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\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2.\left(1-\frac{1}{101}\right)\)
\(=2.\frac{100}{101}=\frac{200}{101}\)
Đặt \(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+..+\frac{4}{99.101}\)
\(A=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(A=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=2.\left(1-\frac{1}{101}\right)\)
\(A=\frac{2.100}{101}=\frac{200}{101}\)
Ủng hộ mk nha !!! ^_^
1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6+1/7-1/7+1/8-1/8+1/9+1/9-1/10
=1/2-1/10
=5/10-1/10
=4/10=2/5
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{8x9}+\frac{1}{9x10}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2}-\frac{1}{10}\)
\(\frac{2}{5}\)
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
A và B dễ
Bài 2:
sai đề bài vì ngay từ cái phép tính đầu đã ko theo quy luật rồi
\(A=\frac{-3}{5}-\frac{2}{5}+2\)
\(A=-1+2=1\)
\(B=\left(6-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}=\frac{1}{4}\)
nÀ NÍ sao lại = đây là dấu trừ hay cộng 1/4
\(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}\cdot\frac{17}{4}-28\cdot\frac{4}{3}\right):\frac{7}{4}\)
\(=\frac{59}{15}-\frac{29}{4}:\frac{7}{4}=\)\(\frac{59}{15}-\frac{29}{7}=\frac{-22}{105}\)
B = \(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}x\frac{17}{4}-2x\frac{4}{3}\right):\frac{7}{4}\)
= \(\frac{59}{10}x\frac{2}{3}-\left(\frac{119}{12}-\frac{8}{3}\right)x\frac{4}{7}\)
= \(\frac{59}{15}-\frac{29}{4}x\frac{4}{7}=\frac{59}{15}-\frac{29}{7}\)
= \(\frac{-22}{105}\)
C = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)
= \(1-\frac{1}{7}=\frac{6}{7}\)
=>A=\(\frac{7}{2}\)(\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+...+\(\frac{1}{99}\)-\(\frac{1}{101}\))
=>A=\(\frac{7}{2}\)(1-\(\frac{1}{101}\))
=>A=\(\frac{350}{101}\)
7/2 ( \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{99}-\frac{1}{101}\))
7/2 ( 1 - 1/101 )
7/2 x 100/101
=350/101
Xét Sn = 1+2+3+4+...+n (1)
=> Sn= n+(n-1)+...+2+1 (2)
Thấy 1+n = 2+(n-1) = 3+(n-2) = n-1+2=n+1
Lấy (1);(2) và chú ý trên ta có:
2.Sn = (n+1)+(n+1)+(n+1)+...+(n+1)=n(n+1) (vì n số hạng giống nhau)
=> Sn= n(n+1)/2 => Sn/n = (n+1)/2
=> P= 1+ S2/2 + S3/3 + S4/4 +...+ Sn/n
P= 1+3/2+4/2+5/2+...+(n+1)/2
P= 2(2+3+4+...+n+n+1) = 2(1+2+...n+n+1) - 2 = 2.S(n+1) - 2
P= 2.(n+1)(n+2)/2 -2 = (n+1)(n+2) -2 = n2+3n
Bài toán chỉ đến S2016/2016 (tức n=2016)
Vậy S= 20162+3.2016=2016.(2016+3)=2016.2019=4070304
E = 1 + 1/2.(1 + 2) + 1/3.(1 + 2 + 3) + 1/4.(1 + 2 + 3 + 4) + ... + 2016.(1 + 2 + 3 + ... + 2016)
E = 1 + 1/2.(1 + 2).2:2 + 1/3.(1 + 3).3:2 + 1/4.(1 + 4).4:2 + ... + 2016.(1 + 2016).2016:2
E = 2/2 + 3/2 + 4/2 + 5/2 + ... + 2017/2
E = 2+3+4+5+...+2017/2
E = (2 + 2017).2016/2
E = 2019.1008
E = 2 035 152
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{3.5.7}+...+\frac{1}{45.47.49}\)
\(\Rightarrow4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{45.47.49}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{45.47}-\frac{1}{47.49}\)
\(=\frac{1}{3}-\frac{1}{47.49}\)
\(\Rightarrow A=\frac{\frac{1}{3}-\frac{1}{47.49}}{4}=\frac{575}{6909}\)
\(\frac{4}{1.3}+\frac{4}{3.5}+........+\frac{4}{2011.2013}\)
\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+........+\frac{2}{2011.2013}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+........+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(=2\left(1-\frac{1}{2013}\right)\)
\(=2.\frac{2012}{2013}\)
\(=\frac{4024}{2013}\)
trả lời cho tui nha!