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Lời giải:
\(2^{27}.3^{18}=(2^3)^9.(3^2)^9=8^9.9^9=(8.9)^9=72^9\)
\(25^4.2^8=25^4.(2^2)^4=25^4.4^4=(25.4)^4=100^4\)
\((-9)^4.27^2=9^4.27^2=(3^2)^4.(3^3)^2=3^8.3^6=3^{8+6}=3^{14}\)
\(-5^4:25=-5^4:(5^2)=-5^{4-2}=-5^2\)
Bài 1 :
a) A = \(8^2\) . \(32^4\) = \(\)(2\(^3\))\(^2\) . ( \(2^5\))\(^4\) = 2\(^6\) . 2\(^{20}\) = 2\(^{26}\)
b) B = 27\(^3\) . 9\(^4\) . 243 = ( \(3^3\))\(^3\) . ( \(3^2\) )\(^4\) . 3\(^5\) = 3\(^9\) . \(3^8\) . 3\(^5\) = 3\(^{22}\)
Bài 2 : So sánh
a) A = 27\(^5\) và B =2433
Ta có : 27\(^5\) =(3\(^3\))\(^5\) = 3\(^8\) = 6561
Vì 6561 > 2433 nên A > B .
b) A = 2300 và B = 3\(^{200}\)
Ta có : B = \(3^{200}\) = 3\(^8\) . 3\(^{192}\) = 6561 . 3\(^{192}\)
Vậy chắc chắn rằng B > A .
Nữ hoàng cảm ơn nhưng vị thám tử tài ba có thể viết cách giải ra đc không ạ
\(16^6:4^2=2^{24}:2^4=2^{24-4}=2^{20}\)
\(12^n:2^{2n}=12^n:4^n=3^n\)
\(27^8:9^4=3^{24}:3^8=3^{24-8}=3^{16}\)
\(125^5:25^3=5^{15}:5^6=5^{15-6}=5^9\)
\(4^{14}.5^{28}=4^{14}.25^{14}=\left(4.25\right)^{14}=100^{14}\)
\(64^4.16^5:4^{20}=2^{24}.2^{20}:2^{40}=2^{24+20-40}=2^4=16\)
Bạn ấy lm câu a rùi mik lm 3 câu còn lại nha
\(27^{16}:9^{10}\)
\(=3^{16}\cdot9^{16}:9^{10}\)
\(=9^8\cdot9^6=9^{14}\)
\(125^3:25^4\)
\(=5^3\cdot25^3:25^4\)
\(=5\cdot25\cdot25^3:25^4\)
\(=5\cdot25^4:25^4\)
\(=5\cdot1=5\)
\(24^4:3^4-32^{12}:16^{12}\)
\(=8^4\cdot3^4:3^4-2^{12}\cdot16^{12}:16^{12}\)
\(=8^4-2^{12}\)
\(=4096-4096\)
\(=0\)
1,
\(A=2^0+2^1+2^2+..+2^{2006}\)
\(=1+2+2^2+...+2^{2016}\)
\(2A=2+2^2+2^3+..+2^{2007}\)
\(2A-A=\left(2+2^2+2^3+..+2^{2007}\right)-\left(1+2+2^2+..+2^{2006}\right)\)
\(A=2^{2017}-1\)
\(B=1+3+3^2+..+3^{100}\)
\(3B=3+3^2+3^3+..+3^{101}\)
\(3B-B=\left(3+3^2+..+3^{101}\right)-\left(1+3+..+3^{100}\right)\)
\(2B=3^{101}-1\)
\(\Rightarrow B=\frac{3^{100}-1}{2}\)
\(D=1+5+5^2+...+5^{2000}\)
\(5D=5+5^2+5^3+...+5^{2001}\)
\(5D-D=\left(5+5^2+..+5^{2001}\right)-\left(1+5+...+5^{2000}\right)\)
\(4D=5^{2001}-1\)
\(D=\frac{5^{2001}-1}{4}\)
Bài 1:
\(\text{a) }x.x^2.x^3.x^4.x^5.....x^{49}.x^{50}\)
\(=x^{1+2+3+4+5+...+49+50}\)
\(=x^{\frac{51.50}{2}}\)
\(=x^{1275}\)
\(\text{b) Ta có:}\)
\(4^{15}=\left(2^2\right)^{15}=2^{2.15}=2^{30}\)
\(8^{11}=\left(2^3\right)^{11}=2^{3.11}=2^{33}\)
\(\text{Vì }2^{30}< 2^{33}\text{ nên }4^{15}< 8^{11}\)
Bài 2: Tìm x
\(\left(x-1\right)^4:3^2=3^6\)
\(\Rightarrow\left(x-1\right)^4=3^6\times3^2\)
\(\Rightarrow\left(x-1\right)^4=3^8\)
\(\Rightarrow\left(x-1\right)^4=3^{2.4}\)
\(\Rightarrow\left(x-1\right)^4=\left(3^2\right)^4\)
\(\Rightarrow x-1=9\)
\(\Rightarrow x=10\)
Bài 3 và bài 4 mk làm sau
Bài 1 : a) \(x.x^2.x^3.x^4.....x^{49}.x^{50}=x^{1+2+3+...+49+50}\) (Dễ rồi tự tính)
b) \(\hept{\begin{cases}4^{15}=\left(2^2\right)^{15}=2^{30}\\8^{11}=\left(2^3\right)^{11}=2^{33}\end{cases}}\)Rồi tự so sánh đi
Bài 2 :
\(\left(x-1\right)^4\div3^2=3^6\Leftrightarrow\left(x-1\right)^4=3^8=\left(3^2\right)^4=9^4\Leftrightarrow x-1=9\Leftrightarrow x=10\)
Bài 3 :
\(\hept{\begin{cases}27^{15}=\left(3^3\right)^{15}=3^{45}\\81^{11}=\left(3^4\right)^{11}=3^{44}\end{cases}}\) nt
a. \(5^{10}.125^2.625^3=5^{10}.\left(5^3\right)^2.\left(5^4\right)^3=5^{10}.5^6.5^{12}=5^{10+6+12}=5^{28}\)
a: \(=5^{10}\cdot5^6\cdot5^{12}=5^{28}\)
b: \(=10^3\cdot10^8\cdot10^{15}=10^{26}\)
c: \(=2^{20}\cdot2^{20}=2^{40}\)
d: \(=2^{16}\cdot2^{16}\cdot3^8=2^{32}\cdot3^8\)
e: \(=\dfrac{3^{24}}{3^8}=3^{16}\)
f: \(=2^{12}\cdot2^{20}\cdot2^5=2^{37}\)
A = 8\(^2\).32\(^4\)
A = (2\(^3\))\(^2\) .(2\(^5)^4\)
A = 2\(^6\).2\(^{20}\)
A = 2\(^{26}\)
B = 27\(^3\).9\(^4\).243
B = (3\(^3\))\(^3\).(3\(^2\))\(^4\).3\(^5\)
B = 3\(^9\).3\(^8\).3\(^5\)
B = 3\(^{17}.\) 3\(^5\)
B = \(3^{22}\)