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\(A\frac{27^4.8^{17}}{9^6.32^3}=\frac{\left(3^3\right)^4.\left(2^3\right)^{17}}{\left(3^2\right)^6.\left(2^5\right)^3}=\frac{3^{12}.2^{51}}{3^{12}.2^{15}}=\frac{3^{12}.2^{15}.2^{36}}{3^{12}.2^{15}}=2^{36}\)
\(B=\frac{72^3.54^3:8^3}{108^5:4^5}=\frac{\left(72.54:8\right)^3}{\left(108:4\right)^5}=\frac{486^3}{27^5}=\frac{\left(3^5.2\right)^3}{\left(3^3\right)^5}=\frac{3^{15}.2^3}{3^{15}}=2^3=8\)
Bài 2
A = 2 +22 + 23 + 24 + ....+ 2100
A = ( 2+22 ) + (23 + 24 ) + ....+ (299 + 2100 )
A = 2(1+2 ) + 23 (1+2 ) + ...+ 299(1+2)
A = 2.3 + 23.3 + ....+ 299 .3
A = 3(2+23 + ...+ 299 )
=> A \(⋮\) 3 ( đpcm )
Bài 3
a, 2.3x = 312 .34 + 20 .274
2.3x = 312 . 34 + 20 . (33 ) 4
2.3x = 312 .34 + 20 .312
2.3x = 312(34+20 )
2.3x = 312 . 54
2.3x = 312 . 27 .2
2.3x = 312 . 33 .2
2.3x = 315 .2
=> x=15
b , (2x +1 ) 2 + 3.(22 + 1 ) = 22 .10
(2x +1 ) 2 + 3.(4+1 ) = 4.10
(2x +1 ) 2 + 3.5 = 40
(2x +1 ) 2 + 15 = 40
(2x +1 ) 2 = 40-15
(2x +1 ) 2 = 25
(2x +1 ) 2 = 52
=> 2x + 1 = 5
2x = 5-1
2x = 4
2x = 22
=> x=2
A=4+(22+23+24+...+220)
A-4=22+23+24+...+220
2(A-4)=23+24+25+...+221
A-4=2(A-4)-(A-4)=(23+24+25+...+221)-(22+23+24+...+220)
A-4=(23-23)+(24-24)+(25-25)+...+(220-220)+(221-22)
A-4=221-4
A =221-4+4
A =221
Bạn làm tiếp nha .
Câu 4
Đặt \(A=3+3^2+...+3^{20}\)
\(\Rightarrow A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{19}+3^{20}\right)\)
\(\Rightarrow A=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{19}\left(1+3\right)\)
\(\Rightarrow A=3.4+3^3.4+...+3^{19}.4\)
\(\Rightarrow A=\left(3+3^3+...+3^{19}\right).4⋮4\)
\(\Rightarrow A⋮4\left(đpcm\right)\)
\(A=3+3^2+...+3^{20}\)
\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{17}+3^{18}+3^{19}+3^{20}\right)\)
\(\Rightarrow A=3\left(1+3+3^2+3^3\right)+...+3^{17}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=3.40+...+3^{17}.40\)
\(\Rightarrow A=\left(3+...+3^{17}\right).40⋮40\)
\(\Rightarrow A⋮40\left(đpcm\right)\)
Câu 3:
Giải:
a) \(5⋮x-5\)
\(\Rightarrow x-5\in\left\{1;5\right\}\)
+) \(x-5=1\Rightarrow x=6\)
+) \(x-5=5\Rightarrow x=10\)
Vậy \(x\in\left\{6;10\right\}\)
b) Ta có: \(x+3⋮x-3\)
\(\Rightarrow\left(x-3\right)+6⋮x-3\)
\(\Rightarrow6⋮x-3\)
\(\Rightarrow x-3\in\left\{1;2;3;6\right\}\)
\(\Rightarrow x\in\left\{4;5;6;9\right\}\)
Vậy \(x\in\left\{4;5;6;9\right\}\)
\(M=2+2^3+2^5+2^7+....+2^{51}\)
\(=\left(2+2^3\right)+\left(2^5+2^7\right)+....+\left(2^{49}+2^{51}\right)\)
\(=10+2^4\left(2+2^3\right)+....+2^{48}\left(2+2^3\right)\)
\(=10+2^4.10+...+2^{48}.10\)
\(=10\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮10\)
\(=2.5.\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮5\)
\(M=2+2^3+2^5+2^7+....+2^{51}.\)
\(M+2^{ }=2+2+2^3+2^5+2^7+.....+2^{51}\)
\(=\left(2+2+2^3\right)+\left(2^5+2^7+2^9\right)+....+\left(2^{47}+2^{49}+2^{51}\right)\)
\(=12+2^4\left(2+2^3+2^5\right)+......+2^{46}\left(2+2^3+2^5\right)\)
\(=12+2^4.42+....+2^{46}.42\)
\(=12+7.3.2\left(2^4+...+2^{46}\right)\)
\(\Rightarrow M=\left[12+7.3.2\left(2^4+.....+2^{46}\right)\right]-2\)
\(=10+7.3.2\left(2^4+....+2^{46}\right)\)
Ta có: \(7.3.2\left(2^4+...+2^{46}\right)⋮7\)mà 10 không chia hết cho 7
Suy M không chia hết cho 7
Bài 1:
Ta có: \(\overline{ababab}=10101.\overline{ab}⋮3\)
\(\Rightarrow\overline{ababab}\in B\left(3\right)\left(đpcm\right)\)
Bài 3:
Đặt \(A=\frac{1}{2^2}+...+\frac{1}{2^n}\)
\(\Rightarrow2A=\frac{1}{2}+...+\frac{1}{2^{n-1}}\)
\(\Rightarrow2A-A=\frac{1}{2}-\frac{1}{2^n}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^n}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
Câu 1 :
\(2^{x+1}+2^{x+2}+2^{x+3}=112\)
\(2^x\cdot\left(2+2^2+2^3\right)=112\)
\(2^x\cdot\left(2+2^2+2^3\right)=112\)
\(2^x\cdot14=112\)
\(2^x=8\)
\(2^x=2^3\)
=> x = 3
Câu 2 :
Ta có :
B = 3 + 3 2 + 3 3 + 3 4 + ... + 3 20
B = ( 3 + 3 2 ) + ( 3 3 + 3 4 ) + ... + ( 3 19 + 3 20 )
B = ( 3 + 3 2 ) + ( 3 + 3 2 ) . 3 2 + ... + ( 3 + 3 2 ) . 3 18
B = 12 + 12 . 3 2 + ... + 12 . 3 18
B = 12 ( 1 + 3 2 + .... + 3 18 )
Vì 12 chia hết cho 12
=> B = 12 ( 1 + 3 2 + .... + 3 18 ) chia hết cho 12
Vậy B là bội của 12