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12 tháng 2 2017

a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)

= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)

= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)

= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)

b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)

\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)

\(=66+44+33+22+12=177\)

c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)

= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)

= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)

= còn lại làm nốt nha! bận ròy

12 tháng 2 2017

gidkjbibvvfrxdrfdfsddf

30 tháng 1 2020

Câu 1 dễ mà :

1.2.3...9 - 1.2.3...8 - 1.2.3...7.82

= 1.2.3...8.9 - 1.2.3...8.1 - 1.2.3...7.8.8

= 1.2.3...8.( 9 - 1 - 8 )

= 1.2.3...8.0

= 0

30 tháng 1 2020

còn câu 2 và 3 thì sao

21 tháng 10 2018

ban nao co chuyen shin ko cho minh muon minh giai cho 10 bai nhu the i love pac pac

28 tháng 1 2019

\(A=\frac{2^{12}.3^4-4^5.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)

\(A=\frac{2^{12}.3^4-2^{10}.3^4}{2^{12}.3^6+2^{12}.3^5}\)

\(A=\frac{2^{10}.3^4\left(2^2-1\right)}{2^{10}.3^4\left(2^2.3^2+2^2.3\right)}\)

\(A=\frac{2^2-1}{2^2.3^2+2^2.3}\)

\(A=\frac{4-1}{36+12}\)

\(A=\frac{3}{48}=\frac{1}{16}\)

14 tháng 4 2019

\(2.THPT\)

\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)

\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=9\left(1-\frac{1}{100}\right)\)

\(A=9.\frac{99}{100}\)

\(A=\frac{891}{100}\)

\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)

\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)

\(B=\frac{1}{5}-\frac{1}{95}\)

\(B=\frac{18}{95}\)

\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)

\(D=\frac{1}{2}-\frac{1}{28}\)

\(D=\frac{13}{28}\)

11 tháng 7 2015

\(A=1+5+5^2+..+5^{49}+5^{50}\)

\(5A=5+5^2+5^3+...+5^{50}+5^{51}\)

\(5A-A=\left(5+5^2+5^3+...+5^{51}\right)-\left(1+5+5^2+...+5^{50}\right)\)

\(4A=\left(5-5\right)+\left(5^2-5^2\right)+...+\left(5^{50}+5^{50}\right)+5^{51}-1\)

\(4A=0+0+...+0+5^{51}-1\)

\(4A=5^{51}-1\)

\(A=\frac{5^{51}-1}{4}\)

2 tháng 4 2021

a)

4 . 25 – 12 . 25 + 170 : 10 

= (4 . 25) – (12 . 25) + (170 : 10) 

= 100 - 300 + 17 

= -183

b)

(7 + 33 + 32) . 4 – 3 

= (7 + 27 + 9) .4 – 3

= 43 . 4 – 3

= (43 . 4) – 3

= 45

c)

12 : {400 : [500 – (125 + 25 . 7)}

= 12 : {400 : [500 – (125 + 175)}

= 12 : (400: 200)

= 12 : 2

= 6

d)

168 + {[2.(24 + 32) - 2560] : 72}.

= 168 + [2 . (16 + 9) – 1] : 49

= 168 + 49: 49

= 168 + 1 

= 167

a) 

4 . 25 – 12 . 25 + 170 : 10 

= (4 . 25) – (12 . 25) + (170 : 10) 

= 100 - 300 + 17 

= -183

b)

(7 + 33 + 32) . 4 – 3 

= (7 + 27 + 9) .4 – 3

= 43 . 4 – 3

= (43 . 4) – 3

= 45

c)

12 : {400 : [500 – (125 + 25 . 7)}

= 12 : {400 : [500 – (125 + 175)}

= 12 : (400: 200)

= 12 : 2

= 6

d)

168 + {[2.(24 + 32) - 2560] : 72}.

= 168 + [2 . (16 + 9) – 1] : 49

= 168 + 49: 49

= 168 + 1 

= 167