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a, 3.5^2+15.2^2-26:2
=3.25+15.4-13
=75+60-13
=135-13
=122
b, 5^3.2-100:4+2^3.5
=125.2-25+8.5
=250-25+40
=225+40
=265
a, A = 3 . 42 - 81 : 32
A = 3 . 16 - 81 : 9
A = 48 - 9
A = 39
b, B = 1999 + [ 100 : ( 8 - 6 )2 - 22 . 3 ]
B = 1999 + [ 100 : 22 - 4 . 3 ]
B = 1999 + [ 100 : 4 - 12 ]
B = 1999 + [ 25 - 12 ]
B = 1999 + 13
B = 2012
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
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
\(\left(1.2.3.........100\right)\left(1^2+2^2+.....+100^2\right).\left(2^4-4^2\right)\)
\(=\left(1.2.3.4......100\right)\left(1^2+2^2+3^2+....+100^2\right)\left(16-16\right)\)
\(=\left(1.2.3.4......100\right)\left(1^2+2^2+3^2+...+100^2\right).0\)
\(=0\)