a) (102+112+122) : (132+142)
b) 1x2x3....x9-1x2x3...x8-1x2x3....x7x82
1x2x3...x9-1x2x3...x9-1x2x3...x7x8^2
1x2x3+4x5x6+...+100x101x 102
1x2x3...9-1x2x3...8-1x2x3...82
Đáp số và cách làm
1x2+1x2x3+1x2x3x4+1x2x3x4x5+....+1x2x3...99x100
tính nhanh
a) 1-2+3-4+5-6+7 ..............+101-102+103
b)(11x9-100+1):(1x2x3...x10)
tính hợp lý:
( 102 + 112 + 122 ) : ( 132 + 142 )
(102 + 112 + 122) : (132 + 142)
= (100 + 121 + 144) :( 169 + 196)
= 365: 365
= 1
(102+112+122) : (132+142)
các bn giúp với ạ :(
(102 + 112 + 122) : (132 + 142)
= (100 + 121 + 144) :( 169 + 196)
= 365: 365
= 1
Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)
Ta có: \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)
\(\Leftrightarrow2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{98\cdot99\cdot100}\)
\(\Leftrightarrow2A=-\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}-\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}-\dfrac{1}{4\cdot5}+...-\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
\(\Leftrightarrow2A=-\dfrac{1}{2}+\dfrac{1}{99\cdot100}\)
\(\Leftrightarrow2A=\dfrac{-1}{2}+\dfrac{1}{9900}\)
\(\Leftrightarrow2A=\dfrac{-4950}{9900}+\dfrac{1}{9900}=\dfrac{-4949}{9900}\)
hay \(A=\dfrac{-4949}{19800}\)
Xét xem các biểu thức sau có bằng nhau hay không? 102+112+122 và 132+142
Ta có: 102+112+122 = 100 + 121 + 144 = 365
132+142 = 169 + 196 = 365
Vậy 102+112+122 = 132+142