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Bài 1:
Ta thấy A < 1
=> A = \(\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}=B\)
Vậy A < B
Bài 2:
Ta thấy C < 1
=> C = \(\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=D\)
Vậy C < D
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+..............+\frac{1}{99^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+................+\frac{1}{98.99}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+............+\frac{1}{98}-\frac{1}{99}\)
\(=1-\frac{1}{99}=\frac{98}{99}< 1\)
\(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.............+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...............+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
Vậy \(\frac{49}{100}< A< 1\)
\(1:\frac{99}{100}:\frac{98}{99}:\frac{97}{98}:.........:\frac{2}{3}:\frac{1}{2}\)
\(=1.\frac{100}{99}.\frac{99}{98}.\frac{98}{97}......\frac{3}{2}.\frac{2}{1}\)
\(=\frac{1.100.99.98....3.2}{99.98.97......2.1}\)
\(=100\)
E = 12 + 22 + 32 + ... + 982 + 992
E = 1 . ( 2 - 1 ) + 2 . ( 3 - 1 ) + 3 . ( 4 - 1 ) + ... + 98 . ( 99 - 1 ) + 99 . ( 100 - 1 )
E = 1 . 2 - 1 + 2 . 3 - 2 + 3 . 4 - 3 + ... + 98 . 99 - 98 + 99 . 100 - 99
E = 1 . 2 + 2 . 3 + 3 . 4 + ... + 98 . 99 + 99 . 100 - 1 - 2 - 3 - ... - 98 - 99
E = ( 1 . 2 + 2 . 3 + 3 . 4 + ... + 98 . 99 + 99 . 100 ) - ( 1 + 2 + 3 + ... + 98 + 99 )
Biểu thức trong ngoặc đầu bằng 333300, biểu thức trong dấu ngoặc sau bằng : 99 . 100 : 2 = 4950
=> E = 333300 - 4950 = 328350
=1.1+2.2+3.3+.......+99.99
\(=1\left(2-1\right)+2\left(3-1\right)+.....+99\left(100-1\right)\)\(=1.2-1+2.3-2+....+99.100-99\)
\(=1.2+2.3+...+99.100+1-2-3-...-99\)
\(=1.2.3+2.3.4+...+99.100.101+1-\left(2+3+...+99\right)\)
\(=99.100.101+1-\left(\frac{\left(2+99\right)98}{2}\right)\)
=99.100.101+(-4948)
=999900-4948
=994952
1. a) 2B = 1 + 1/2 + 1/22+...+1/298
B - B = (1+1/2+...+1/298) - (1/2+....+1/299)
B = 1 - 299 => B < 1
b) Làm tương tự như câu a, ra là (1 - 1/399) : 2 = 1/2 - 1/2.399(C bé hơh 1/2)
1. a). Theo đầu bài ta có:
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
\(\Leftrightarrow B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
\(\Leftrightarrow B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\)
\(\Leftrightarrow B=1-\frac{1}{2^{99}}< 1\)( đpcm )
Ta có:
2A = 2 + 22 +...+2100
2A - A = (2-2) + (22 - 22) +....+2100 - 1
A = 2100 - 1= 4.298 - 1
Ta so sánh: 5.298 và 4.298 - 1
Vì 298 = 298 ; 5 > 4
=> 5.298 > A