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\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{19\cdot20}\right)\div x=\frac{9}{10}\)
\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\div x=\frac{9}{10}\)
\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{20}\right)\div x=\frac{9}{10}\)
\(\Leftrightarrow\frac{19}{20}\div x=\frac{9}{10}\)
\(\Leftrightarrow x=\frac{19}{18}\)
Sửa đề : \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right):x=\frac{9}{10}\)
\(\Leftrightarrow VT=\frac{9}{10}x\)
\(\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)=\frac{9}{10}x\)
\(\Leftrightarrow\left(1-\frac{1}{20}\right)=\frac{9}{10}x\Leftrightarrow\frac{19}{20}=\frac{9}{10}x\)
\(\Leftrightarrow\frac{19}{20}=\frac{18x}{20}\) Khử mẫu ta đc : \(\Leftrightarrow18x=19\Leftrightarrow x=\frac{19}{18}\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}=\frac{9}{10}\)
\(\Rightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(\Rightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)
\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}:1=5\)
\(\Rightarrow x=5-\frac{206}{100}=\frac{147}{50}\)
Vậy \(x=\frac{147}{50}.\)
số số hạng là : ( 99.100 - 1.2 ) / 1.1 +1=90 ( số )
tổng của S là : ( 1.2 + 99.100 ) * 90 / 2=4513.5
\(3A=1.2.3+2.3.\left(4-1\right)+...+100.101.\left(102-99\right)\)
\(3A=1.2.3+2.3.4-1.2.3+.......+100.101.102-99.100.101\)
\(3A=100.101.102\)
\(A=\frac{100.101.102}{3}\)
\(A=343400\)
3=1.2.3+2.3(4-1)+...+100.101(102-99)
3=1.2.3+2.3.4-1.2.3+.....+100.101.102-99.100.101
3=100.101.101
=100.101.102/3
=343400
mn ủng hộ ^--^
Ta có: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Leftrightarrow100\left(\dfrac{1}{1}-\dfrac{1}{10}\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]\cdot2=89\)
\(\Leftrightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow x+\dfrac{103}{50}=5\)
hay \(x=\dfrac{147}{50}\)
Gọi A là biểu thức ta có:
A = 1.2+2.3+3.4+......+99.100
Gấp A lên 3 lần ta có:
A . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
A . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98)
A . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100
A . 3 = 99.100.101
A = 99.100.101 : 3
A = 33.100.101
A = 333 300
A= 1.2+2.3+3.4+...+99.100
3A= 1.2.3+2.3.4+3.4.3+...+99.100.3
3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+ 99.100(101-98)
3A= (1.2.3+2.3.4+3.4.5+...+ 99.100.110)-(0.1.2+1.2.3+2.3.4+3.4.5+...+98.99.100)
3A= 99.100.101- 0.1.2
3A= 999 900-0
3A= 999 900
A= 999 900:3
A= 333 300
GOOD LUCK !!!! ^ ^
A = 1.2 + 2.3 + 3.4 + ... + 2005.2006
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2005.2006.3
3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2005.2006.(2007 - 2004)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2005.2006.2007 - 2004.2005.2006
3A = 2005.2006.2007
A = 2690738070
Có A=1.2+2.3+3.4+...+2005.2006
3A=1.2.3+2.3.3+3.4.3+....+2005.2006.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+....+2005.2006.(2007-2004)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+2005.2006.2007-2004.2005.2006
3A=2005.2006.2007
A=(2005.2006.2007):3
Vậy A=....
đề sai rùi hay sao í
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+............+99.100.(101-98)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+............+99.100.101-98.99.100
3A=99.100.101
A=99.100.101:3
A=333300