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P =2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10
= ( 2+2^2 ) + ( 2^3 + 2^4 ) + ( 2^5 + 2^6 ) + ( 2^7 + 2^8 ) + ( 2^9 + 2^10 )
= 2.( 1+2 ) + 2^3.( 1+2 ) + 2^5.( 1+2 ) + 2^7.( 1+2 ) + 2^9.( 1 + 2 )
= 2.3+2^3.3+2^5.3+2^7.3+2^9.3
= ( 2+2^3+2^5+2^7+2^9).3 chia hết cho 3
=> P chia hết cho 3
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abc + acb = bca
abc
acb
-----
bca
----
(1)c+b=a+10.t (t=0hoăc1)
(2)b+c+t=c+10k (k=0 hoặc 1)
(3)a+a+k=b
--------
(2)=>b+t=10k
b<>0=>k=1
b<10=>t=1=>b=9
(3)=>a=4
(1)=>c=5
abc=495
495+459=954
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Gọi d là \(ƯC\left(2n+3;3n+4\right)\)
Ta có: \(2n+3⋮d\Rightarrow3\left(2n+3\right)⋮d\Leftrightarrow6n+9⋮d\)
\(3n+4⋮d\Rightarrow2\left(3n+4\right)⋮d\Rightarrow6n+8⋮d\)
\(\Rightarrow\left(6n+9\right)-\left(6n+8\right)⋮d\)
\(\Rightarrow6n+9-6n-8⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d\inƯ\left(1\right)=\left\{1\right\}\)
Vậy \(ƯCLN\left(2n+3;3n+4\right)=1\left(đpcm\right)\)
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|-x| < 5
=> |-x| \(\in\){0;1;2;3;4}
|-x| = 0
=> x = 0
|-x| = 1
=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
|-x| = 2
=> \(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
|-x| = 3
=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
|-x| = 4
=> \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
Vậy x \(\in\){0;-1;1;-2;2;-3;3;-4;4}
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1 + 2 + 3 +... + 100 = (100 + 1) x 100 : 2 = 4950
x + 4950 = 5056
x = 5056 - 4950 = 106
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\(48+172+2015+\left(-220\right)\)
\(=220+2015+\left(-220\right)\)
\(=2235+\left(-220\right)\)
\(=2015\)
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Đây là câu trả lời của mình:
3 x ( 4x -1 )2 = 675
=> ( 4x - 1 )2 = 675 : 3
=> ( 4x - 1 )2 = 225
=> ( 4x - 1 )2 = 152
=> 4x - 1 = 15
=> 4x = 15 + 1
=> 4x = 16
=> x = 16 : 4
=> x = 4
Vậy x = 4
\(3\left(4x-1\right)^2=675\)
\(\left(4x-1\right)^2=675:3\)
\(\left(4x-1\right)^2=225\)
\(\left(4x-1\right)^2=15^2\)
\(4x=15+1\)
\(4x=16\)
\(x=16:4\)
\(x=4\)
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