A=\(\frac{1}{1x2}+\frac{1}{1x3}+\frac{1}{1x4}+\frac{1}{1x5}+\frac{12}{10}\)
Tìm giá trị của A
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7 tháng 2 2015
\(2A=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{999x1001}\)
\(2A=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{1001-999}{999x1001}\)
\(2A=\frac{3}{1x3}-\frac{1}{1x3}+\frac{5}{3x5}-\frac{3}{3x5}+\frac{7}{5x7}-\frac{5}{5x7}+...+\frac{1001}{999x1001}-\frac{999}{999x1001}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\)
\(2A=1-\frac{1}{1001}=\frac{1000}{1001}\)=> A = 500/1001
16 tháng 6 2017
\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{10}\)\(+\)\(\frac{1}{10}\)\(-\)\(\frac{1}{11}\)\(+\)\(\frac{1}{11}\)\(-\)\(\frac{1}{12}\)\(+\)\(\frac{1}{12}\)\(-\)\(\frac{1}{13}\)\(+\)\(\frac{1}{13}\)\(-\)\(\frac{1}{14}\)\(+\)\(\frac{1}{14}\)\(-\)\(\frac{1}{15}\)
\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{15}\)
\(A\)\(=\)\(\frac{2}{45}\)
16 tháng 6 2017
\(A=\left(\frac{1}{9}.\frac{1}{10}+\frac{1}{10}.\frac{1}{11}\right)+\left(\frac{1}{11}.\frac{1}{12}+\frac{1}{12}.\frac{1}{13}\right)+\left(\frac{1}{13}.\frac{1}{14}+\frac{1}{14}.\frac{1}{15}\right)\)
Sau đó nhân phân phối ra là xong nhé bạn
9 tháng 6 2018
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow M=1-\frac{1}{100}\)
\(\Rightarrow M=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
\(b,N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\Rightarrow N=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}\)
\(\Rightarrow N=\frac{1.98}{2.99}=\frac{49.2}{2.99}=\frac{49}{99}\)
8 tháng 6 2018
\(a,M=1-\frac{1}{100}=\frac{99}{100}\)
\(b=2N=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)
\(=1-\frac{1}{99}=\frac{98}{99}\)
=>\(N=\frac{98}{99}:2=\frac{49}{99}\)
26 tháng 3 2017
a) Đặt \(A=\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}\)
\(\Rightarrow A=\left(1^2+2^2+..........+100^2\right)\)\(.\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+.....+100^2\right).\left(\frac{100}{101}\right)\)(a)
Đặt \(M=\left(1^2+2^2+........+100^2\right)\)
\(\Rightarrow M=1.1+2.2+.....+100.100\)
\(\Rightarrow M=1.\left(2-1\right)+2.\left(3-1\right)+....+100.\left(101-1\right)\)
\(\Rightarrow M=\left(1.2-1\right)+\left(2.3-2\right)+.....+\left(100.101-100\right)\)
\(\Rightarrow M=\left(1.2+2.3+.....+100.101\right)-\left(1+2+......+100\right)\)
\(\Rightarrow M=\left(1.2+2.3+......+100.101\right)-5050\)(1)
Đặt \(N=1.2+2.3+....+100.101\)
\(\Rightarrow3.N=1.2.3+2.3.3+......+100.101.3\)
\(\Rightarrow3N=1.2.\left(3-0\right)+2.3.\left(4-1\right)+......+100.101.\left(102-99\right)\)
\(\Rightarrow3N=\left(1.2.3-0\right)+\left(1.2.3-2.3.4\right)+.......+\left(100.101.102-100.101.99\right)\)
\(\Rightarrow3N=100.101.102-0\)
\(\Rightarrow N=343400\)
Thay N = 343400 vào 1) ta được:
M = 343400 - 5050
=> M = 338350
Thay M = 338350 Vào (a) ta được:
A = 338350 . \(\frac{100}{101}\)
=> \(A=\frac{33835000}{101}\)
Vậy \(\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}=\frac{33835000}{101}=335000\)
b) Đặt \(B=\frac{2^2}{1.3}+\frac{3^2}{2.4}+..........+\frac{59^2}{58.60}\)
\(\Rightarrow B=\left(2^2+3^2+........+59^2\right).\left(\frac{1}{1.3}+\frac{1}{2.4}+.....+\frac{1}{58.60}\right)\)
Đặt \(G=2^2+3^2+.........+59^2\)VÀ \(H=\frac{1}{1.3}+\frac{1}{2.4}+.........+\frac{1}{58.60}\)
\(\Rightarrow G=2.2+3.3+.......+59.59\) VÀ \(2.H=\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{58.60}\)
Rồi bạn làm như ở phần a) ý
21 tháng 7 2016
a.\(\frac{3x-1}{3x+1}+\frac{x-3}{x+3}=2\)
\(\frac{\left(3x-1\right)\left(x+3\right)+\left(3x+1\right)\left(x-3\right)}{\left(3x+1\right)\left(x+3\right)}=\frac{3x^2+8x-3+3x^2-8x-3}{\left(3x+1\right)\left(x+3\right)}=\frac{6x^2-6}{\left(3x+1\right)\left(x+3\right)}=2\)
\(6x^2-6=2\left(3x^2+10x+3\right)\)
\(6x^2-6=6x^2+20x+6\)
-20x-12=0
x=\(\frac{-3}{5}\)
\(\frac{149}{60}\)
k mk nha mk đag âm
\(\frac{1}{1\times2}\) + \(\frac{1}{1\times3}\) + \(\frac{1}{1\times4}\) + \(\frac{1}{1\times5}\) + \(\frac{12}{10}\)
= \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\) + \(\frac{12}{10}\)
= \(\frac{149}{60}\)