tìm B biết
B= (1-1/4)(1-1/9)(1-1/16).........(1-1/400)
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tìm B=(1/4-1)(1/9-1)(1/16-1)...(1/400-1
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-399}{400}\)
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Tìm đa thức b biết
B-(3\(x^6\)-4\(xy^5\)+\(\dfrac{1}{3}\)\(xy^2\)-\(\dfrac{3}{2}\))=(7\(x^6\)-\(\dfrac{1}{2}xy^5-xy^2-\dfrac{1}{3}\))
B-(\(3x^6-4xy^5+\dfrac{1}{3}xy^2\))=
B= \(\left(7x^6-\dfrac{1}{2}xy^5-xy^2-\dfrac{1}{3}\right)+\left(3x^6-4xy^5+\dfrac{1}{3}xy^2-\dfrac{3}{2}\right)\)
B= \(7x^6-\dfrac{1}{2}xy^5-xy^2-\dfrac{1}{3}+3x^6-4xy^5+\dfrac{1}{3}xy^2-\dfrac{3}{2}\)
B= \(7x^6+3x^6-\dfrac{1}{2}xy^5-4xy^5-xy^2+\dfrac{1}{3}xy^2-\dfrac{1}{3}+\dfrac{2}{3}\)
B= \(10x^6-\dfrac{9}{2}xy^5-\dfrac{2}{3}xy^2+\dfrac{1}{3}\)
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A=(1-1/4)×(1-1/9)×(1-1/16)×...×(1-1/400)
Thực hiện phép tính
B = ( 1/4-1) .( 1/9-1).(1/16-1)x....x(1/400-1)
7 tháng 3 2021 lúc 9:35
Tính
B = \(\left(\dfrac{1}{4}-1\right)+\left(\dfrac{1}{9}-1\right)+\left(\dfrac{1}{16}-1\right)....\left(\dfrac{1}{400}-1\right)\)
Sửa đề: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\cdot...\cdot\left(\dfrac{1}{400}-1\right)\)
Ta có: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\cdot...\cdot\left(\dfrac{1}{400}-1\right)\)
\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-399}{400}\)
\(=\dfrac{-3\cdot8\cdot15\cdot...\cdot399}{4\cdot9\cdot16\cdot...\cdot400}\)
\(=\dfrac{-3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot19\cdot21}{2^2\cdot3^2\cdot4^2\cdot...\cdot20^2}\)
\(=\dfrac{-2\cdot3\cdot4\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot21}{2\cdot3\cdot4\cdot20}\)
\(=\dfrac{-1}{20}\cdot\dfrac{21}{2}\)
\(=\dfrac{-21}{40}\)
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1-3+5-7+9-...-1999-2000+2001
1+2+3-4+5+6+7-8+...+1999-2000 + 2001
B= 1-7+13-19+25-31+....; biếtB có 2007 số hạng
a=2020. 2020-2022.2018
b= (1/4 -1).(1/9-1).(1/16-1)....(1/400-1)
a) $2020.2020-2022.2018$
$ = 2020^2-(2020+2).(2020-2)$
$ = 2020^2 - (2020^2-2^2)$
$ = 4$
b) \(\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)...\left(\dfrac{1}{400}-1\right)\)
\(=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{20^2}-1\right)\)
\(=\dfrac{\left(-1\right)\cdot3\cdot\left(-2\right)\cdot4\cdot\left(-3\right)\cdot5\cdot\cdot\cdot\left(-19\right)\cdot21}{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot20^2}\)
\(=-\dfrac{1}{20}\cdot\dfrac{21}{2}=-\dfrac{21}{40}\)
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Giải:
a) 2020.2020−2022.2018
=20202−(2020+2).(2020−2)
=20202−(20202−22)
=4
b) =(1/22−1)(1/32−1)(1/42−1)...(1/202-1)
=−1/20⋅21/2
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(1/4-1)nhân(1/9-1)nhân(1/16-1)nhân....nhân(1/400-1)
(1/4-1)(1/9-1)(1/16-1)..........(1/400-1)
tính
=-3/4×(-8/9)×(-15/16)×....×(-399/400)
=-3×8×15×....×399/4×9×16×...×400
=-3×2×4×3×5×.....×21×19/2^2×3^2×4^2×....×20^2
=-2×3×4×...×19/2×3×4×...×20 × 3×4×5×....×21/2×3×4×....×20
=-1/20×21/20
=-21/40
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