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\(\dfrac{2}{5}\)= \(\dfrac{6}{15}\)> \(\dfrac{6}{16}\) = \(\dfrac{3}{8}\) > \(\dfrac{3}{9}\) = \(\dfrac{1}{3}\) = \(\dfrac{1\times5}{3\times5}\) = \(\dfrac{5}{15}\) > \(\dfrac{5}{16}\) vậy \(\dfrac{2}{5}\) > \(\dfrac{3}{8}\) > \(\dfrac{1}{3}\) > \(\dfrac{5}{16}\)
\(\dfrac{5}{16}\) = \(\dfrac{5\times4}{16\times4}\) = \(\dfrac{20}{64}\) > \(\dfrac{20}{65}\) = \(\dfrac{4}{13}\)
Các phân số đã cho được sắp xếp theo thứ tự tăng dần là:
\(\dfrac{4}{13}\); \(\dfrac{5}{16}\); \(\dfrac{1}{3}\); \(\dfrac{3}{8}\); \(\dfrac{2}{5}\)
Bài 3 :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)
\(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)
Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)
\(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)
\(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)
\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
3.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)
\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
3 - ( 5\(\frac{3}{8}\)+ x - 7\(\frac{5}{24}\)) : 16\(\frac{2}{3}\)= 2
3 - ( \(\frac{43}{8}\)+ x - \(\frac{173}{24}\)) : \(\frac{50}{3}\)= 2
3 - ( \(\frac{43}{8}\)+ x - \(\frac{173}{24}\)) = 2 x \(\frac{50}{3}\)
3 - ( \(\frac{43}{8}\)+ x - \(\frac{173}{24}\)) = \(\frac{100}{3}\)
\(\frac{43}{8}\)+ x - \(\frac{173}{24}\)= 3 - \(\frac{100}{3}\)
\(\frac{43}{8}\)+ x - \(\frac{173}{24}\)= \(-\frac{97}{3}\)
\(\frac{43}{8}\)+ x = \(-\frac{97}{3}+\frac{173}{24}\)
\(\frac{43}{8}\)+ x = \(-\frac{201}{8}\)
x = \(-\frac{201}{8}-\frac{43}{8}\)
x = \(=-\frac{61}{2}\)
\(2-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)\)\(:16\frac{2}{3}=0\)
\(\Rightarrow\)\(2-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)=0\)
\(\Rightarrow\)\(5\frac{3}{8}+x-7\frac{5}{24}=2\)
\(\Rightarrow5\frac{3}{8}+x=9\frac{5}{24}\)
\(\Rightarrow x=3\frac{5}{6}\)
( Cách làm thì đúng rồi nhưng đáp án tớ ko chắc chắn )
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\)
\(=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{3.7}+\frac{1}{7.8}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}+0+0+...+0\)
\(=\frac{7}{8}\)
TS :
3 + 5 + 7 + ... + 2015
SSH là : ( 2015 - 3 ) : 2 + 1 = 1007 ( số )
Tổng là : ( 2015 + 3 ) . 1007 : 2 = 4064252
→Vì TS : MS = 1 => TS = MS
Ta có : 2 + 4 + 6 + ... + 2014 + x = 4 064 252
2 + 4 + 6 + ... + 2014 = 4 064 252 - x
SSH ở vế trái là : ( 2014 - 2 ) : 2 + 1 = 1007 ( số )
Tổng là : ( 2014 + 2 ) . 1007: 2 = 4 060 224
Vậy x là : 4 064 252 - 4 060 224 = 4028
Vậy x là 4028
\(5+\frac{x}{8}=\frac{14}{16}\)
\(\frac{x}{8}=\frac{14}{16}-5\)
\(\frac{x}{8}=\frac{-33}{8}\)
\(x=-33\)
\(5+\frac{x}{8}=\frac{14}{16}\)
\(\Rightarrow\frac{x}{8}=5-\frac{14}{16}\)
\(\Rightarrow\frac{x}{8}=\frac{33}{8}\)
\(\Rightarrow x=33\)
tíc mình nha