Tính:
\(\left(30^2+28^2+29^2+....+4^2+2^2\right)-\left(29^2+27^2+25^2+....+3^2+1^2\right)\)
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31 tháng 10 2017
Ta có:
\(T\left(-2\right)=a_0-2a_1+2^2a_2-...-2^{29}a_{29}+2^{30}a_{30}=a_0+H=\left(1+4\right)^{15}\)
\(\Leftrightarrow1+H=5^{15}\)
\(\Leftrightarrow H=5^{15}-1\)
16 tháng 7 2017
1/ \(\frac{9.5^{20}.27^9-3.9^{15}.25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)
\(=\frac{5^{20}.3^{29}-3^{31}.5^{18}}{7.3^{29}.5^{18}-3^{29}.5^{19}}=\frac{3^{29}.5^{18}.\left(25-9\right)}{3^{29}.5^{18}.\left(7-5\right)}=\frac{16}{2}=8\)
CÁC BÀI CÒN LẠI TƯƠNG TỰ HẾT NHÉ E
19 tháng 5 2022
23: \(=\left(2a-b\right)^2-\left(2a-2b\right)^2\)
\(=\left(2a-b-2a+2b\right)\left(2a-b+2a-2b\right)\)
\(=b\left(4a-3b\right)\)
24: \(=\left(3a+3b\right)^2-\left(2a-4b\right)^2\)
\(=\left(3a+3b-2a+4b\right)\left(3a+3b+2a-4b\right)\)
\(=\left(a+7b\right)\left(5a-b\right)\)
25: \(=\left(4a-2b\right)^2-\left(4a-4b\right)^2\)
\(=\left(4a-2b-4a+4b\right)\left(4a-2b+4a-4b\right)\)
\(=2b\left(8a-6b\right)\)
=4b(4a-3b)
1 tháng 2 2021
Ta có: \(A=\dfrac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{18}\cdot3^{18}\cdot2^{12}-2\cdot2^{28}\cdot3^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{28}\cdot3^{18}\cdot2}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2\cdot2^{28}\cdot3^{18}}{2^{28}\cdot3^{18}\cdot\left(5-7\cdot2\right)}\)
\(=\dfrac{2^{28}\cdot3^{18}\cdot\left(5\cdot2^2-2\right)}{2^{28}\cdot3^{18}\cdot\left(5-14\right)}\)
\(=\dfrac{20-2}{-9}=\dfrac{18}{-9}=-2\)
1 tháng 2 2019
\(5^6-25^3=\left(5^2\right)^3-25^3=25^3-25^3=0\)
\(\Rightarrow\frac{\left(1^6-29^3\right)\left(2^6-28^3\right)\left(3^6-27^3\right)\left(4^6-26^3\right)\left(5^6-25^3\right).....\left(10^6-20^3\right)}{\left(1^6+29^3\right)\left(2^6+28^3\right)\left(3^6+27^3\right)\left(4^6+26^3\right)\left(5^6+25^3\right).....\left(10^6+20^3\right)}=0\)
HH
15 tháng 11 2020
\(A=\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^{14}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{5.\left(2^2\right)^9.\left(3^2\right)^9.\left(2^2\right)^6-2.\left(2^2\right)^{14}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{18}}{5.2^{28}.3^{18}-7.2.2^{28}.3^{18}}\)
\(A=5.\frac{\left(2^{18}.2^{12}\right).3^{18}.2^{29}.3^{18}}{2^{28}.3^{18}.\left(5-7-2\right)}\)
\(A=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(-9\right)}\)
\(A=\frac{5.2.2^{29}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(-9\right)}\)
\(A=\frac{2^{20}.2^{18}.\left(5.2-1\right)}{2^{28}.3^{18}.\left(-9\right)}\)
\(A=\frac{2^{29}.3^{18}.9}{2^{28}.3^{18}.\left(-9\right)}\)
\(A=\frac{2.11}{11.\left(-2\right)}\)
\(A=-2\)
Vậy : A = -2