Ta có  \(\frac{52}{9}=5+\frac{7}{9}=5+\frac{1}{\frac{9}{7}}\)

                     =\(5+\frac{1}{1+\frac{2}{7}}=5+\frac{1}{1+\frac{1}{\frac{7}{2}}}\)

                     \(=5+\frac{1}{1+\frac{1}{3+\frac{1}{2}}}\)

                           Vậy a = 1

                                   b=3

                                   c=2

giúp mình nhé mình đang gấp lắm

\(\frac{52}{9}=5+\frac{7}{9}=5+\frac{1}{1+\frac{2}{7}}=5+\frac{1}{1+\frac{1}{3+\frac{1}{2}}}\\ \)

(a;b;c) =(1;3;2)

\(\frac{a}{2}+\frac{b}{3}=\frac{a+b}{5}\Leftrightarrow\frac{3a+2b}{6}=\frac{a+b}{5}\\ \Rightarrow15a+10b=6a+6b\Rightarrow9a+4b=0\)

mà a,b là số tự nhiên nên \(a,b\ge0\)

nên \(9a+4b\ge0\)

dấu bằng xảy ra khi a=b=0

mk làm sai nha bạn

sr bạn

a \(\frac{a}{3}+\frac{b}{4}=\frac{a+b}{3+4}\Leftrightarrow\frac{4a+3b}{12}=\frac{a+b}{7}\Leftrightarrow28a+21b=12a+12b\)

\(\Leftrightarrow\left(16a+9b\right)+\left(12a+12b\right)=12a+12b\)

\(\Leftrightarrow16a+9b=0\)

Vì \(16a\ge0;9b\ge0\) ( vì a;b là số TN )

=> \(16a+9b\ge0\)

Dấu "=" xảy ra <=> a = b = 0

b) \(\frac{52}{9}=5+\frac{7}{9}=5+\frac{1}{\frac{9}{7}}=5+\frac{1}{1+\frac{2}{7}}=5+\frac{1}{1+\frac{1}{\frac{7}{2}}}=5+\frac{1}{1+\frac{1}{3+\frac{1}{2}}}\)

\(\Rightarrow a=1;b=3;c=2\)

mik ko bít

Ta có : 

\(\frac{52}{9}=5+\frac{7}{9}\)

\(\frac{7}{9}=\frac{1}{\frac{9}{7}}=\frac{1}{1+\frac{2}{7}}\)

\(\frac{2}{7}=\frac{1}{\frac{7}{2}}=\frac{1}{1+\frac{5}{2}}\)

\(\frac{5}{2}=\frac{1}{\frac{2}{5}}\)

\(\Rightarrow\frac{52}{9}=5+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{2}{5}}}}\)

\(\Rightarrow\hept{\begin{cases}a=1\\b=1\\c=\frac{2}{5}\end{cases}}\)