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\(A=\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+............+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{49.50}\)
A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+ \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)
A=1-\(\frac{1}{50}\)
A=\(\frac{49}{50}\)
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)
\(\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\Rightarrow x+\frac{266}{100}=5\Rightarrow x=\frac{117}{50}\)
Vậy x = 117/50
Ta có:
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right).100\\ =\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100\)
\(=\left(1-\frac{1}{10}\right).100\)
\(=\frac{9}{10}.100\)
= 90
Khi đó đề bài sẽ thành : \(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\Rightarrow\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)
\(\Rightarrow\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\)
\(\Rightarrow x+\frac{266}{100}=5\)
\(\Rightarrow x=\frac{117}{50}\)
Vậy \(x=\frac{117}{50}\)
bài A: áp dụng công thức: 1 + 2 + 3 + ... + n = n x (n + 1) : 2 tính được 5050
bài B: áp dụng công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) rồi triệt tiêu gần hết, qui đồng mẫu số tính được B = 99/100
A = 1 + 2 + 3 + 4 + 5 + ... + 99 + 100
= ( 100 + 1 ) x 100 : 2 = 5050
Vậy A = 5050
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Vậy \(B=\frac{99}{100}\)
Học tốt #
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\)
A=\(1-\frac{1}{4}\)
A=\(\frac{3}{4}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}=\frac{6}{12}+\frac{2}{12}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}\)
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50
= 1 - 1/50
= 49/50
ỦNG HỘ NHA
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(x-1\right)\times x}=\frac{15}{16}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x-1}-\frac{1}{x}=\frac{15}{16}\)
\(1-\frac{1}{x}=\frac{15}{16}\)
\(\frac{1}{x}=\frac{1}{16}\)
\(\Rightarrow x=16\)
Ta có:A: 1/1.2 +1/2.3 +1/3.4+...+1/18.19+1/19.20
=> A= 1-1/2 +1/2-1/3+1/3-1/4+...+1/18-1/19+1/19-1/20
=>A= 1-1/20=19/20
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
thấy đúng thì k cho mk nha mấy bạn
Vì 2-1=1; 3-2=1; 4-3=1; ...
\(\Rightarrow=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow=\frac{1}{1}-\frac{1}{100}\)
\(\Rightarrow=\frac{99}{100}\)
`1/(1*2)+1/(2*3)+1/(3*4)+,.,,+1/(n(n+1))`
`=(2-1)/(1*2)+(3-2)/(2*3)+(4-3)/(3*4)+...+((n+1)-n)/(n(n+1))`
`=2/(1*2)-1/(1*2)+3/(2*3)-2/(2*3)+4/(3*4)-3/(3*4)+...+(n+1)/(n(n+1))-n/(n(n+1))`
`=1-1/2+1/2-1/3+1/3-1/4+...+1/n-1/(n+1)`
`=1+(1/2-1/2)+(1/3-1/3)+(1/4-1/4)+...+(1/n-1/n)+1/(n+1)`
`=1+0+0+...+0-1/(n+1)`
`=1-1/(n+1)`
`=(n+1)/(n+1)-1/(n+1)`
`=(n+1-1)/(n+1)`
`=n/(n+1)`
Vậy: `1/(1*2)+1/(2*3)+1/(3*4)+...+1/(n(n+1))=n/(n+1)`
1/(1∗2)+1/(2∗3)+1/(3∗4)+,.,,+1/(n(n+1))
\(= \left(\right. 2 - 1 \left.\right) / \left(\right. 1 * 2 \left.\right) + \left(\right. 3 - 2 \left.\right) / \left(\right. 2 * 3 \left.\right) + \left(\right. 4 - 3 \left.\right) / \left(\right. 3 * 4 \left.\right) + . . . + \left(\right. \left(\right. n + 1 \left.\right) - n \left.\right) / \left(\right. n \left(\right. n + 1 \left.\right) \left.\right)\)
\(= 2 / \left(\right. 1 * 2 \left.\right) - 1 / \left(\right. 1 * 2 \left.\right) + 3 / \left(\right. 2 * 3 \left.\right) - 2 / \left(\right. 2 * 3 \left.\right) + 4 / \left(\right. 3 * 4 \left.\right) - 3 / \left(\right. 3 * 4 \left.\right) + . . . + \left(\right. n + 1 \left.\right) / \left(\right. n \left(\right. n + 1 \left.\right) \left.\right) - n / \left(\right. n \left(\right. n + 1 \left.\right) \left.\right)\)
\(= 1 - 1 / 2 + 1 / 2 - 1 / 3 + 1 / 3 - 1 / 4 + . . . + 1 / n - 1 / \left(\right. n + 1 \left.\right)\)
\(= 1 + \left(\right. 1 / 2 - 1 / 2 \left.\right) + \left(\right. 1 / 3 - 1 / 3 \left.\right) + \left(\right. 1 / 4 - 1 / 4 \left.\right) + . . . + \left(\right. 1 / n - 1 / n \left.\right) + 1 / \left(\right. n + 1 \left.\right)\)
\(= 1 + 0 + 0 + . . . + 0 - 1 / \left(\right. n + 1 \left.\right)\)
\(= 1 - 1 / \left(\right. n + 1 \left.\right)\)
\(= \left(\right. n + 1 \left.\right) / \left(\right. n + 1 \left.\right) - 1 / \left(\right. n + 1 \left.\right)\)
\(= \left(\right. n + 1 - 1 \left.\right) / \left(\right. n + 1 \left.\right)\)
\(= n / \left(\right. n + 1 \left.\right)\)
Vậy: \(1 / \left(\right. 1 * 2 \left.\right) + 1 / \left(\right. 2 * 3 \left.\right) + 1 / \left(\right. 3 * 4 \left.\right) + . . . + 1 / \left(\right. n \left(\right. n + 1 \left.\right) \left.\right) = n / \left(\right. n + 1 \left.\right)\)