Giải:

\(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\) 

\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\) 

\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\) 

\(=\dfrac{1}{100}\)

#)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 !!!

\(\left(1-\frac{3}{4}\right)x\left(1-\frac{3}{7}\right)x\left(1-\frac{3}{10}\right)x\left(1-\frac{3}{13}\right)x...x\left(1-\frac{3}{97}\right)x\left(1-\frac{3}{100}\right)\)

\(=\frac{1}{4}x\frac{4}{7}x\frac{7}{10}x\frac{10}{13}x...x\frac{94}{97}x\frac{97}{100}\)

\(=\frac{1}{100}\)

\(\left(1-\frac{3}{4}\right)\times\left(1-\frac{3}{7}\right)\times\left(1-\frac{3}{10}\right)...\times\left(1-\frac{3}{97}\right)\times\left(1-\frac{3}{100}\right)\)

\(=\frac{1}{4}\times\frac{4}{7}\times\frac{7}{10}\times...\times\frac{94}{97}\times\frac{97}{100}\)

\(=\frac{1\times4\times7\times10\times...\times97}{1\times4\times7\times10\times...\times97\times100}\)

\(=\frac{1}{100}\)

\(\frac{1}{4}\times\frac{4}{7}\times.....\times\frac{94}{97}\times\frac{97}{100}\)

\(=\frac{1\times4\times7\times.....\times97}{1\times4\times7\times.....\times97\times100}\)

\(=100\)

`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`

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: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)

\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)

\(=\dfrac{4}{7}\)

1) Tính biểu thức:

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!!!

\(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{3}{7}\right)\times\left(1-\dfrac{3}{10}\right)\times\left(1-\dfrac{3}{97}\right)\times\left(1-\dfrac{3}{100}\right)\)

\(=\dfrac{2}{3}\times\dfrac{4}{7}\times\dfrac{7}{10}\times...\times\dfrac{94}{97}\times\dfrac{97}{100}\)

\(=\dfrac{2\times4\times7\times...\times94\times97}{3\times7\times10\times...\times97\times100}\)

\(=\dfrac{2\times4}{3\times100}=\dfrac{8}{300}=\dfrac{2}{75}\)

\(A=\dfrac{2}{3}\cdot\dfrac{4}{7}\cdot\dfrac{7}{10}\cdot...\cdot\dfrac{94}{97}\cdot\dfrac{97}{100}=\dfrac{2}{3}\cdot\dfrac{1}{25}=\dfrac{2}{75}\)

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:

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