làm bừa thui,ai tích mình mình tích lại

Số số hạng là : 

Có số cặp là :

50 : 2 = 25 ( cặp )

Mỗi cặp có giá trị là :

99 - 97 = 2 

Tổng dãy trên là :

25 x 2 = 50

Đáp số : 50

ta co:A = 11.3²⁹-9¹⁵/ (2.3¹⁴)²

            =11.3²⁹-(3²)¹⁵/2².3²⁸

           =11.3²⁹-3³⁰/2².3²⁸

            =3²⁹.(11-3)/2².3²⁸

           = 3.7/4

           =21/4

=6 bạn à

\(\frac{11.3^{29}-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.8}{2^2.2^{28}}=\frac{3^{29}.2^3}{3^{28}.2^2}=\frac{3.2}{1.1}=6\)

a) (2/5 x 25/29) + (3/5 x 25/29)
= (50/145) + (75/145)
= 125/145

b) (5/2 x 3/7) - (3/14 : 6/7)
= 15/14 - (3/14 x 7/6)
= 15/14 - 1/2
= (30/28) - (14/28)
= 16/28
= 4/7

c) (15/4 : 5/12) - (6/5 : 11/15)
= (15/4 x 12/5) - (6/5 x 15/11)
= 180/20 - 90/55
= 9 - 18/11
= (99/11) - (18/11)
= 81/11
= 7 4/11

a) (2/3) + (20/21 x 3/2 x 7/5)
= 2/3 + (60/210)
= 2/3 + 2/7
= (14/21) + (6/21)
= 20/21

b) (5/17 x 21/32 x 47/24 x 0)
= 0

c) (11/3 x 26/7) - (26/7 x 8/3)
= (286/21) - (208/21)
= 78/21
= 3 9/21
= 3 3/7

a) (25/8) : x = 5/16
=> (25/8) x (16/5) = x
=> 4 = x

b) x + (7/15) = 6/15
=> x = (6/15) - (7/15)
=> x = -1/15

c) x : (28/49) = 7/12
=> x x (49/28) = 7/12
=> x = (7/12) x (28/49)
=> x = 1/2

a) 6 x x = (5/8) : (3/4)
=> 6x = (5/8) x (4/3)
=> 6x = 20/24
=> 6x = 5/6
=> x = (5/6) / 6
=> x = 5/36

câu,b,không,đủ,thông,tin,nhan,bạn.

\(\frac{11.3^{25}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{32}-\left(3^2\right)^{15}}{2^2.\left(3^{14}\right)^2}=\frac{11.3^{32}-3^{30}}{4.3^{28}}=\frac{3^{30}.\left(11.3^2-1\right)}{4.3^{28}}=\frac{3^2.98}{4}=\frac{9.98}{4}=\frac{882}{4}=\frac{441}{2}\)

\(A=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.\left(3^{14}\right)^2}=\frac{11.3^{29}-3^{30}}{4.3^{28}}\) \(=\frac{3^{29}.\left(11-3\right)}{4.3^{28}}=\frac{3.8}{4}=3.2=6\)

Ta có:\(A=\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{98.99}+\frac{9}{99.100}\)

\(=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

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

\(=9.\frac{99}{100}=\frac{891}{100}\)