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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
\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)
\(B=\frac{2^{19}.3^9+3^9.5.2^{18}}{2^{19}.3^9+3^{10}.2^{20}}\)
\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+3.2\right)}\)
\(B=\frac{7}{2.7}\)
\(B=\frac{1}{2}\)
\(C=\frac{2^{13}.4^{11}-16^9}{\left(3.2^{17}\right)^2}\)
\(C=\frac{2^{13}.2^{22}-2^{36}}{3^2.2^{34}}\)
\(C=\frac{2^{35}-2^{36}}{3^2.2^{34}}\)
\(C=\frac{2^{35}\left(1-2\right)}{3^2.2^{34}}\)
\(C=\frac{-2}{9}\)
\(D=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)
\(D=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^2.2^{20}}\)
\(D=\frac{2^{14}.2^8}{3.2^{23}-5.2^{22}}\)
\(D=\frac{2^{22}}{2^{22}\left(3.2-5\right)}\)
\(D=1\)
\(H=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(H=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(H=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(H=\frac{2^{19}.3^9+2^{18}.5.3^9}{2^{19}.3^9+2^{20}.3^{10}}\)
\(H=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9.\left(1+2.3\right)}\)
\(H=\frac{2^{18}.3^9.7}{2^{19}.3^9.\left(1+6\right)}\)
\(H=\frac{2^{18}.3^9.7}{2^{19}.3^9.7}=\frac{1}{2}\)
\(K=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)
\(K=\frac{\left(2^2\right)^7.2^8}{3.2^{15}.\left(2^4\right)^2-5.2^2.2^{20}}\)
\(K=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^{22}}\)
\(K=\frac{2^{22}}{3.2^{23}-5.2^{22}}\)
\(K=\frac{2^{22}}{2^{22}.\left(3.2-5\right)}=\frac{2^{22}}{2^{22}.1}=1\)
\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
= \(\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{\left(5-7.2\right).2^{28}.3^{18}}\)
= \(\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{-9.2^{28}.3^{18}}\)
= \(\frac{\left(5.2^2-2\right).2^{28}.3^{18}}{-9.2^{28}.3^{18}}\)
= \(\frac{18.2^{28}.3^{18}}{-9.2^{28}.3^{18}}\)
= 18/-9 = -2
\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(=\frac{2^{28}.3^{18}.\left(5.2^2-2\right)}{2^{28}.3^{18}.\left(5-7.2\right)}\)
\(=\frac{5.4-2}{5-14}\)\(=\frac{18}{-9}=-2\)
T**k mik nha!
d)
\(\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}\\ =\dfrac{3^{29}.2^6.2^2}{3^{24}.3^5.2^6}\\ =\dfrac{3^{29}.2^6.4}{3^{29}.2^6}\\ =4\)
e)
\(\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}\\ =\dfrac{2^{21}.5^3.3^4}{2^3.2^{18}3^4.5}\\ =\dfrac{2^{21}.5.5^2.3^4}{2^{21}.3^4.5}\\ =5^2\\ =25\)
f)
\(=\dfrac{24\left(315+561+124\right)}{\dfrac{\left(1+99\right).50}{2}-500}\\ =\dfrac{24.1000}{2500-500}\\ =12\)
\(a,\dfrac{-14.15}{21.\left(-10\right)}=\dfrac{-7.2.3.5}{7.3.\left(-2\right).5}=1\)
\(b,\dfrac{5.7-7.9}{7.2+6.7}=\dfrac{7\left(5-9\right)}{7\left(2+6\right)}=\dfrac{-4}{8}=-\dfrac{1}{2}\)
\(c,\dfrac{\left(-7\right).3+2.\left(-14\right)}{\left(-5\right).7-2.7}=\dfrac{-7.\left(3+4\right)}{7\left(-5-2\right)}\)
\(=\dfrac{\left(-7\right).7}{7.\left(-7\right)}=1\)
\(d,\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}=\dfrac{3^{29}.2^8}{3^{24}.3^5.2^6}=\dfrac{3^{29}.2^8}{3^{29}.2^6}=2^2=4\)
\(e,\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}=\dfrac{2^{21}.3^4.5^3}{2^{18}.2^3.3^4.5}=\dfrac{2^{21}.3^4.5^3}{2^{21}.3^4.5}=5^2=25\)
\(f,\dfrac{24.315+3.561.8+4.124.6}{1+3+5+...+97+99-500}\)
\(=\dfrac{24.315+24.561+24.124}{1+3+5+...+97+99-500}\)
\(=\dfrac{24\left(315+561+124\right)}{1+3+5+...+97+99-500}\)
\(=\dfrac{24.1000}{1+3+5+...+97+99-500}\) (1)
Đặt A = 1 + 3 + 5 + ... + 97 + 99
Số số hạng trong A là: (99 - 1) : 2 + 1 = 50 (số)
Tổng A bằng: (99 + 1) . 50 : 2 = 2500
Thay A = 2500 vào biểu thức (1), ta được:
\(\dfrac{24.1000}{2500-500}=\dfrac{24.1000}{2.1000}=12\)
:
\(=\frac{2^{14}.2^8}{3.2^{15}-2^8-5.2^2.2^{20}}=\frac{2^{22}}{3.2^{15}-2^8-5.2^{22}}=\frac{2^{22}}{2^{22}.\left(3.2^{-7}-2^{-14}-5\right)}=\frac{1}{3.2^{-7}-2^{-14}-5}\)
\(\frac{4^7\times2^8}{3\times2^{15}-16^2-5\times2^2\times\left(2^{10}\right)^2}=\frac{\left(2^2\right)^7\times2^8}{3\times2^{15}-\left(2^4\right)^2-5\times2^2\times2^{20}}=\frac{2^{14}\times2^8}{3\times2^{15}-2^8-5\times2^{22}}=\frac{2^{14}\times2^8}{2^8\times\left(3\times2^7-1-2^{14}\times5\right)}\)
\(=\frac{2^{22}}{3.2^{33}-5.2^{22}}=\frac{2^{22}}{2^{22}.\left(32-5\right)}=1\)