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Ta có:
\(8^5+4^7-16^3=\left(2^3\right)^5+\left(2^2\right)^7-\left(2^4\right)^3\)
\(=2^{15}+2^{14}-2^{12}\)
\(=2^4.\left(2^{11}+2^{10}-2^8\right)=16.\left(2^{11}+2^{10}-2^8\right)\)
Vì 16 chia hết cho 16 nên \(16.\left(2^{11}+2^{10}-2^8\right)\) chia hết cho 16
Do đó \(8^5+4^7-16^3\)chia hết cho 16 (đpcm)
Chúc bạn học tốt!!!
\(=2^{15}+2^{14}-2^{12}=2^4\left(2^{11}+2^{10}-2^8\right).\) chia hết cho 16
Ta có \(8^5+4^7-16^3\)
\(=\left(2^3\right)^5+\left(2^2\right)^7-\left(2^4\right)^3\)
\(=2^{15}+2^{14}-2^{12}\)
\(=2^8\left(2^7+2^6-2^4\right)\)
\(=256\left(2^7+2^6-2^4\right)⋮256\)
Vậy \(8^5+4^7-16^3⋮256\left(đpcm\right)\)
Chúc bn học tốt
2) \(\dfrac{x}{y}=\left(\dfrac{x}{y}\right)^2\)
\(\Rightarrow\left(\dfrac{x}{y}\right)^2-\dfrac{x}{y}=0\)
\(\Rightarrow\dfrac{x}{y}\left(\dfrac{x}{y}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{y}=0\Rightarrow x=0;y\in R\\\dfrac{x}{y}-1=0\Rightarrow\dfrac{x}{y}=1\Rightarrow x=y\end{matrix}\right.\)
3) \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}.2^5+2^{15}.1=2^{15}.33⋮33\rightarrowđpcm\)
4)\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\\\left(y+2\right)^2=0\Rightarrow y+2=0\Rightarrow y=-2\end{matrix}\right.\)
\(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\\\left(x-4-y\right)^{200}\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-12+y\right)^{200}+\left(x-y-4\right)^{200}\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-y-4\right)^{200}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-12+y=0\Rightarrow x+y=12\\x-y-4=0\Rightarrow x-y=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+y\right)+\left(x-y\right)=12+4\Rightarrow x+y+x-y=16\Rightarrow2x=16\Rightarrow x=8\\y=8-4=4\end{matrix}\right.\)
Bài 2:
\(A=\frac{8^5(-5)^8+(-2)^5.10^9}{2^{16}.5^7+20^8}\) \(=\frac{(2^3)^5(-5)^8+(-2)^5.2^9.5^9}{2^{16}.5^7+(2^2.5)^8}\)
\(=\frac{2^{15}.5^8-2^5.2^9.5^9}{2^{16}.5^7+2^{16}.5^8}\)
\(=\frac{2^{14}.5^8(2-5)}{2^{16}.5^7(1+5)}\)
\(=\frac{5(-3)}{2^2.6}=\frac{-5}{8}\)
Bài 3:
Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\)
Thay vào:
\(\frac{5a+3b}{5a-3b}=\frac{5bt+3b}{5bt-3b}=\frac{b(5t+3)}{b(5t-3)}=\frac{5t+3}{5t-3}\)
\(\frac{5c+3d}{5c-3d}=\frac{5dt+3d}{5dt-3d}=\frac{d(5t+3)}{d(5t-3)}=\frac{5t+3}{5t-3}\)
Do đó: \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) (đpcm)
Bài 4:
Ta có:
\(A=3+3^2+3^3+3^4+...+3^{100}\)
\(=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+....+(3^{97}+3^{98}+3^{99}+3^{100})\)
\(=3(1+3+3^2+3^3)+3^5(1+3+3^2+3^3)+...+3^{97}(1+3+3^2+3^3)\)
\(=3.40+3^5.40+....+3^{97}.40\)
\(=120(1+3^4+....+3^{96})\vdots 120\)
Ta có đpcm.
a ) 76 + 75 - 74
= 74 ( 72 + 7 - 1 )
= 74. 55 chia hết cho 55
b ) 165 + 215
= ( 24 ) 5 + 215
= 220 + 215
= 215 ( 25 + 1 )
= 215 . 33 chia hết cho 33
c ) 817 - 279 - 913
= ( 34 )7 - ( 33 )9 - ( 32 )13
= 328 - 327 - 326
= 326 ( 32 - 3 - 1 )
= 326 . 5
= 322 . 34 . 5
= 322 . 81 . 5
= 322 . 405 chia hết cho 405
\(8^5+4^7-16^3=2^{15}+2^{14}-2^{12}=2^4.2^{11}+2^4.2^{10}-2^4.2^8\)
\(=2^4\left(2^{11}+2^{10}-2^8\right)=16\left(2^{11}+2^{10}-2^8\right)⋮16\) (đpcm)
Vậy \(8^5+4^7-16^3⋮16\)
=45056
ko chia được cho 6 chỉ có thể viết dưới dạng phân số