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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:
\(\dfrac{24.1000}{2500-500}=\dfrac{24.1000}{2.1000}=12\)
a, \(\frac{24.315+3.561.8+4.124.6}{1+3+5+7+...+97+99-500}\) (1)
Đặt : S = 1 + 3 + 5 + 7 + ... + 97 + 99
SSH của S là : (99 -1) : 2 + 1 = 50(sh)
Tổng của S là : \(\frac{\left(99+1\right).50}{2}=\frac{100.50}{2}=\frac{5000}{2}=2500\)
Thay S vào biểu thức (1) Ta có :
\(\frac{24.315+3.561.8+4.124.6}{2500-500}\)
\(=\frac{3.8.315+3.561.8+4.2.124.3}{2000}\)
\(=\frac{3.8.315+3.561.8+8.124.3}{2000}\)
\(=\frac{\left(3.8\right).\left(315+561+124\right)}{2000}=\frac{24.1000}{2000}=\frac{24000}{2000}=12\)
b, \(\frac{3^9.3^{20}.2^8}{3^{24}.243.2^6}=\frac{3^{29}.2^8}{3^{24}.3^5.2^6}=\frac{3^{29}.2^6.2^2}{3^{29}.2^6}=2^2=4\)
lời giải nè:
= 39+20.28/324.243.26
=329.28/324.27.9.26
=329.28/324.3332.26
=1.22/1.1.1.1
=4
\(P=\frac{3^9\cdot3^{20}\cdot2^8}{3^{24}\cdot243\cdot2^6}\)
\(P=\frac{3^{29}\cdot2^8}{3^{29}\cdot2^6}\)
\(P=2^2=4\)
\(Q=\frac{2^{15}\cdot5^3\cdot2^6\cdot3^4}{8\cdot2^{18}\cdot81\cdot5}\)
\(Q=\frac{2^{21}\cdot5^3\cdot3^4}{2^{21}\cdot3^4\cdot5}\)
\(Q=5^2=25\)
a. \(\dfrac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}\) = \(\dfrac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}\) = \(\dfrac{-5}{3^2}\)= \(\dfrac{-5}{9}\)
b. \(\dfrac{-11^5.13^7}{11^5.13^8}\) = \(\dfrac{-11^5.13^7}{11^5.13^7.13}\)= \(\dfrac{-1}{13}\)
c. \(\dfrac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}\)= \(\dfrac{2^{10}\left(3^{10}-3^9\right)}{2^9.3^{10}}\)= \(\dfrac{2^{10}.3}{2^9.3^{10}}\)= \(\dfrac{2^9.2.3}{2^9.3.3^9}\)= \(\dfrac{2}{3^9}\)=\(\dfrac{2}{19683}\)
a 25 phần 41
b 9
c 1 phần 12
d 12
e 14 phần 15
f 24 phần 7
`(3^9. 3^20. 2^8)/(3^24. 243. 2^6)`
`=(3^29. 2^8)/(3^24. 3^5. 2^6)`
`=(3^29. 2^8)/(3^29. 2^6)`
`=2^2=4`