1/2.3+4 1/3.+.....+1/99.100=
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1 + 1/3 + 1/6 + 1/10 + .......... + 1/x.(x+1):2 =1 + 1991/1993
1/2.(1 + 1/3 + 1/6 + 1/10+........+ 1/x.(x+1):2=3984/3986
1/2 + 1/6 +1/12 + .......... +1/x.(x+1)=3984/3986
1/1.2 + 1/2.3 + 1/3.4 +..........+.1/x.(x+1)=3984/3986
2-1/1.2 + 3-2/2.3 + 4-3/3.4 +..........+ x + 1 - x/x.(x+1)
1-1/2+1/2-1/3+1/3-1/4+..........+1/x -1/x+1 =3984/3986
1-1/x+1=3984/3986
1/x+1=1-3984/3986
1/x+1=2/3986=1/1993
x+1=1993
x =1993-1
x =1992
Bài 3 :
b) Ta có 1+ 2 + 3 +4 + ...+ x =15
Nên \(\frac{x\left(x+1\right)}{2}=15\)
\(x\left(x+1\right)=30\)
=> \(x\left(x+1\right)=5.6\)
=> x = 5
Bài 2:
h; \(\dfrac{2}{3}\)\(x\) + 50% + \(x\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{2}\) + \(x\) = \(\dfrac{1}{10}\)
(\(\dfrac{2}{3}\)\(x\) + \(x\)) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) (\(\dfrac{2}{3}\) + 1) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{1}{10}\) - \(\dfrac{1}{2}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{-2}{5}\)
\(x\) = \(\dfrac{-2}{5}\): \(\dfrac{5}{3}\)
\(x\) = - \(\dfrac{6}{25}\)
Lớp 5 chưa học số âm em nhé.
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}=\frac{127}{128}\)
Ủng hộ mk nha ^_-
\(2\cdot A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
\(A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}=\frac{128}{128}-\frac{1}{128}=\frac{127}{128}\)
5/6+3/4 < 19/11 4/9 x 6/5 = 8/15
17/12-9/8 < 1/3 15/4:3/8 > 8
Ta có: \(\frac{5}{6}+\frac{3}{4}=\frac{10}{12}+\frac{9}{12}=\frac{19}{12}\)
Vì \(12>11\)
\(\Rightarrow\frac{19}{12}< \frac{19}{11}\)
\(\Rightarrow\frac{5}{6}+\frac{3}{4}< \frac{19}{11}\)
Ta có: \(\frac{17}{12}-\frac{9}{8}=\frac{34}{24}-\frac{27}{24}=\frac{7}{24}\)
Ta lại có: \(\frac{1}{3}=\frac{8}{24}\)
Vì \(8>7\)
\(\Rightarrow\frac{7}{24}< \frac{8}{24}\)
\(\Rightarrow\frac{17}{12}-\frac{9}{8}< \frac{1}{3}\)
Ta có: \(\frac{4}{9}\times\frac{6}{5}=\frac{24}{45}=\frac{8}{15}\)
Vì \(\frac{8}{15}=\frac{8}{15}\)
\(\Rightarrow\frac{4}{9}\times\frac{6}{5}=\frac{8}{15}\)
Ta có: \(\frac{15}{4}:\frac{3}{8}=\frac{15}{4}\times\frac{8}{3}=10\)
Vì \(10>8\)
\(\Rightarrow\frac{15}{4}:\frac{3}{8}>8\)
HOK TOT
Ta có : A = 1/1.2 + 1/2.3 + .... + 1/98.99 + 1/99.100 .
=> A = 1 - 1/2 + 1/2 - 1/3 + .... + 1/98 - 1/99 + 1/99 - 1/100 .
=> A = 1 - 1/100 .
=> A = 99/100 .
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A=1-\frac{1}{100}\)
\(\Rightarrow A=\frac{99}{100}\)
=1/2-1/3+1/3-1/4+...+1/99-1/100
=1/2-1/100=49/100
k nhe
=1/2 -1/3 +1/3 -1/4 +......+1/99 -1/100
=1/2 -1/100
=49/100