gọi số đó là a => a:2=3 (dư 1 ) => a= 2.3+1=7

Đề là gì vậy bạn????

\(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)

\(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}=\dfrac{7}{4}\)

TH1 : \(2.\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{29}{24}\)

\(\Rightarrow x=\dfrac{29}{24}:\dfrac{1}{2}=\dfrac{29}{12}\)

Trường hợp 2 tương tự

b) \(\dfrac{3}{x+1}=\dfrac{4}{y-2}=\dfrac{5}{z-3}=\dfrac{3+4+5}{\left(1-2-3\right)+\left(x+y+z\right)}=\dfrac{12}{14}=\dfrac{6}{7}\)

Ta có: \(\dfrac{3}{x+1}=\dfrac{6}{7}\Rightarrow x+1=\dfrac{7}{2}\Rightarrow x=\dfrac{5}{2}\)

\(\dfrac{4}{y-2}=\dfrac{6}{7}\Rightarrow y-2=\dfrac{14}{3}\Rightarrow y=\dfrac{20}{3}\)

\(\dfrac{5}{z-3}=\dfrac{6}{7}\Rightarrow z-3=\dfrac{35}{6}\Rightarrow z=\dfrac{53}{6}\)

Vậy...............

\(\dfrac{12}{6.7}+\dfrac{12}{7.22}+\dfrac{12}{22.15}+\dfrac{12}{15.38}+...+\dfrac{12}{97.202}\)

\(=12.\left(\dfrac{1}{6.7}+\dfrac{1}{7.22}+\dfrac{1}{22.15}+\dfrac{1}{15.38}...+\dfrac{1}{97.202}\right)\)

\(=12.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{15}+...+\dfrac{1}{97}-\dfrac{1}{202}\right)\)

\(=12.\left(\dfrac{1}{6}-\dfrac{1}{202}\right)\)

\(=\dfrac{196}{101}\)

\(D=\left(\dfrac{32}{15}.\dfrac{9}{17}.\dfrac{3}{32}\right):\left(\dfrac{-3}{17}\right)\)

\(D=\dfrac{9}{85}:\left(\dfrac{-3}{17}\right)\)

\(D=\dfrac{9}{85}.\left(\dfrac{-17}{3}\right)\)

\(D=-\dfrac{3}{5}\)

Ta có: \(8^{34}-8^{33}-8^{32}=8^{32}\left(8^2-8-1\right)\)

\(=8^{32}.55\)

Mà 55 chia hết cho 11 nên \(8^{32}.55⋮11\)

\(\Rightarrow\)đpcm

e thik cái đề này nè,tiếc là ko đc lm.Bị loại từ vòng 1 rồi còn đâu

dựa vào hình vẽ bạn suy ra!