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3 tháng 7 2018
a)\(\dfrac{2^{15}.3^8}{2^6.3^6.2^9}\)\(\dfrac{ }{ }\)=\(^{3^2}\)=9
b)\(\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.15}{-2^{12}.3^{12}-2^{11}.3^{11}}\)=\(\dfrac{2^{11}.3^{11}.\left(1+15\right)}{2^{11}.3^{11}\left(-2.3-1\right)}\)
=\(\dfrac{32}{-21}\)
c)\(\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)=\(\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)=\(-\dfrac{1}{3}\)
em dựa vào vd \(\dfrac{4^{16}}{2^8}\)= \(\dfrac{\left(2^2\right)^{16}}{2^8}=\dfrac{2^{16\cdot2}}{2^8}=2^4=16\)
NN
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Y
20 tháng 7 2019
Bài 2
| x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | ( -3,2) + \(\frac{2}{5}\)|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | -2,8|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= -2,8
=> | x - \(\frac{1}{3}\)| = -2,8 - \(\frac{4}{5}\)
=> | x - \(\frac{1}{3}\)| = - 3,6
=> x - \(\frac{1}{3}\)= -3,6
=> x = -3,6 + \(\frac{1}{3}\)
=> x = \(\frac{-49}{15}\)
21 tháng 7 2019
Bài 3 :
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a_1-1}{9}=\frac{a_2-2}{8}=...=\frac{a_9-9}{1}=\frac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}\)
\(=\frac{\left[a_1+a_2+...+a_9\right]-\left[1+2+...+9\right]}{9+8+...+1}=\frac{90-45}{45}=1\)
Ta có : \(\frac{a_1-1}{9}=1\Rightarrow a_1=10\)
Tương tự : \(a_1=a_2=....=a_9=10\)
IM
22 tháng 7 2016
a)
\(5A=5+5^2+.....+5^{101}\)
\(\Rightarrow5A-A=\left(5+5^2+.....+5^{101}\right)-\left(1+5+.....+5^{100}\right)\)
\(\Rightarrow4A=5^{101}-1\)
\(\Rightarrow A=\frac{5^{101}-1}{4}\)
b)
\(2B=1+\left(\frac{1}{2}\right)^2+....+\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2^2}+.....+\frac{1}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{99}}\right)\)
\(\Rightarrow B=1-\frac{1}{2^{100}}\)
NG
2
S
5 tháng 2 2023
\(A=\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(=-\dfrac{2}{6}=-\dfrac{1}{3}\)
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=-\frac{2}{6}=-\frac{1}{3}\)
\(B=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{11}.2.3^{10}.\left(1+5\right)}{2^{11}.3^{10}.3.\left(6-1\right)}=\frac{12}{15}=\frac{4}{5}\)
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2^2.3^9}{2^{10}.3^8-\left(-\left(2^2.3^8.5\right)\right)}=\frac{2^2.3^9}{-\left(2^2.3^8.5\right)}=-\frac{3}{5}\)