đây là tính nhanh à nếu tính bình thường thì tính may tính là ra

a) 17/23 . 8/16 . 23/17. (-80) . 3/4

= (17/23 . 23/17) . (8/16 . 3/4) . (-80)

= 1 . 3/8 . (-80)

= 3/8 . (-80)

= -30

b) 5/11 . 18/29 - 5/11 . 8/29 + 5/11 . 19/29

= 5/11 . (18/29 - 8/29 + 19/29)

= 5/11 . 1

= 5/11

c)(13/23 + 1313/2323 - 131313/232323).(1/3+1/4 -7/12)

= (13/23 + 1313/2323 - 131313/232323).0

= 0

d) 1²/2x2 . 2²/2x3 . 3²/3x4 . 4²/4x5 . 5²/5x6 . 6²/6x7 . 7²/7x8 . 8²/8x9 . 9²/9x10

= 1/2 . 2/3 . 3/4 . 4/5 . 5/6 . 6/7 . 7/8 . 8/9 .9/10

= 1/10

Khó nhìn quá. Bạn thông cảm nhé!

a: \(=\left(\dfrac{5^4}{5^2\cdot7^2}\right)^{15}:\left(\dfrac{5^6}{2\cdot7\cdot31}\right)^7\)

\(=\dfrac{5^{30}}{7^{30}}:\dfrac{5^{42}}{2^7\cdot7^7\cdot31^7}\)

\(=\dfrac{5^{30}}{7^{30}}\cdot\dfrac{2^7\cdot7^7\cdot31^7}{5^{42}}=\dfrac{2^7\cdot31^7}{7^{23}\cdot5^{12}}\)

b: \(=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\)

\(=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-7\cdot4\right)}{5^{29}\cdot2^8\cdot7^{48}}=5\cdot\left(-27\right)=-135\)

\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)

\(B=\frac{300}{343}:\frac{1347}{343}\)

\(B=\frac{100}{449}\)

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)

\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)

\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)

\(A=\frac{-1}{2}+\frac{1710}{9}\)

\(A=\frac{-1}{2}+190\)

\(A=\frac{-1}{2}+\frac{380}{2}\)

\(A=\frac{379}{2}\)

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^3+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(2.7\right)^3}\)

\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^6.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

\(A=\frac{2^{12}\left(3^5-3^4\right)}{2^6\left(3^6-3^5\right)}-\frac{5^{10}\left(7^3-7^4\right)}{5^9\left(7^3+2^3.7^3\right)}\)

\(A=\frac{2^6\left(3^5-3^4\right)}{3^6-3^5}-\frac{5\left(7^3-7^4\right)}{7^3+2^3.7^3}\) ( \(\frac{5\left(7^3-7^4\right)}{7^3.1+2^3.7^3}=\frac{5\left(7^3-7^4\right)}{7^3\left(1+2^3\right)}\))

\(A=\frac{2^6\left(3^5-3^4\right)}{3^6-3^5}-\frac{5\left(7^3-7^4\right)}{7^3\left(1+2^3\right)}\)

\(A=\frac{2^6.162}{486}-\frac{5.\left(-2058\right)}{7^3.9}\)

\(\Rightarrow A=\frac{648}{486}-\frac{-12540}{3087}\)

bạn tự tính nha máy tính mình mất r

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

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

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

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

tick cho mk

\(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(=\dfrac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\dfrac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(2.7\right)^3}\)

\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

= \(\dfrac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\dfrac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}\)

= \(\dfrac{2}{3.4}-\dfrac{5\left(-6\right)}{9}\)

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

`a)4/5+5 1/2 xx (4,5-2)+7/10`

`=4/5+11/2*2,5+7/10`

`=0,8+2,2+0,7`

`=3+0,7=3,7`

`b)125%xx 17/4:(1 5/16-0,5)+2008`

`=1,25xx4,25:13/16+2008`

`=85/13+2008`

`=2014 7/13`

`c)5/11+(16/11+1)`

`=5/11+1+5/11+1`

`=2+10/11=32/11`

`d)3/17+11/4+5/8+14/17+3/8`

`=3/17+14/17+5/8+3/8+11/4`

`=1+1+11/4`

`=19/4`

a) 

\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)

<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)

<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)

b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)

= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)

c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)

= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)

d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)

= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)

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

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

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

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

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

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

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

a,

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

\(=2^2-\left(\sqrt{3}\right)^2+\dfrac{\sqrt{2}\cdot\sqrt{2^5}}{1-3}=4-3+\dfrac{\sqrt{2^6}}{-2}=1+\dfrac{8}{-2}=1+\left(-4\right)=-3\)

b,

\(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\right)\cdot\dfrac{49}{50}\)

\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\cdot\dfrac{49}{50}\)

\(=\left(1-\dfrac{1}{50}\right)\cdot\dfrac{49}{50}=\dfrac{49}{50}\cdot\dfrac{49}{50}=\dfrac{49^2}{50^2}=\dfrac{2401}{2500}\)