152 . 164 - 153. 163
122 . 203 - 202 . 123
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Xét B = \(\frac{201+202+203}{202+203+204}\)
= \(\frac{201}{202+203+204}\)\(+\)\(\frac{202}{202+203+204}\)\(+\)\(\frac{203}{202+203+204}\)
Vì 202 < 202 + 203 + 204
=> \(\frac{201}{202}\)> \(\frac{201}{202+203+204}\)( 1 )
Vì 203 < 202 + 203 + 204
=> \(\frac{202}{203}\)>\(\frac{202}{202+203+204}\)( 2 )
Vì 204 < 202 + 203 + 204
=> \(\frac{203}{204}\)> \(\frac{203}{202+203+204}\)( 3 )
Cộng vế với vế của ( 1 ), ( 2 ) và ( 3 )
=> \(\frac{201}{202}+\frac{202}{203}+\frac{203}{204}\)> \(\frac{201+202+203}{202+203+204}\)
=> A > B
Vậy A > B
Xét B = \(\frac{201+202+203}{202+203+204}\)
= \(\frac{201}{202+203+204}\)+ \(\frac{202}{202+203+204}\)+ \(\frac{203}{202+203+204}\)
Vì 202 < 202 + 203 + 204 nên \(\frac{201}{202}\)>\(\frac{201}{202+203+204}\)(1)
Vì 203 < 202 + 203 + 204 nên \(\frac{202}{203}\)> \(\frac{202}{202+203+204}\)(2)
Vì 204 < 202 + 203 + 204 nên \(\frac{202}{203}\)>\(\frac{202}{202+203+204}\)(3)
Cộng vế vơi vế của (1) , (2) và (3)
=>\(\frac{201}{202}+\frac{202}{203}+\frac{203}{204}\)> \(\frac{201+202+203}{202+203+204}\)
=> A > B
Vậy A > B
Ta có :
202203 = 8 242 408101 ( 1 )
203202 = 42 209101 ( 2 )
Từ ( 1 ) và ( 2 ) suy ra 202203 < 203202
..................tên em là jullei Trinh...........................
Ta có : \(202^{203}=(2\cdot101)^{3\cdot101}=(1^3\cdot101^3)^{101}=(8\cdot101\cdot10^{12}\cdot101)=(808\cdot1012)^{101}\)
\(303^{202}=(3\cdot101)^{2\cdot101}=(32\cdot101^2)^{101}=(9\cdot101^2)^{101}\)
\(\Rightarrow(808\cdot101^2)>(9\cdot101^2)\)
Vậy :
\(S=1\cdot2+2\cdot3+3\cdot4+...+39\cdot40\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+39\cdot40\cdot\left(41-38\right)\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+39\cdot40\cdot41-38\cdot39\cdot40\)
\(3S=39\cdot40\cdot41\)\(\Rightarrow S=\dfrac{39\cdot40\cdot41}{3}=21320\)
Ta có : \(S=1.2+2.3+3.4+...+38.39+39.40\)
\(3S=1.2.3+2.3.3+3.4.3+...+38.39.3+39.40.3\)
\(3S=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+39.40.\left(41-38\right)\) \(3S=39.40.41\)
\(S=\dfrac{39.40.41}{3}\)\(=21320\)
S = 1 x 2 + 2 x 3 + 3 x 4 + ... + 38 x 39 + 39 x 40
3S = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 38 x 39 x 3 + 39 x 40 x 3
3S = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + ... + 38 x 39 x ( 40 - 37 ) + 39 x 40 x ( 41 - 38 )
3S = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + ... + 38 x 39 x 40 - 37 x 38 x 39 + 39 x 40 x 41 - 38 x 39 x 40
S = 39 x 40 x 41 : 3
S = 21320
S=1x2+2x3+3x4+...+38x39+39x40
3S=(1x2+2x3+3x4+...+38x39+39x40)x3
3S=1x2x3+2x3x3+3x4x3+....+39x40x3
3S=1x2x3+2x3x(4-1)+3x4x(5-2)+...+39x40x(41-38)
3S=1x2x3+2x3x4-1x2x3+3x4x5-2x3x4+...+39x40x41-38x39x40
3S=39x40x41
3S=63960
S=63960:3
S=21320
\(202^{303}=\left(2.101\right)^{3.101}=\left(2^3.101^3\right)^{101}=\left(8.101^3\right)^{101}\)
\(303^{202}=\left(3.101\right)^{2.101}=\left(3^2.101^2\right)^{101}=\left(9.101^2\right)^{101}\)
Mà \(8.101^3>9.101^2\)
\(\Rightarrow202^{303}>303^{202}\)
S = 1.2 + 2.3 + 3.4 + ... + 38.39 + 39.40
3S = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 38.39.(40-37) + 39.40.(41-38)
3S = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 38.39.40 - 37.38.39 + 39.40.41 - 38.39.40
3S = 39.40.41
S = 13.40.41
S = 21320
\(\frac{15^2.16^4-15^3.16^3}{12^2.20^3-20^2.12^3}\)
\(=\frac{15^2.16^3.\left(16-15\right)}{12^2.20^2.\left(20-12\right)}\)
\(=\frac{15^2.16^3}{12^2.20^2.8}\)
\(=2\)
học tốt
152 . 164 - 153. 163
= 152.162.(162-15.16)
=(15.16)2.16.(16-15)
=2402.16
=57600.16
= 921600