Tính gía trị của các biểu thức sau :
\(a,\frac{205.5^{10}}{100^5} \)
\(b,\frac{\left(0,9\right)^5}{\left(0,3\right)^5}\)
\(c,\frac{6^3+3.6+3^3}{-13}\)
\(d,\frac{4^6.9^5.6^9.120}{8^43^{12}-6^{11}}\)
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M
20 tháng 7 2017
a, 205.510/1005
=205.55.55/1005
=1005.55/1005
=55
=3125
b, (0,9)5/(0,3)6
=(0,3.3)5/0,36
=0,55.35/0,36
=35/0,3
=810
c, 63+3.62+33/-13
=(2.3)3+3.(3.2)2+33/-13
=23.33+3.32.22+33/-13
=33.23+33.22+33/-13
=33(23+22+1)/-13
=27.13/-13
=-27
d, 46.95+69.120/84.312-611
=(22)6.(32)5+(2.3)9.3.23.5/(23)4.312-(2.3)11
=212.310+29.39.3.23.5/212.312-211.311
=212.310+212.310.5/211.311.2.3-211.311
=212.310.(1+5)/211.311(6-1)
=212.310.6/211.311.5
=2.6/3.5
=12/15
=4/5
XC
0
XC
0
1 tháng 8 2018
a. \(\frac{20^5.5^{10}}{100^5}\)
\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)
\(=\frac{20^5.25^5}{100^5}\)
\(=\frac{500^5}{100^5}\)
\(=\left(\frac{500}{100}\right)^5\)
\(=5^5=3125\)
b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)
\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)
\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)
\(=3^5.\frac{1}{0,3}\)
\(=810\)
c. \(\frac{6^3+3.6^2+3^3}{-13}\)
\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)
\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\frac{3^3.13}{-13}\)
\(=\left(-3\right)^3\)
\(=-27\)
a) \(\frac{205.5^{10}}{100^5}=\frac{41.5^{11}}{5^2.4}=\frac{41.5^9}{4}\)
b) \(\frac{\left(0,9\right)^5}{\left(0,3\right)^5}=\left(0,9:0,3\right)^5=3^5=243\)
c) \(\frac{6^2+3.6+3^2}{-13}=\frac{2^2.3^2+3.6+3^2}{-13}=\frac{3\left(2^2.3+6+3\right)}{-13}=\frac{3.21}{-13}=\frac{63}{-13}\)
d) \(\frac{4^6.9^5.6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5.\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2^{24}.3^{20}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{24}.3^{20}.5}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{2^{13}.3^9.5}{5}=2^{13}.3^9\)
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