Chứng minh rằng:
\(81^7-27^9-9^{13}⋮405\)
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Ta có:817-279-913
=(34)7-(33)9-(32)13
=328-327-326=326.(32-3-1)=326.5=322.34.5=322.405 luôn chia hết cho 405
=>đpcm
7^6+7^5-7^4 = 7^4*(7^2+7-1) = 7^4*55
mình học lớp 5 mong bạn thông cảm và
Ta có : \(\left(x-\frac{1}{2}\right)^2+\left|y+\frac{1}{3}\right|=0\)
Mà \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\left|x+\frac{1}{3}\right|\ge0\forall x\)
Nên : \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left|x+\frac{1}{3}\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\x+\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{3}\end{cases}}\)
81^7 - 27^9 - 9^13
=(34)7-(33)9-(32)13
=328-327-326
=322.36-322.35-322.34
=322.(36-35-34)
=322.405
=> chứng minh rằng: 81^7 - 27^9 - 9^13 chia hết cho 405
\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-2^{27}-3^{26}\)
=\(3^{26}.3^2-3^{26}.3-3^{26}=3^{26}.\left(3^2-3-1\right)=3^{26}.5=3^{22}.3^4.5\)
=3^22.405 chia het ch 405
nho tick cho tớ nha
a) 87-218
=(23)7-218
=221-218
=218.(23-1)
=218. 7
=217.2.7
=217.14 chia het cho 14
81^7-27^9-9^13
=(3^4)^7-(3^3)^9-(3^2)^13
=3^28-3^27-3^26
=(3^26.3^2)-(3^26.3^1)-(3^26.1)
=3^26.(9-3-1)
=3^22.(3^4.5)
=3^22.405 chia het cho 405
=> 81^7-27^9-9^13 chia het cho 405
a) ta có : \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.\left(49+7-1\right)=7^4.55⋮55\)
\(\Rightarrow7^4.55\) chia hết cho \(55\) \(\Leftrightarrow7^6+7^5-7^4\) chia hết cho \(55\)
vậy \(7^6+7^5-7^4\) chia hết cho \(55\) (đpcm)
b) ta có \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}.\left(32+1\right)=2^{15}.33⋮33\)
\(\Rightarrow2^{15}.33\) chia hết cho \(33\) \(\Leftrightarrow16^5+2^{15}\) chia hết cho \(33\)
vậy \(16^5+2^{15}\) chia hết cho \(33\) (đpcm)
c) ta có \(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}\)
\(=3^{22}\left(3^6-3^5-3^4\right)=3^{22}\left(729-243-81\right)=3^{22}.405⋮405\)
\(\Rightarrow3^{22}.405\) chia hết cho \(405\) \(\Leftrightarrow81^7-27^9-9^{13}\) chia hết cho \(405\)
vậy \(81^7-27^9-9^{13}\) chia hết cho \(405\) (đpcm)
\(a.\)
\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55⋮55\)
\(b.\)
\(16^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}.33⋮33\)
\(c.\)
Ta có : \(405=3^4.5\)
\(\Rightarrow81^7-27^9-9^{13}=3^{28}-3^{27}-3^{26}=3^{26}\left(3^2-3-1\right)=3^{26}.5⋮405\)
81^7-27^9-9^13
=(3^4)^7-(3^3)^9-(3^2)^13
=3^28-3^27-3^26
=(3^26.3^2)-(3^26.3^1)-(3^26.1)
=3^26.(9-3-1)
=3^22.(3^4.5)
=3^22.405 chia het cho 405
=> 81^7-27^9-9^13 chia het cho 405
81^7-27^9-9^13
=(3^4)^7-(3^3)^9-(3^2)^13
=3^28-3^27-3^26
=(3^26.3^2)-(3^26.3^1)-(3^26.1)
=3^26.(9-3-1)
=3^22.(3^4.5)
=3^22.405 chia het cho 405
=> 81^7-27^9-9^13 chia het cho 405
\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)\(=3^{28}-3^{27}-3^{26}=3^{26}\times\left(3^2-3-1\right)=3^{26}\times5\div3^4\times5\)
Nguyễn huy hoàng ơi, bn giải thích cho mk: 326x5:34 x5 vs....
\(81^7-27^9-9^{13}\)
\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}\left(3^2-3-1\right)\)
\(=3^{26}.5=3^{22}.405⋮405\)