Kết quả của dãy tính sau bằng cách nhanh nhất :
( 1 - 3/4 ) x ( 1 - 3/7 ) x ( 1 - 3/10 ) x ( 1 - 3/13 ) x ......x ( 1 - 3/97 ) x ( 1 - 3/100 )
Tính bằng cách thuận tiện:
(1-\(\dfrac{3}{4}\)) x (1 - \(\dfrac{3}{7}\)) x (1 - \(\dfrac{3}{10}\)) x (1 - \(\dfrac{3}{13}\)) x ....... x (1 - \(\dfrac{3}{97}\)) x (1 - \(\dfrac{3}{100}\))
Giải:
\(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\)
\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\)
\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\)
\(=\dfrac{1}{100}\)
Bài 12 : Tính nhanh
a) (1-3/4) x (1-3/7) x (1-3/10) x (1-1/13) x ........ x (1- 3/97) x ( 1-3/100)
#)Giải :
\(\left(1-\frac{3}{4}\right)x\left(1-\frac{3}{7}\right)x\left(1-\frac{3}{10}\right)x\left(1-\frac{1}{13}\right)x...x\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}x\frac{4}{7}x\frac{7}{10}x...x\frac{94}{97}x\frac{97}{100}\)
\(=\frac{1x4x7x...x94x100}{4x7x10x...x97x100}\)
\(=\frac{1}{100}\)
#~Will~be~Pens~#
\(\left(1-\frac{3}{4}\right)\left(1-\frac{3}{7}\right)\left(1-\frac{3}{10}\right)\left(1-\frac{1}{13}\right)...\left(1-\frac{1}{97}\right)\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}.\frac{4}{7}.\frac{7}{10}.\frac{10}{13}...\frac{94}{97}.\frac{97}{100}\)
\(=\frac{1}{100}\)
\(\left(1-\frac{3}{4}\right)\left(1-\frac{3}{7}\right)\left(1-\frac{3}{10}\right)....\left(1-\frac{3}{100}\right)\)
\(\frac{1}{4}.\frac{4}{7}.\frac{7}{10}....\frac{94}{97}.\frac{97}{100}\)
Rút gọn đi (gạch chéo ý)
\(=\frac{1}{100}\)
Chúc bạn học tốt !!!
1. tính nhanh:
( 1 - 3/4 ) x ( 1 - 3/7 ) x ( 1 - 3/10) x ( 1 - 3/13) x......x ( 1 - 3/97 ) x ( 1 - 100)
\(\left(1-\frac{3}{4}\right)x\left(1-\frac{3}{7}\right)x\left(1-\frac{3}{10}\right)x\left(1-\frac{3}{13}\right)x...x\left(1-\frac{3}{97}\right)x\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}x\frac{4}{7}x\frac{7}{10}x\frac{10}{13}x...x\frac{94}{97}x\frac{97}{100}\)
\(=\frac{1}{100}\)
\(\left(1-\frac{3}{4}\right)\times\left(1-\frac{3}{7}\right)\times\left(1-\frac{3}{10}\right)...\times\left(1-\frac{3}{97}\right)\times\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}\times\frac{4}{7}\times\frac{7}{10}\times...\times\frac{94}{97}\times\frac{97}{100}\)
\(=\frac{1\times4\times7\times10\times...\times97}{1\times4\times7\times10\times...\times97\times100}\)
\(=\frac{1}{100}\)
\(\frac{1}{4}\times\frac{4}{7}\times.....\times\frac{94}{97}\times\frac{97}{100}\)
\(=\frac{1\times4\times7\times.....\times97}{1\times4\times7\times.....\times97\times100}\)
\(=100\)
tính nhanh
a (1-1/2) x (1-1/3) x (1-1/4) x (1-1/5)
b (1-3/4)x ( 1-3/7) x ( 1-3/10) x (1-3/13) x...x (1-3/97) x (1-3/100)
`a)(1-1/2)xx(1-1/3)xx(1-1/4)xx(1-1/5)`
`=1/2xx2/3xx3/4xx4/5`
`=[1xx2xx3xx4]/[2xx3xx4xx5]`
`=1/5`
`b)(1-3/4)xx(1-3/7)xx(1-3/10)xx(1-3/13)xx .... xx(1-3/97)xx(1-3/100)`
`=1/4xx4/7xx7/10xx10/13xx .... xx94/97xx97/100`
`=[1xx4xx7xx10xx...xx94xx97]/[4xx7xx10xx13xx....xx97xx100]`
`=1/100`
Tính bằng cách thuận tiên nhất:
1):(3/4 x 5/97 + 1/9 x 13/47) x (1/5 - 7/25 x 5/7)
2): 8/17 x 4/15 + 8/17 x 22/15 - 8/15 x 9/17
3): 2021/2 x 1/3 + 4042/4 x 1/5 + 6063/3 x 22/15
4); 4/7 x 2/13 + 8/13 :7/4 + 4/7 : 13/2 + 4/7 x 1/13
5): 2022 x 2021 - 1/ 2021 + 2022 x 2020
6): 18 x 123 + 9 x 4567 x 2 + 3 x 5310 x 6 / (2 + 4 + 6 + 8 + ...+20 + 22) + 48
7): A= 2021 x 2021 x 202020 - 2020 x 2020 x 20212021 / 2020 x 20192019
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
Bài 1: Tính nhanh
a) (1-1/2) x (1-1/3)x (1-1/4) x (1-1/5)
b) (1-3/4) x (1-3/7) x (1-3/10) x (1-3/13)x...x (1-3/97) x (1-3/100)
Các anh các cj giúp em vs ạ
Các anh các cj giúp em nhé
Em cảm ơn trước
Giải:
a) \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)
\(=\dfrac{1.2.3.4}{2.3.4.5}\)
\(=\dfrac{1}{5}\)
b) \(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\)
\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\)
\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\)
\(=\dfrac{1}{100}\)
Chúc bạn học tốt!
A)
\(\left(1-\dfrac{1}{2}\right)x\left(1-\dfrac{1}{3}\right)x\left(1-\dfrac{1}{4}\right)x\left(1-\dfrac{1}{5}\right)\)
\(=\dfrac{1}{2}x\dfrac{1}{3}x\dfrac{1}{4}x\dfrac{1}{5}\)
\(=\dfrac{1x2x3x4}{2x3x4x5}\)
\(=\dfrac{1}{5}\)
B)
\(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right)......\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\)
\(\dfrac{1}{4}.\dfrac{1}{7}.....\dfrac{94}{97}.\dfrac{97}{100}\)
\(\dfrac{1}{100}\)
học tốt nhé!!!
Bài 1:(ghi rõ ra nhé) Tính nhanh
A, (1-1/3)x(1-3/7)x(1-3/10)x(1-3/13)x...x(1-3/97)x(1-3/100)
x là nhân nhá
\(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{3}{7}\right)\times\left(1-\dfrac{3}{10}\right)\times\left(1-\dfrac{3}{97}\right)\times\left(1-\dfrac{3}{100}\right)\)
\(=\dfrac{2}{3}\times\dfrac{4}{7}\times\dfrac{7}{10}\times...\times\dfrac{94}{97}\times\dfrac{97}{100}\)
\(=\dfrac{2\times4\times7\times...\times94\times97}{3\times7\times10\times...\times97\times100}\)
\(=\dfrac{2\times4}{3\times100}=\dfrac{8}{300}=\dfrac{2}{75}\)
\(A=\dfrac{2}{3}\cdot\dfrac{4}{7}\cdot\dfrac{7}{10}\cdot...\cdot\dfrac{94}{97}\cdot\dfrac{97}{100}=\dfrac{2}{3}\cdot\dfrac{1}{25}=\dfrac{2}{75}\)
1 tính nhanh
a, 7/19 x 1/3 + 7/19 x 2/3 b , 15 x 2121/4343 + 15 x 212121/434343 c, 1991/1990 x 1992/1991 x 1993/1992 x 1994/1993 x 995/994
b, [ 1 - 3/4 ] x [ 1 - 3/7 ] x [ 1 -3/10 ] x [ 1 - 13 ] x .........x [ 1 - 3/97 ] x [ 1 - 3/100 ]
có lời giải nha các bạn
Bài 1
a; \(\dfrac{7}{19}\) x \(\dfrac{1}{3}\) + \(\dfrac{7}{19}\) x \(\dfrac{2}{3}\)
= \(\dfrac{7}{19}\) x (\(\dfrac{1}{3}+\dfrac{2}{3}\))
= \(\dfrac{7}{19}\) x 1
= \(\dfrac{7}{19}\)
b; 15 x \(\dfrac{2121}{4343}\) + 15 x \(\dfrac{212121}{434343}\)
= 15 x \(\dfrac{21}{43}\) + 15 x \(\dfrac{21}{43}\)
= 15 x \(\dfrac{21}{43}\) x (1 + 1)
= 15 x \(\dfrac{21}{43}\) x 2
= (15 x 2) x \(\dfrac{21}{43}\)
= 30 x \(\dfrac{21}{43}\)
= \(\dfrac{630}{43}\)
C; \(\dfrac{1991}{1990}\) x \(\dfrac{1992}{1991}\) x \(\dfrac{1993}{1992}\) x \(\dfrac{1994}{1993}\) x \(\dfrac{995}{994}\)
= \(\dfrac{1991\times1992\times1993\times1994\times995}{1991\times1992\times1993\times994\times995\times2}\)
= \(\dfrac{1994}{994\times2}\)
= \(\dfrac{997}{994}\)
Bài 1 : Tính bằng cách thuận tiện :
a) 2/19 + 2/10 + 2/22 + 17/19 + 2/11 + 4/5+ 8/11
b) 3/9 + 4/12 + 6/18 + 1/3 + 5/15 + 7/21
c) 100 + ( 125 x3 - 125 x2 - 125 ) x (1 + 3 + 5 + 7 + ...+ 97 + 99)
Bài 2 : Tìm số có 2 chữ số , biết rằng số đó lớn gấp 6 lần tổng các chữ số của nó
Bài 3 : Không làm phép tính , hãy cho biết kết quả phép tính sau có chữ số tận cùng bằng chữ số nào? Vì sao ?
P= 11 X 13 X 15 X 17 + 23 X 25 X 27 X 29 + 33 X 35 X 37 X 39
Bài 1:
a) 2/19 + 2/10 + 2/22 + 17/19 + 2/11 + 4/5 + 8/11
=(2/19 +17/19) + 1/5 + 1/11 + 2/11 + 4/5 + 8/11
= 1 + (1/5 + 4/5) + (2/11 + 8/11 + 1/11)
= 1 + 1 + 1 = 3
b) 3/9 + 4/12 + 6/18 + 1/3 + 5/15 + 7/21
= 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3
= 1/3 x 6 = 2
c) 100 + (125x3-125x2-125) x (1 + 3 + 5 + 7 + ...+ 97 + 99)
= 100 + [125x(3-2-1)] x A
= 100 + (125x0) x A
= 100 + 0 x A
= 100 + 0
= 100
Bài 2:
Gọi số đó là ab
(a+b) x 6 = ab
a x 6 + b x 6= a x 10 + b
b x 5 = a x 4
suy ra a=5; b=4; ab=54
Bài 3:
Vì các số lẻ x 5 đều có tận cùng là 5 nên các tích đều có tận cùng là 5.
Mà 5x3=15 nên P có tận cùng là 5
Bài 1:
a) 2/19 + 2/10 + 2/22 + 17/19 + 2/11 + 4/5 + 8/11
=(2/19 +17/19) + 1/5 + 1/11 + 2/11 + 4/5 + 8/11
= 1 + (1/5 + 4/5) + (2/11 + 8/11 + 1/11)
= 1 + 1 + 1 = 3
b) 3/9 + 4/12 + 6/18 + 1/3 + 5/15 + 7/21
= 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3
= 1/3 x 6 = 2
c) 100 + (125x3-125x2-125) x (1 + 3 + 5 + 7 + ...+ 97 + 99)
= 100 + [125x(3-2-1)] x A
= 100 + (125x0) x A
= 100 + 0 x A
= 100 + 0
= 100
Bài 2:
Gọi số đó là ab
(a+b) x 6 = ab
a x 6 + b x 6= a x 10 + b
b x 5 = a x 4
suy ra a=5; b=4; ab=54
Bài 3:
Vì các số lẻ x 5 đều có tận cùng là 5 nên các tích đều có tận cùng là 5.
Mà 5x3=15 nên P có tận cùng là 5