\(A=\left(\dfrac{5.5^3}{125^2.49}\right)^{15}:\left(\dfrac{5^6}{434}\right)^7\)

\(=\left(\dfrac{5^4}{5^6.7^2}\right)^{15}.\dfrac{434^7}{5^{42}}\)

\(=\left(\dfrac{1}{5^2.7^2}\right)^{15}.\dfrac{434^7}{5^{42}}\)

\(=\dfrac{62^7.7^7}{5^{30}.7^{30}.5^{42}}\)

\(=\dfrac{62^7}{5^{72}.7^{23}}\)

Sao mình không thấy câu hỏi ?

Tuấn Anh Phan Nguyễn ( tui k giỏi = ông , lm ik )

Đoàn Đức Hiếu ( pé lm hộ cj nha , cj đag hc lại lp 4 )

a) S = 1.2 + 2.3 + 3.4 + ... + 99.100

S có thể được viết lại thành:

S = 1(2 - 0) + 2(3 - 1) + 3(4 - 2) + ... + 99(100 - 98)

= 1.2 - 0 + 2.3 - 1 + 3.4 - 2 + ... + 99.100 - 98

= (1.2 + 2.3 + 3.4 + ... + 99.100) - (0 + 1 + 2 + ... + 98)

Để tính tổng 1.2 + 2.3 + 3.4 + ... + 99.100, ta sử dụng công thức:

S = n(n+1)(2n+1)/6

Với n = 99, ta có:

S = 99.100.199/6 = 331650

Tính tổng 0 + 1 + 2 + ... + 98, ta sử dụng công thức:

S = n(n+1)/2

Với n = 98, ta có:

S = 98.99/2 = 4851

Do đó, S = 331650 - 4851 = 326799

b) B = 4924.12517.28−530.749.45529.162.748

B có thể được viết lại thành:

B = (4924.12517.28) / (530.749.45529.162.748)

B = (4924 / 530) . (12517 / 749) . (28 / 45529) . (162 / 162) . (748 / 748)

B = 9.17.28/45529 = 2^2 . 3^2 . 17 / 45529

B = 108 / 45529

c) C = (13+132+133+134).35+(135+136+137+138).39+...+(1397+1398+1399+13100).3101

C = (13(1 + 13 + 13^2 + 13^3)) . 3^5 + (13^5(1 + 13 + 13^2 + 13^3)) . 3^9 + ... + (13^97(1 + 13 + 13^2 + 13^3)) . 3^101

C = (1 + 13 + 13^2 + 13^3) . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)

C = 80 . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)

C = 80 . (13^5 . 3^4 . 3 + 13^9 . 3^8 . 3 + ... + 13^97 . 3^96 . 3)

C = 80 . (13^6 . 3^5 + 13^10 . 3^9 + ... + 13^98 . 3^97)

C = 80 . 3^5 (13^6 + 13^10 + ... + 13^98)

d) D = 3 - 3^2 + 3^3 - 3^4 + ... + 3^2017 - 3^2018

D = (3 - 3^2) + (3^3 - 3^4) + ... + (3^

a)\(\left(4-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=\left(\dfrac{4}{1}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=\left(\dfrac{20}{5}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=\dfrac{8}{5}.\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=1-\dfrac{2}{5}.\dfrac{25}{-4}\)

\(=1-\dfrac{-5}{2}\)

\(=\dfrac{2}{2}-\dfrac{-5}{2}\)

\(=\dfrac{7}{2}\)

dài quá nên mik sẽ giải lần lượt mỗi câu trả lời là một câu nhá bạn!!

Giải:

a)(4-12/5).25/8-2/5:-4/25

=8/5.25/8-(-5/2)

=5+5/2

=15/2

b)(-5/24+3/4-7/12):(-5/16)

=-1/24:(-5/16)

=2/15

c)6/7+5/4:(-5)-(-1/28).(-2)²

=6/7+(-1/4)-(-1/28).4

=6/7-1/4-(-1/7)

=6/7-1/4+1/7

=(6/7+1/7)-1/4

=1-1/4

=3/4

Chúc bạn học tốt!

A -\(\dfrac{24}{25}\)

B -\(\dfrac{5}{21}\)

C -\(\dfrac{24}{47}\)

D -\(\dfrac{19}{42}\)

tick cho mk

a, \(125^3:5^7=\left(5^3\right)^3:5^7=5^9:5^7=5^2\)

b, \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{4}{49}\right)^5:\left(\dfrac{8}{343}\right)^2\)

= \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2^2}{7^2}\right)^5:\left(\dfrac{2^3}{7^3}\right)^2\)

= \(\left(\dfrac{2}{7}\right)^{18}:\left[\left(\dfrac{2}{7}\right)^2\right]^5:\left[\left(\dfrac{2}{7}\right)^3\right]^2\)

=\(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2}{7}\right)^{10}:\left(\dfrac{2}{7}\right)^6\)

= \(\left(\dfrac{2}{7}\right)^{18-10-6}=\left(\dfrac{2}{7}\right)^2\)

c, \(3-\left(\dfrac{-7}{9}\right)^0+\left(\dfrac{1}{3}\right)^5.3^5\)

= 3 - 1 +\(\left[\left(\dfrac{1}{3}\right)^5.3^5\right]\)

= 2 + 1=3

d, \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(9.5\right)^{10}.5^{20}}{\left(25.3\right)^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}\)

= \(\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5\)

a) \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{4}{-9}+\dfrac{7}{15}\)

=\(\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{-2}{11}\)

=\(\left(-1\right)+1+\dfrac{-2}{11}\)

=\(\dfrac{-2}{11}\)

b) \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)

=\(\dfrac{-5}{12}+\dfrac{6}{11}+\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\)

=\(\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{7}{17}\)

=\(0+0+\dfrac{7}{17}\)

=\(\dfrac{7}{17}\)

c) A= \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

A=\(49\dfrac{8}{23}-5\dfrac{7}{32}-14\dfrac{8}{23}\)

A=\(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)-5\dfrac{7}{32}\)

A=\(35-5\dfrac{7}{32}\)

A=\(35-\dfrac{167}{32}=\dfrac{953}{32}\)

d) C=\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)

C=\(\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{7}\)

C=\(\dfrac{-3}{7}.1+\dfrac{17}{7}\)

C=\(\dfrac{-3}{7}+\dfrac{17}{7}=2\)

a, `(-5)/9+8/15+(-2)/11+4/(-9)+7/15`

`=-5/9+8/15-2/11-4/9+7/15`

`=(-5/9-4/9)+(8/15+7/15)-2/11`

`=-9/9+15/15-2/11`

`=-1+1-2/11`

`=-2/11`

b, `((-5)/12+6/11)+(7/17+5/11+5/12)`

`=-5/12+6/11+7/17+5/11+5/12`

`=(-5/12+5/12)+(6/11+5/11)+7/17`

`=0+11/11+7/17`

`=1+7/17`

`=17/17+7/17`

`=24/17`

c, `A=49 8/23 - (5 7/32 + 14 8/23)`

`A=49 8/23 - 5 7/32 - 14 8/23`

`A=(49 8/23 - 14 8/23)-5 7/32`

`A=35 - 167/32`

`A=953/32`

d, `C=(-3)/7.5/9+4/9.(-3)/7+2 3/7`

`C=-3/7 . 5/9-4/9 . 3/7+17/7`

`C=-3/7.(5/9+4/9)+17/7`

`C=-3/7 . 1+17/7`

`C=2`

a) Ta có: \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{-4}{9}+\dfrac{7}{15}\)

\(=\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)-\dfrac{2}{11}\)

\(=-1+1-\dfrac{2}{11}\)

\(=-\dfrac{2}{11}\)

b) Ta có: \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{11}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)

\(=\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{7}{11}+\dfrac{5}{11}\right)\)

\(=\dfrac{18}{11}\)

c) Ta có: \(A=49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

\(=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}\)

\(=30-\dfrac{7}{32}=\dfrac{953}{32}\)

a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

=0

\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)

\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)

\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)

\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)

\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)