Ta có: \(\dfrac{4^7\cdot2^8}{3\cdot2^{15}\cdot16^2-5\cdot2^2\cdot\left(2^{10}\right)^2}\)

\(=\dfrac{2^{22}}{3\cdot2^{15}\cdot2^8-5\cdot2^2\cdot2^{20}}\)

\(=\dfrac{2^{22}}{2^{22}\left(3\cdot2-5\right)}\)

\(=1\)

ta có: 4^7.2^8 / 3.2^15.16^2 - 5.2^2.(2^10)^2

      = 2^14.2^8 / 3.2^15.2^8 - 5.2^2.2^20

      = 2^22 / 6.2^22 - 5.2^22

      = 2^22 / 2^22 ( 6 - 5 )

      = 1

\(H=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(H=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(H=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)

\(H=\frac{2^{19}.3^9+2^{18}.5.3^9}{2^{19}.3^9+2^{20}.3^{10}}\)

\(H=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9.\left(1+2.3\right)}\)

\(H=\frac{2^{18}.3^9.7}{2^{19}.3^9.\left(1+6\right)}\)

\(H=\frac{2^{18}.3^9.7}{2^{19}.3^9.7}=\frac{1}{2}\)

\(K=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)

\(K=\frac{\left(2^2\right)^7.2^8}{3.2^{15}.\left(2^4\right)^2-5.2^2.2^{20}}\)

\(K=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^{22}}\)

\(K=\frac{2^{22}}{3.2^{23}-5.2^{22}}\)

\(K=\frac{2^{22}}{2^{22}.\left(3.2-5\right)}=\frac{2^{22}}{2^{22}.1}=1\)

bạn phân tích hết ra rồi rút gọn

C=\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{1}{2}=0,5\)

D=\(\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)\(=1\)

Đúng,Đúng,Đúng,...!^-^

\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)

\(B=\frac{2^{19}.3^9+3^9.5.2^{18}}{2^{19}.3^9+3^{10}.2^{20}}\)

\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+3.2\right)}\)

\(B=\frac{7}{2.7}\)

\(B=\frac{1}{2}\)

\(C=\frac{2^{13}.4^{11}-16^9}{\left(3.2^{17}\right)^2}\)

\(C=\frac{2^{13}.2^{22}-2^{36}}{3^2.2^{34}}\)

\(C=\frac{2^{35}-2^{36}}{3^2.2^{34}}\)

\(C=\frac{2^{35}\left(1-2\right)}{3^2.2^{34}}\)

\(C=\frac{-2}{9}\)

\(D=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)

\(D=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^2.2^{20}}\)

\(D=\frac{2^{14}.2^8}{3.2^{23}-5.2^{22}}\)

\(D=\frac{2^{22}}{2^{22}\left(3.2-5\right)}\)

\(D=1\)

\(=\frac{2^{22}}{3.2^{33}-5.2^{22}}=\frac{2^{22}}{2^{22}.\left(32-5\right)}=1\)

Lời giải:

Gọi tổng trên là $A$. Ta có:

$A=-2.2+2(3+5+7+....+105+107)$

$=-4+2(3+5+7+....+105+107)$

Đặt $M=3+5+7+...+105+107$

Số số hạng của $M$: $(107-3):2+1=53$

$M=53(107+3):2=2915$ 

$A=-4+2M=-4+2915.2=5826$