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b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
1.
1+2+3+...+99+100
=[(100-1):1+1]x[(100+1):2]
=100x50,5
=5050
2.
a, x2017=x
=> x=1 hoặc x=-1
b, 2x+2=250:8
=> 2x+2=250:23
=> 2x+2=247
=> x+2=47
=> x= 45
c, 3x+3x+2=810
=> 3x+3x+2=34+36
=> x=4
chúc bạn học tốt k mình nha .
Bài 2
a) Ta có
S = \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)
S = \(\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)
Vì \(\dfrac{1}{13}< \dfrac{1}{12}\)
\(\dfrac{1}{14}< \dfrac{1}{12}\)
\(\dfrac{1}{15}< \dfrac{1}{12}\)
=> \(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}.3\)
Lại có
\(\dfrac{1}{61}< \dfrac{1}{60}\)
\(\dfrac{1}{62}< \dfrac{1}{60}\)
\(\dfrac{1}{63}< \dfrac{1}{60}\)
=> \(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}.3\)
=> S = \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) < \(\dfrac{1}{5}+\dfrac{1}{12}.3+\dfrac{1}{60}.3\)
= \(\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\) = \(\dfrac{1}{2}\)
=> đpcm
Ta có
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{2015}{2016}\)
\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{2015}{2016}\)
\(\dfrac{1}{1}-\dfrac{1}{x+2}=\dfrac{2015}{2016}\)
\(\dfrac{1}{x+2}=\dfrac{1}{1}-\dfrac{2015}{2016}\)
\(\dfrac{1}{x+2}=\dfrac{1}{2016}\)
2016 = x + 2
x = 2016 - 2
x = 2014
Vậy x = 2014 là giá trị cần tìm
Gọi A là biểu thức ta có:
CÂU1 :A = 1.2+2.3+3.4+......+99.100
3A = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
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 = 33.100.101
A = 333 300
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
cho mình **** đi
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\text{…}+\frac{1}{2^{n-1}}\)
\(2A-A=1+\frac{1}{2}+\frac{1}{2^2}+\text{…}+\frac{1}{2^{n-1}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-\text{…}-\frac{1}{2^n}\)
\(A=1-\frac{1}{2^n}\)
Vậy A < 1 với n thuộc N*
a) \(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{201.203}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{201}-\frac{1}{203}\)
\(A=\left(\frac{1}{3}-\frac{1}{203}\right):2=\frac{100}{609}\)
Các ý còn lại cx tách như vật nha
CT chung này \(\frac{x}{n\left(n+x\right)}=\frac{1}{n}-\frac{1}{n+x}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{201.203}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{201.203}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{201}-\frac{1}{203}\)
\(2A=\frac{1}{3}-\frac{1}{203}=\frac{200}{609}\)
\(A=\frac{100}{609}\)
Tương tự với b thôi.