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\(S=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\\ =\left(3+3^2+3^3\right)+3^3.\left(3+3^2+3^3\right)+3^6.\left(3+3^2+3^3\right)\\ =39+3^3.39+3^6.39\\ =-39.\left(-1-3^3-3^6\right)⋮\left(-39\right)\)
S = 3 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39
S = ( 3 + 32 + 33 ) +34 + 35 + 36 + 37 + 38 + 39
S = 39 + 34 + 35 + 36 + 37 + 38 + 39
Vì 39 ⋮ -39
<=> S ⋮ -39
C = 3 - 32 + 33 - 34 + 35 - 36 +...+ 323 - 324
3C = 32 - 33 + 34 - 35 + 36-...- 323 + 324 - 325
3C - C = -325 - 3
2C = -325 - 3
2C = - ( 325 + 3) = - [(34)6. 3 + 3] = - [\(\overline{...1}\)6.3+3] = -[ \(\overline{..3}\) + 3]
2C = - \(\overline{..6}\)
⇒ \(\left[{}\begin{matrix}C=\overline{..3}\\C=\overline{..8}\end{matrix}\right.\)
⇒ C không thể chia hết cho 420 ( xem lại đề bài em nhé)
b, (\(x+1\))2022 + (\(\sqrt{y-1}\) )2023 = 0
Vì (\(x+1\))2022 ≥ 0
\(\sqrt{y-1}\) ≥ 0 ⇒ (\(\sqrt{y-1}\))2023 ≥ 0
Vậy (\(x\) + 1)2022 + (\(\sqrt{y-1}\))2023 = 0
⇔ \(\left\{{}\begin{matrix}\left(x+1\right)^{2022}=0\\\sqrt{y-1}=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x+1=0\\y-1=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Kết luận: cặp (\(x,y\)) thỏa mãn đề bài là:
(\(x,y\)) = (-1; 1)
Ta có: A=32.32+25.2-32
=32.32+32.2-32
=32(32+2-1)
=32.33 chia hết cho 33(đpcm)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
= ( 72024 + 32 ). ( 71012 . 71012 + 34 )
= ( 72024 + 32 ) . ( 72024 + 34 )
= 72024 ( 32 + 34 )
= 72024 . 66 ⋮ 6
\(16^{10}+32^7=\left(2^4\right)^{10}+\left(2^5\right)^7=2^{40}+2^{35}=2^{35}.2^5+3^{35}=2^{35}.\left(2^5+1\right)=2^{35}.33\)
chia hết cho 33
tick nhé
Ta có : 32 = 25
1610 = (24)10 = 240 chia hết cho 25
347 = (17.2)7 = 177.27 mà 27 chia hết cho 25 nên 177.27 chia hết cho 25
=> 1610 + 347 chia hết cho 32
+) Vì \(3⋮3\); \(3^2⋮3\); \(3^3⋮3\); \(3^4⋮3\); .............. ; \(3^{119}⋮3\); \(3^{120}⋮3\)
\(\Rightarrow3+3^2+3^3+3^4+.........+3^{119}+3^{120}⋮3\)
hay \(A⋮3\)
+) \(A=3+3^2+3^3+3^4+..........+3^{119}+3^{120}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+..........+\left(3^{119}+3^{120}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+.........+3^{119}\left(1+3\right)\)
\(=3.4+3^3.4+........+3^{119}.4=4.\left(3+3^3+.......+3^{119}\right)⋮4\)
+) \(A=3+3^2+3^3+3^4+...........+3^{119}+3^{120}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+........+\left(3^{118}+3^{119}+3^{120}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+..........+3^{118}\left(1+3+3^2\right)\)
\(=3.13+3^4.13+.......+3^{118}.13=13.\left(3+3^4+........+3^{118}\right)⋮13\)
Vậy \(A⋮3,4,13\)
A = 3 + 32 + 33 + ... + 3120
= 3 (1 + 3 + 32 + ... + 3119)
Vì 3 chia hết cho 3 nên 3 (1 + 3 + 32 + ... + 3119) chia hết cho 3
=> A chia hết cho 3 (đpcm)
A = 3 + 32 + 33 + ... + 3120
= (3 + 32) + (33 + 34) + ... + (3119 + 3120)
= 3 (1 + 3) + 33 (1 + 3) + ... + 3119 (1 + 3)
= 3 . 4 + 33 . 4 + ... + 3119 . 4
Vì 4 chia hết cho 4 nên 3 . 4 + 33 . 4 + ... + 3119 . 4 chia hết cho 4
=> A chia hết cho 4 (đpcm)
A = 3 + 32 + 33 + ... + 3120
= (3 + 32 + 33) + (34 + 35 + 36) + ... + (3118 + 3119 + 3120)
= 3 (1 + 3 + 32) + 34 (1 + 3 + 32) + ... + 3118 (1 + 3 + 32)
= 3 . 13 + 34 . 13 + ... + 3118 . 13
Vì 13 chia hết cho 13 nên 3 . 13 + 34 . 13 + ... + 3118 . 13 chia hết cho 13
=> A chia hết cho 13 (đpcm)