < b
thank nhìu :3