(1-3/4)x(1-3/7)x(1-1/13)x.......x(1-3/97)x(1-3/100)
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`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`
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~#
\(\left(1-\frac{3}{4}\right)\left(1-\frac{3}{7}\right)\left(1-\frac{3}{10}\right)\left(1-\frac{1}{13}\right)...\left(1-\frac{1}{97}\right)\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}.\frac{4}{7}.\frac{7}{10}.\frac{10}{13}...\frac{94}{97}.\frac{97}{100}\)
\(=\frac{1}{100}\)
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!
( 1 - 3/4 ) x ( 1 - 3/7 ) x ( 1 - 3/10 ) x ( 1 - 3/13 ) x ......x ( 1 - 3/97 ) x ( 1 - 3/100 ) .
= 1/4 x 4/7 x 7/10 x ... x 97/100 .
Khử đi các số giống nhau .
= 1/100 nha bạn .
1 − 4 3 1 − 7 3 1 − (10 3 ... 1 − 97 3 1 − 100 3 = 4 1 . 7 4 . 10 7 ..... 97 94 . 100 97 = 4.7.10.....97.100 1.4.7.....94.97 = 100 1
\(\left(1-\frac{3}{4}\right)x\left(1-\frac{3}{7}\right)x\left(1-\frac{3}{10}\right)x\left(1-\frac{3}{13}\right)x...x\left(1-\frac{3}{97}\right)x\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}x\frac{4}{7}x\frac{7}{10}x\frac{10}{13}x...x\frac{94}{97}x\frac{97}{100}\)
\(=\frac{1}{100}\)
\(\left(1-\frac{3}{4}\right)\times\left(1-\frac{3}{7}\right)\times\left(1-\frac{3}{10}\right)...\times\left(1-\frac{3}{97}\right)\times\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}\times\frac{4}{7}\times\frac{7}{10}\times...\times\frac{94}{97}\times\frac{97}{100}\)
\(=\frac{1\times4\times7\times10\times...\times97}{1\times4\times7\times10\times...\times97\times100}\)
\(=\frac{1}{100}\)
\(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{3}{7}\right)\times\left(1-\dfrac{3}{10}\right)\times\left(1-\dfrac{3}{97}\right)\times\left(1-\dfrac{3}{100}\right)\)
\(=\dfrac{2}{3}\times\dfrac{4}{7}\times\dfrac{7}{10}\times...\times\dfrac{94}{97}\times\dfrac{97}{100}\)
\(=\dfrac{2\times4\times7\times...\times94\times97}{3\times7\times10\times...\times97\times100}\)
\(=\dfrac{2\times4}{3\times100}=\dfrac{8}{300}=\dfrac{2}{75}\)
\(A=\dfrac{2}{3}\cdot\dfrac{4}{7}\cdot\dfrac{7}{10}\cdot...\cdot\dfrac{94}{97}\cdot\dfrac{97}{100}=\dfrac{2}{3}\cdot\dfrac{1}{25}=\dfrac{2}{75}\)
1) 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
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 \(\dfrac{21}{43}\)
= \(\dfrac{630}{43}\)
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