ta có A=-3/2

        B= 2

nên -3/2 x 2 = -3

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

\(A=\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)

\(B=\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=2\)

\(C=\frac{4^5\cdot9^{4-2\cdot6^9}}{2^{10}\cdot3^8+6^8\cdot20}=0\)

A=\(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)

Mình đồng ý cách làm của bạn trần như

Bạn Trương Diệu Ngọc làm theo cách bạn ấy nha

Đảm bảo 100%

1/

\(\frac{2n+1}{n-3}+\frac{3n-5}{n-3}-\frac{4n-5}{n-3}=\frac{2n+1+\left(3n-5\right)-\left(4n-5\right)}{n-3}=\frac{2n+1+3n-5-4n+5}{n-3}=\frac{n+1}{n-3}=\frac{n-3+4}{n-3}=\frac{n-3}{n-3}+\frac{4}{n-3}=1+\frac{4}{n-3}\)

Để S là số nguyên <=> n - 3 thuộc Ư(4) = {1;-1;2;-2;4;-4}

Vậy...

câu 2 dễ ẹt

2/

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

...........

\(\frac{1}{99^2}< \frac{1}{98.99}=\frac{1}{98}-\frac{1}{99}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{99^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}=1-\frac{1}{99}< 1\)

=> ĐPCM

3/

\(\frac{m}{9}-\frac{3}{n}=\frac{1}{18}\)

=> \(\frac{3}{n}=\frac{m}{9}-\frac{1}{18}\)

=> \(\frac{3}{n}=\frac{2m}{18}-\frac{1}{18}\)

=> \(\frac{3}{n}=\frac{2m-1}{18}\)

=> n(2m - 1) = 3.18 = 54

=> n và 2m - 1 thuộc Ư(54) = {1;-1;2;-2;3;-3;6;-6;9;-9;18;-18;27;-27;54;-54}

Mà 2m - 1 là số lẻ => 2m - 1 thuộc {1;-1;3;-3;9;-9;27;-27}

                                  n thuộc {2;-2;6;-6;18;-18;54;-54}

Ta có bảng:

Vậy các cặp (m;n) là (1;54) ; (0;-54) ; (2;18) ; (-1;-18) ; (5;6) ; (-4;-6) ; (14;2) ; (-13;-2)

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

còn câu 2 và 3 thì sao

K ghi lại đề câu 2 nha :

\(=\frac{3^2.\left(2^2\right)^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)

\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)

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

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

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

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

\(=1.2=2\)

\(B=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{3.2^2.2^{16.2}}{11.2^{13}.2^{22}-2^{36}}=\frac{3.2^{34}}{11.2^{35}-2^{36}}=\frac{3.2^{34}}{2^{34}\left(11.2-2^2\right)}\)

\(=\frac{3}{11.2-2^2}=\frac{3}{11.2-4}=\frac{3}{22-4}=\frac{3}{18}=\frac{1}{6}\)

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