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ta có : \(A=1+3+3^2+...+3^{99}\)
\(\Rightarrow3A=3\left(1+3+3^2+...+3^{99}\right)\)
\(\Leftrightarrow3A=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A-A=2A=\left(3+3^2+3^3+...+3^{100}\right)-\left(1+3+3^2+...+3^{99}\right)\)
\(\Leftrightarrow2A=3^{100}-1\)
\(\Rightarrow2A+1=3^{100}-1+1=3^{100}=\left(3^{25}\right)^4\)
vậy \(2A+1=\left(3^{25}\right)^4\)
`2^5 . 8^4 = 2^5 . (2^3)^4 = 2^5 . 2^12 = 2^17`
`25^6 . 125^3 = (5^2)^6 . (5^3)^3 = 5^12 . 5^9 = 5^21`
`625^5 : 25^7 = (5^4)^7 : (5^2)^7 = 5^28 : 5^14 = 5^14`
`12^3 . 3^3 = (12 . 3)^3 = 36^3`
\(A=1+3+3^2+...+3^{41}\)
\(3A=3+3^2+3^3+...+3^{42}\)
\(3A-A=3+3^2+...+3^{42}-1-3-...-3^{41}\)
\(2A=3^{42}-1\)
\(A=\dfrac{3^{42}-1}{2}\)
Ta có: \(2A+1\)
\(=2\cdot\dfrac{3^{42}-1}{2}+1\)
\(=3^{42}-1+1\)
\(=3^{42}\)
\(=\left(3^2\right)^{21}\)
\(=9^{21}\)
Ta coi \(B=2^1+2^2+2^3+...+2^{100}\)
\(\Rightarrow A=B+2\)
Ta có:
\(B=2^1+2^2+2^3+...+2^{100}\)
\(\Rightarrow2B=\left(2^1+2^2+2^3+...+2^{100}\right).2\)
\(=2^2+2^3+2^4+...+2^{101}\)
\(\Rightarrow B=2B-B=\left(2^2+2^3+2^4+...+2^{101}\right)-\left(2^1+2^2+2^3+...+2^{100}\right)\)
\(=2^{101}-2\)
\(\Rightarrow A=2^{101}-2+2=2^{101}\)
a)\(4^3.2^4\div\left(4^2.\frac{1}{32}\right)\)
\(=\left(2^2\right)^3.2^4\div\left(2^2\right)^2\div32\).
\(=2^{\left(2.3\right)}.2^4\div2^{\left(2.2\right)}\div2^5\)
\(=2^6.2^4\div2^4\div2^5\)
\(=2^{6+4-4-5}=2^1\)
b)\(\left(\frac{1}{5}\right)^5=\frac{1}{5^5}=\left|5^5\right|=5^{-5}\)
\(\frac{1}{125}=\frac{1}{5^3}=\left|5^3\right|=5^{-3}\)
c)\(\frac{4}{25}=\frac{2^2}{5^2}=\left(\frac{2}{5}\right)^2=0,4^2\)
\(\frac{-8}{125}=\frac{-2^3}{5^3}=\left(\frac{-2}{5}\right)^2=-0,4^3=0,4^{-3}\)
\(\frac{16}{625}=\frac{2^4}{5^4}=\left(\frac{2}{5}\right)^4=0,4^4\)
Ta có 2A=2+22+23+.............+ 22016 (1)
A=1+22+........+ 22015 (2)
LẤY (1)-(2) ta đc : A= 22016-1
A+1=22016=(21008)2 là số chính phương
phần còn lai em tự là nhé dễ rồi
A=1+3+3^2+...+3^99
3A=3+3^2+3^3+...+3^100
3A-A=3^100-1
2A=3^100-1
A=(3^100-1):2
mik chỉ làm được đến đó thôi
A=1+3+3^2+...+3^99
3A=3+3^2+3^3+...+3^100
3A-A=2A=3^100-1
\(\Rightarrow\)2A+1=3^100
Khong viet dc vi 3^100 le ma 4^n chan