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18 tháng 8 2024

\(\left(3\cdot4\cdot2^{16}\right)^2:\left(11\cdot2^{13}\cdot4^{11}-16^9\right)\\ =\left(3\cdot2^2\cdot2^{16}\right)^2:\left(11\cdot2^{13}\cdot2^{22}-2^{36}\right)\\ =3^2\cdot2^4\cdot2^{32}:\left(11\cdot2^{35}-2^{36}\right)\\ =3^2\cdot2^{36}:\left[2^{35}\cdot\left(11-2\right)\right]\\ =9\cdot2^{36}:\left(2^{35}\cdot9\right)\\ =9\cdot2^{36}:2^{35}:9\\ =2\)

\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=2\)

21 tháng 5 2017

d)

đặt A = 1 + 2 + 22 + ... + 280 

2A = 2 + 22 + 23 + ... + 281

2A - A = ( 2 + 22 + 23 + ... + 281 ) - ( 1 + 2 + 22 + ... + 280 )

A = 281 - 1 > 281 - 2

e) 

đặt \(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{899}{900}\)

\(A=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{900}\right)\)

\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\right)\)

\(A=29-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\right)\)

đặt \(B=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\)

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}\)

\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{29.30}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{29}-\frac{1}{30}\)

\(=1-\frac{1}{30}=\frac{29}{30}< 1\)

\(\Rightarrow A< 29\)

30 tháng 9 2020

So sánh C và D biết
C=1+13+13^2+...+13^13/1+13+13^2+...+13^12
D=1+11+11^2+...+11^13/1+11+11^2+...+11^12

10 tháng 3 2020

a) Đặt \(A=\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)

- Ta có: \(A=\left(100+121+144\right)\div\left(169+196\right)\)

      \(\Leftrightarrow A=365\div365=1\)

Vậy \(A=1\)

b) Đặt \(B=1.2.3.....9-1.2.3.....8-1.2.3.....8^2\)

- Ta có: \(B=1.2.3.....8.\left(9-1\right)-1.2.3.....8^2\)

     \(\Leftrightarrow B=1.2.3.....8.8-1.2.3.....8.8=0\)

Vậy \(B=0\)

c) Đặt \(C=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)

- Ta có: \(C=\frac{3^2.4^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)

      \(\Leftrightarrow C=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}\)

      \(\Leftrightarrow C=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)

      \(\Leftrightarrow C=\frac{9.2^{36}}{2^{35}.9}\)

      \(\Leftrightarrow C=2\)

Vậy \(C=2\)

d) Đặt \(D=1152-\left(374+1152\right)+\left(-65+374\right)\)

- Ta có: \(D=1152-374-1152-65+374\)

      \(\Leftrightarrow D=\left(1152-1152\right)+\left(374-374\right)-65\)

      \(\Leftrightarrow D=-65\)

Vậy \(D=-65\)

1 tháng 2 2019

a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)

   =1

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

   =(1.2.3....8).0

   =0

mik chỉ giải được zậy thôi.

t mik nha.

4 tháng 11 2016

a, A = \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)

\(A=\frac{2^{10}\left(13+65\right)}{2^8.2^2.26}=\frac{2^{10}.78}{2^{10}.26}=\frac{78}{26}=3\)

Vậy A = 3

b, \(B=\frac{72^3.54^2}{108^4}=\frac{72^3.54^2}{\left(54.2\right)^4}=\frac{72^3.54^2}{54^4.2^4}=\frac{72^3}{54^2.2^4}=\frac{\left(8.9\right)^3}{\left(6.9\right)^2.2^4}\)

\(=\frac{\left(2^3\right)^3.9^3}{6^2.9^2.2^4}=\frac{2^9.9^3}{2^2.3^2.9^2.2^4}=\frac{2^9.9^3}{2^6.9^3}=\frac{2^9}{2^6}=2^3=8\)

Vậy B = 8

c, \(C=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}.3^{30}}{2^2.3^{28}}=\frac{11.3^{29}.3.3^{29}}{2^2.3^{28}}=\frac{\left(11-3\right)3^{29}}{2^2.3^{28}}\)

\(=\frac{2^3.3^{29}}{2^2.3^{28}}=2.3=6\)

Vậy C = 6

d, \(D=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{\left(3.2^{18}\right)^2}{11.2^{35}-\left(2^4\right)^9}=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}=\frac{3^2.2^{36}}{\left(11-2\right)2^{35}}=\frac{3^2.2}{9}=2\)

Vậy D = 2

5 tháng 8 2018

\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)

\(=\frac{\left(3.2^{18}\right)^2}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)

\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)

\(=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)

\(=\frac{3^2.2^{36}}{2^{35}.9}=\frac{3^2.2^{36}}{2^{35}.3^2}=2\)

17 tháng 4 2018

Violympic toán 6