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Bài 1:
a) \(8^5\cdot8^2=8^7\)
b) \(9^3\cdot3^2=\left(3^2\right)^3\cdot3^2=3^6\cdot3^2=3^8\)
c) \(2^7\cdot5^7=10^7\)
d) \(27^6:3^3=\left(3^3\right)^6:3^3=3^{18}:3^3=3^{15}\)
Bài 2:
a) \(x^6:x^3=125\)
\(\Rightarrow x^3=125\)
\(\Rightarrow x=5\)
b) \(x^{20}=x\)
\(\Rightarrow x^{20}-x=0\)
\(\Rightarrow x\left(x^{19}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{19}-1=0\Rightarrow x=1\end{matrix}\right.\)
c) \(3^x\cdot3=243\)
\(\Rightarrow3^x=81\)
\(\Rightarrow x=4\)
d) \(2x-138=2^3\cdot3^2\)
\(\Rightarrow2x-138=72\)
\(\Rightarrow2x=200\)
\(\Rightarrow x=100\)
Giải:
Bài 1:
a) \(8^5.8^2=8^{5+2}=8^7\)
b) \(9^3.3^2=3^6.3^2=3^{6+2}=3^8\)
c) \(2^7.5^7=\left(2.5\right)^7=10^7\)
d) \(27^6:3^3=3^{18}:3^3=3^{18-3}=3^{15}\)
Bài 2:
a) \(x^6:x^3=x^{6-3}=x^3=125\)
\(\Leftrightarrow x=5\)
b) \(x^{20}=x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\)
c) \(3^x.3=243\)
\(\Leftrightarrow3^{x+1}=243\)
\(\Leftrightarrow3^{x+1}=3^5\)
\(\Leftrightarrow x+1=5\Leftrightarrow x=4\)
d) \(2.x-138=2^3.3^2\)
\(\Leftrightarrow2.x-138=8.9\)
\(\Leftrightarrow2.x-138=72\)
\(\Leftrightarrow2.x=72+138\)
\(\Leftrightarrow2.x=210\Leftrightarrow x=105\)
Chúc bạn học tốt!
\(A=1+3+3^2+...+3^{41}\)
\(3A=3+3^2+3^3+...+3^{42}\)
\(3A-A=3+3^2+...+3^{42}-1-3-...-3^{41}\)
\(2A=3^{42}-1\)
\(A=\dfrac{3^{42}-1}{2}\)
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 có: \(A=2+2^2+2^3+...+2^{100}\)
\(2A=2^2+2^3+2^4+...+2^{101}\)
\(2A-A=2^{101}-2\)
Hay \(A=2^{101}-2\)
Vậy \(A=2^{101}-2\)
_Học tốt_
\(A=1+2+2^2+...+2^{30}\)
\(2A=2+2^2+2^3+...+2^{31}\)
\(2A-A=\left(2+2^2+2^3+...+2^{31}\right)-\left(1+2+2^2+...+2^{30}\right)\)
\(A=2^{31}-1\)
\(A+1=2^{31}\)
\(A=1+2+2^2+2^3+....+2^{30}\)
\(2.A=2+2^2+2^3+2^4+...+2^{30}\)
\(2.A-A=\left(2+2^2+2^3+2^4+...+2^{31}\right)-\left(1+2+2^2+2^3+...+2^{30}\right)\)
\(A=2^{31}-1\)
\(\Rightarrow A+1=2^{31}-1+1\)
\(\Rightarrow A+1=2^{31}\)
\(1;4^5.6^5=\left(4.6\right)^5=24^5\)
\(7^2.8^2=\left(7.8\right)^2=56^2\)
\(9^2.2^4=9^2.4^2=\left(9.4\right)^2=36^2\)
\(4^3.7^6=4^3.49^3=\left(49.4\right)^3\)
\(27^4.4^6=\left(27^2\right)^2.64^2=\left(27^2.64\right)^2\)
Bài 1 : Viết tích dưới dạng 1 lũy thừa :
a) 45 . 65 = ( 4 . 6 )5 = 245
b) 72 . 82 = ( 7 . 8 )2 = 562
c) 92 . 24 = ( 32 )2 . 24 = 34 . 24 = ( 3 . 2 )4 = 64
d) 43 . 76 = ( 22 )3 . 76 = 26 . 76 = ( 2 . 7 )6 = 146
e) 274 . 46 = ( 33)4 . ( 22 )6 = 312 . 212 = ( 3 . 2 )12 = 612
Ta có: \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\)
a, \(1^3+2^3+3^3+4^3=\left(1+2+3+4\right)^2=10^2\)
b, \(1^3+2^3+3^3+4^3+5^3=\left(1+2+3+4+5\right)^2=15^2\)
c,
\(3^6:3^2+2^3\cdot2^2-2\\ =3^{6-2}+2^{3+2-1}\\ =3^4+2^4\)
\(a,1^3+2^3+3^3+4^3.\)
\(=\left(1+2+3+4\right)^2.\)
\(=10^2=100.\)
\(b,1^3+2^3+3^3+4^3+5^3.\)
\(=\left(1+2+3+4+5\right)^2.\)
\(=15^2=225.\)
(2 phần a, b thì áp dụng công thức: \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2.\))
a: \(A=8^2\cdot32^4=2^6\cdot2^{20}=2^{26}\)
b: \(B=27^3\cdot9^4\cdot243=3^9\cdot3^8\cdot3^5=3^{22}\)