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a) \(2^{2017}+2^{2014}=2^{2014}\left(2^3+1\right)=2^{2014}.9⋮9\)
b) \(4^{2016}+4^{2014}=4^{2014}\left(4^2+1\right)=4^{2014}.17\)
2) \(3.4^{n+2}+4^n=49\\ \Rightarrow4^n\left(3.4^2+1\right)=49\\ \Rightarrow4^n.33=49\\ \Rightarrow4^n=16\\ \Rightarrow n=2\)
3) \(200-180:\left[36.5-7.25\right]\\ =200-180:\left[180-175\right]\\ =200-180:5\\ =200-36\\ =164\)
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Bài 1: ( sai đề. mình sửa lại là chia hết cho 31)
Ta có:
\(A=1+5+5^2+...+5^{2013}\)
\(A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{2011}+5^{2012}+5^{2013}\right)\)
\(A=5^0\cdot\left(1+5+5^2\right)+5^3\cdot\left(1+5+5^2\right)+...+5^{2011}\cdot\left(1+5+5^2\right)\)
\(A=5^0\cdot31+5^3\cdot31+...+5^{2011}\cdot31\)
\(A=31\cdot\left(5^0+5^3+...+5^{2011}\right)\)
Vì \(31⋮31\)
\(\Rightarrow31\cdot\left(5^0+5^3+...+5^{2011}\right)⋮31\)
hay\(A⋮31\) (đpcm)
Này đề là chia hết cho 13 sao lại làm chia hết cho 31 cô mình ra bài này mà
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c/ 2x - 1 = \(5^{98}:5^{96}\)
2x - 1 = \(5^2\) = 25
2x = 25 + 1 = 26
x = 26 : 2
x = 13
d/ 7x + 3 = \(3^5.2^3.9\)
7x + 3 = \(3^5.3^2.8=3^7.8=2187.8\)
7x + 3 = \(17496\)
7x = 17496 - 3 = 17493
x = 17493 : 7
x = 2499
e/\(2^{2x+6}=1\)
\(2^{2x+6}=2^0\)
2x + 6 = 0
2x = 0 - 6 = - 6
x = - 6 : 2
x = - 3
j/ \(2^x=8\)
\(2^x=2^3\)
x = 3
g/ \(2^x:2^3=16\)
\(2^{x-3}=2^4\)
x - 3 = 4
x = 4 + 3
x = 7
h/ \(2^x+2^{x+1}+2^{x+2}=56\)
\(2^x\left(1+2+2^2\right)\) = 56
\(2^x.7=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
x = 3
Bài a, b thiên phong giải r, mk chỉ làm những bài còn lại thôi. Chúc bạn học tốt!!!
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a) 1 - 2 + 3 - 4 + 5 - 6 + .....+ 25 - 26
= (1 - 2) + (3 - 4) + (5 - 6) + .....+ (25 - 26)
= -1 + (-1) + ( -1 ) +...+ ( -1 ) {có 13 số )
= -13
b) tương tự nhé bn
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a. 52 + (x+3) = 52
=> x + 3 = 52 - 52
=> x + 3 = 0
=> x = -3
b. 23 + (x-32) = 53 - 43
=> 8 + (x-9) = 125 - 64
=> x - 9 = 125 - 64 - 8
=> x - 9 = 53
=> x = 53 + 9
=> x = 62
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a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
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6)\(3.\left(x+1\right)-2^3.2=11\)
\(3\left(x+1\right)-2^4=11\)
\(3.\left(x+1\right)=11+16\)
3.(x+1)=27
x+1=27:3
x+1=9
x=9-1
x=8
Vậy x=8
7) \(180-2.\left(x+5\right)^2=130\)
\(2.\left(x+5\right)^2=180-130\)
2.(x+5)2=50
(x+5)2=50:2
(x+5)2=25
x+5=5 hoặc x+5=-5
x=5-5 x=-5-5
x=0 x=-10
Vậy x=0 hoặc x=-10
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a, \(3^4\div3^2-\left[120-\left(2^6.2+5^2.2\right)\right]\)
\(=3^2-\left\{120-\text{[}2.\left(2^6+5^2\right)\text{]}\right\}\)
\(=3^2-\left(120-2\cdot89\right)\)
\(=9--58=9+58=67\)
1. \(a,3^4:3^2-\left[120-(2^6\cdot2+5^2\cdot2)\right]\)
\(=3^2-\left[120-\left\{(2^6+5^2)\cdot2\right\}\right]\)
\(=3^2-\left[120-\left\{(64+25)\cdot2\right\}\right]\)
\(=9-\left[120-89\cdot2\right]\)
\(=9-\left[120-178\right]=9-(-58)=67\)
b, Tương tự như bài a
2.a,\(4^x\cdot5+4^2\cdot2=2^3\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+16\cdot2=8\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+32=56+56\)
\(\Leftrightarrow4^x\cdot5+32=112\)
\(\Leftrightarrow4^x\cdot5=80\)
\(\Leftrightarrow4^x=16\Leftrightarrow4^x=4^2\Leftrightarrow x=2\)
\(b,24:(2x-1)^3-2=1\)
\(\Leftrightarrow24:(2x-1)^3=3\)
\(\Leftrightarrow(2x-1)^3=8\)
\(\Leftrightarrow(2x-1)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
Làm nốt là xong thôi