\(\frac{11.3^{22}.3^8-9^{15}}{\left(2.3^{14}\right)^2}=\frac{45}{2}\) Nhé

Tk mình nha

\(\frac{45}{2}\) nhé !

Học tốt !

Bài 35 :

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

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

\(A=\frac{2^8.2^2.98}{2^8.104}\)

\(A=\frac{2^8.4.98}{2^8.4.26}\)

\(A=\frac{49}{13}\)

Vậy \(A=\frac{49}{13}\)

\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)

\(B=\frac{11.3^{29}-9^{15}}{2^2.\left(3^{14}\right)^2}\)

\(B=\frac{11.3^{29}-9^{15}}{2^2.3^{28}}\)

\(B=\frac{11.3^{29}-\left(3^2\right)^{15}}{4.3^{28}}\)

\(B=\frac{11.3^{29}-3^{30}}{4.3^{28}}\)

\(B=\frac{11.3^{29}-3^{29}.3}{4.3^{28}}\)

\(B=\frac{3^{29}.\left(11-3\right)}{4.3^{28}}\)

\(B=\frac{3^{29}.8}{4.3^{28}}\)

\(B=\frac{3^{28}.3.4.2}{4.3^{28}}\)

\(B=3.2\)

\(B=6\)

Vậy B = 6 

A = 2^10 . 13 + 2^10 . 65 / 2^8 . 104

   = 2^10 ( 13 + 65 ) / 2^8 . 104 = 2^10 . 78 / 2^8 . 104 = 2^8 . 2^2 . 78 / 2^8 . 104 = 2^8 . 4 . 78 / 2^8 . 104 = 2^8 . 312 / 2^8 . 104

   = 312/104

   = 3

B = 11 . 3^22 . 3^7 - 9^15 / ( 2.3^14)^2

   = 11 . 3^29 - (3^2)^15 / ( 3.2^14)^2

   = 11 . 3^29 - 3^30 / ( 3. 2 )^28

    = ( 8 + 3 ) . 3^29 - 3^30 / ( 3. 2)^28

    = 8 . 3^29 + 3.3^29 - 3^30 / ( 3.2)^28

     = 8 . 3^29 + 3^30 - 3^30 / ( 3 . 2)^28

    = 8 . 3^29 / 3^28 . 2^28

     = 2^3 . 3 / 2^28

     = 3/ 2^25

2 bạn ai làm đúng vậy

\(A=\frac{11.3^{22}.3^7.9^{15}}{\left(2.3^{14}\right)^2}\)

\(A=\frac{11.3^{29}-3^{30}}{2^2.\left(3^{14}\right)^2}\)

\(A=\frac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\)

\(A=\frac{3.2}{1.1}\)

\(A=\frac{6}{1}\)

\(A=6\)

\(\frac{11.3^{32}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{39}-3^{30}}{4.3^{28}}=\frac{3^{30}.\left(11.3^9-1\right)}{4.3^{28}}=\frac{9.\left(11.3^9-1\right)}{4}=487152\)

ta có    \(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.\left(3^{14}\right)^2}\)

=  \(\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)

=   \(\frac{3^{29}.2^3}{2^2.3^{28}}=3.2=6\)

\(=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.2^3}{2^2.3^{28}}=3.2=6.\)

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

\(=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3.8}{4}=\frac{24}{4}=6\)

Đáp số: \(6\)

\(\frac{11.3^{29}-3^{30}}{4.3^{28}}\)=\(\frac{3^{29}.\left(11-3\right)}{4.3^{28}}\)=\(\frac{3^{29}.8}{4.3^{28}}\)=3.2=6.

\(A=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)

\(A=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)

\(A=\frac{11.3^{29}-3^{30}}{4.3^{28}}\)

\(A=\frac{3^{29}.\left(11-3\right)}{4.3^{28}}\)

\(A=\frac{3^{29}.8}{4.3^{28}}=3.2=6\)

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

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