\(10^3\)+ 5.\(10^2\)+ 7.10+

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4 . 103 + 5 . 102 + 7 . 10

= 10 . (4 . 102 + 5 . 10 + 7)

= 10 . (400 + 50 + 7)

= 10 . 457 

= 4570

   4.10+ 5.102 + 7.10

= 4.10.102 + 5.102 + 70

= 40.102 + 5.102 + 70

= 102.(40 + 5) +70

= 100.45 + 70

= 4500 + 70

= 4570

25 tháng 9 2019

2\(^1\)=2

4\(^2\)= 16

8 = 8

10\(^3\)= 1000

3 = 3

5\(^2\)= 25

7\(^2\)= 49

9\(^2\)= 81

xin thank 

học tốt

22 tháng 10 2019

a) \(47-\left[\left(45\cdot2^4-5^2\cdot12\right):14\right]\)

\(=47-\left[\left(45\cdot16-25\cdot12\right):14\right]\)

\(=47-\left[\left(720-300\right):14\right]\)

\(=47-\left[420:14\right]\)

\(=47-30=17\)

b) \(50-\left[\left(20-2^3\right):2+34\right]\)

\(=50-\left[\left(20-8\right):2+34\right]\)

\(=50-\left[12:2+34\right]\)

\(=50-\left[6+34\right]\)

\(=50-40=10\)

c) \(10^2-\left[60:\left(5^6:5^4-3\cdot5\right)\right]\)

\(=100-\left[60:\left(5^{6-4}-15\right)\right]\)

\(=100-\left[60:\left(5^2-15\right)\right]\)

\(=100-\left[60:\left(25-15\right)\right]\)

\(=100-\left[60:10\right]\)

\(=100-6=94\)

d) \(50-\left[\left(50-2^3\cdot5\right):2+3\right]\)

\(=50-\left[\left(50-8\cdot5\right):2+3\right]\)

\(=50-\left[\left(50-40\right):2+3\right]\)

\(=50-\left[10:2+3\right]\)

\(=50-\left[5+3\right]\)

\(=50-8=42\)

e) \(10-\left[\left(8^2-48\right)\cdot5+\left(2^3\cdot10+8\right)\right]:28\)

\(=10-\left[\left(64-48\right)\cdot5+\left(8\cdot10+8\right)\right]:28\)

\(=10-\left[16\cdot5+\left(80+8\right)\right]:28\)

\(=10-\left[80+88\right]:28\)

\(=10-168:28\)

\(=10-6=4\)

f) \(8697-\left[3^7:3^5+2\left(13-3\right)\right]\)

\(=8697-\left[3^{7-5}+2\cdot10\right]\)

\(=8697-\left[3^2+20\right]\)

\(=8697-\left[9+20\right]\)

\(=8697-29=8668\)

CHÚC BẠN HỌC TỐT!!!!!!!!!!!

21 tháng 10 2017

\(\dfrac{5.4^6.9^5+6^{10}.10^2}{8^4.3^{12}-6^{11}}=\dfrac{5.\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^{10}.\left(5.2\right)^2}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)

=\(\dfrac{5.2^{12}.3^{10}+2^{10}.3^{10}.5^2.2^2}{2^{12}.3^{12}-2^{11}.3^{11}}=\dfrac{5.2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3-1\right)}\)

=\(\dfrac{5.2.6}{1.3.5}=4\)

20 tháng 1 2017

\(\frac{7256.4375-725}{4375.7255+3650}=\frac{\left(7255+1\right).4375-725}{4375.7255+3650}=\frac{7255.4375+4375-725}{7255.4375+3650}=\frac{7255.4375+3650}{7255.4375+3650}=1\)

\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{16}=3\)

\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.104}=\frac{2^2.78}{26.2^2}=\frac{78}{26}=3\)

\(\left(125^3.7^5-175^5.5\right):2001^{2002}\) ( bạn xem lại đề xem sai đâu ko nhé )

Để Thiên giải câu 3 cho:

(1253.75 -1755;5):20012001

\(=\left[\left(5^3\right)^3.7^5-175^5:5\right]:2001^{2002}\)

\(=\left(5^9.7^5-175:5\right):2001^{2002}\)

\(=\left(5^5.5^4.7^4.7-175^4.175:5\right):2001^{2002}\)

\(=\left(5^5.35^4.7-175^4.35\right):2001^{2002}\)

\(=\left(5^4.35^4.5.7-175^4.35\right):2001^{2002}\)

\(=\left(175^4.35-175^4.35\right):2001^{2002}\)

\(=0:2001^{2002}\)

\(=0\)

1 tháng 8 2020

thanks friend!vui

a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)

\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)

\(=4-\frac{4}{7}=\frac{24}{7}\)

b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)

\(=7+\frac{4}{9}+4+\frac{7}{11}-3-\frac{4}{9}\)

\(=8+\frac{7}{11}=\frac{95}{11}\)

c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)

\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)

\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)

\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)

d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)

\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)

\(=5\cdot\frac{7}{35}=1\)

e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)

\(=\frac{42}{43}\)

25 tháng 3 2018

Bài nhìn vô muốn xỉu rồi ='((

1. a) \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)

\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}+\frac{3}{94.97}\right)\)

\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\right)\)

\(=\frac{2}{3}\left(1-\frac{1}{97}\right)=\frac{2}{3}.\frac{96}{97}=\frac{64}{97}\)

b) Bạn tự làm, làm nữa chắc xỉu =((( Khi nào rảnh mình sẽ làm, nếu bạn cần

2 ) 

a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:2=\frac{1005}{4022}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{1005}{4022}=\frac{3017}{4020+2}\)

\(\Rightarrow x=4020\)

24 tháng 3 2018

tu ma lam nguoi ta con gap hon min nhieu