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Đáp án:

$E=\dfrac{12}{7}$

Giải thích các bước giải:

Cho `E=(1+\frac{1}{1xx3})xx(1+\frac{1}{2xx4})xx(1+\frac{1}{3xx5})xx(1+\frac{1}{4xx6})xx(1+\frac{1}{5xx7})`

`E=(1+\frac{1}{3})xx(1+\frac{1}{8})xx(1+\frac{1}{15})xx(1+\frac{1}{24})xx(1+\frac{1}{35})`

`E=(\frac{3}{3}+\frac{1}{3})xx(\frac{8}{8}+\frac{1}{8})xx(\frac{15}{15}+\frac{1}{15})xx(\frac{24}{24}+\frac{1}{24})xx(\frac{35}{35}+\frac{1}{35})`

`E=\frac{4}{3}xx\frac{9}{8}xx\frac{16}{15}xx\frac{25}{24}xx\frac{36}{35}`

`E=\frac{4}{3}xx\frac{3xx3}{4xx2}xx\frac{4xx4}{3xx5}xx\frac{5xx5}{4xx2xx3}xx\frac{3xx3xx4}{7xx5}`

`E=\frac{4xx3xx3xx4xx4xx5xx5xx3xx3xx4}{3xx4xx2xx3xx5xx4xx2xx3xx7xx5}`

`E=\frac{4xx4xx3}{2xx2xx7}`

`E=\frac{2xx2xx2xx2xx3}{2xx2xx7}`

`E=\frac{2xx2xx3}{7}`

`E=\frac{12}{7}`

Vậy giá trị của biểu thức $E$ sau khi rút gọn là `\frac{12}{7}`

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Nháp ở dấu $=$ thứ $5:$

`=\frac{\cancel{4}xx\cancel{3}xx\cancel{3}xx\cancel{4}xx4xx\cancel{5}xx\cancel{5}xx\cancel{3}xx3xx4}{\cancel{3}xx\cancel{4}xx2xx\cancel{3}xx\cancel{5}xx\cancel{4}xx2xx\cancel{3}xx7xx\cancel{5}}`

Đáp án:

`E=(12)/7` 

Giải thích các bước giải:

`E=(1+1/(1xx3))xx(1+1/(2xx4))xx(1+1/(3xx5))xx(1+1/(4xx6))xx(1+1/(5xx7))`

`E=(1+1/3)xx(1+1/8)xx(1+1/(15))xx(1+1/(24))xx(1+1/(35))`

`E=(3/3+1/3)xx(8/8+1/8)xx((15)/(15)+1/(15))xx((24)/(24)+1/(24))xx((35)/(35)+1/(35))`

`E=4/3xx9/8xx(16)/(15)xx(25)/(24)xx(36)/(35)`

`E=(4xx9xx16xx25xx36)/(3xx8xx15xx24xx35)`

`E=(2xx2xx3xx3xx4xx4xx5xx5xx6xx6)/(1xx3xx2xx4xx3xx5xx4xx6xx5xx7)`

`E=(2xx6)/7`

`E=(12)/7`

Vậy `E=(12)/7`