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3/ ta để ý thấy ở số mũ sẽ có thừa số 1000-103=0
nên số mũ chắc chắn bằng 0
mà số nào mũ 0 cũng bằng 1 nên A=1
5/ vì |2/3x-1/6|> hoặc = 0
nên A nhỏ nhất khi |2/3x-6|=0
=>A=-1/3
6/ =>14x=10y=>x=10/14y
23x:2y=23x-y=256=28
=>3x-y=8
=>3.10/4y-y=8
=>6,5y=8
=>y=16/13
=>x=10/14y=10/14.16/13=80/91
8/106-57=56.26-56.5=56(26-5)=59.56
có chứa thừa số 59 nên chia hết 59
4/ tính x
sau đó thế vào tinh y,z
3.
\(2^x=256+2^y\\ \Rightarrow2^x-2^y=256\\ \Rightarrow2^y\left(2^{x-y}-1\right)=2^8\)
\(\Rightarrow2^y;2^{x-y}-1\in U\left(2^8\right)\)
Mà \(2^{x-y}-1\) là số lẻ
\(\Rightarrow2^{x-y}-1=1\\ \Rightarrow\left\{{}\begin{matrix}2^y=2^8\\2^{x-y}=2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}y=8\\x-y=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}y=8\\x=9\end{matrix}\right.\)
4.
Gọi d là ƯCLN(2n+5;3n+7)
\(\Rightarrow\left\{{}\begin{matrix}2n+5⋮d\\3n+7⋮d\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}3\left(2n+5\right)⋮d\\2\left(3n+7\right)⋮d\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}6n+15⋮d\\6n+14⋮d\end{matrix}\right.\\ \Rightarrow\left(6n+15\right)-\left(6n+14\right)⋮d\\ \Rightarrow1⋮d\\ \Rightarrow d=1\)
=> đpcm
Nguyễn Huy Tú lê thị hương giang Hồng Phúc Nguyễn
Nguyễn Thanh Hằng Akai Haruma Nam Nguyễn Hà Nam Phan Đình
Aki Tsuki
Đặt \(A=\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-\dfrac{1}{3^4}+......+\dfrac{1}{3^{2015}}-\dfrac{1}{3^{2016}}\)
\(3A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\dfrac{1}{3^4}-......+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}\)
\(3A+A=4A=1-\dfrac{1}{3^{2016}}\)
\(A=\dfrac{1-\dfrac{1}{3^{2016}}}{4}=\dfrac{\dfrac{3^{2016}-1}{3^{2016}}}{4}=\dfrac{3^{2016}-1}{3^{2016}.4}\)
P/s : Chắc là vậy
T lm câu 2 trc nhé
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó \(\frac{2a+3c}{2b+3d}=\frac{2.bk+3.dk}{2b+3d}=\frac{k\left(2b+3d\right)}{2b+3d}=k\left(1\right)\)
\(\frac{2a-3c}{2b-3d}=\frac{2bk-3dk}{2b-3d}=\frac{k\left(2b-3d\right)}{2b-3d}=k\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\) .....đpcm
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó \(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{b^2.k^2+d^2.k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\) ( *1 )
\(\frac{ac}{bd}=\frac{bk.dk}{bd}=\frac{k^2.bd}{bd}=k^2\) ( *2)
Từ (*1) và (*2) \(\Rightarrow\) ...... ( đpcm)
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}=\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\Rightarrow a=b=c\Rightarrow M=1\)
Chứng minh rằng :
\(a.\)
\(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
\(b.\)
\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)
\(.a.\) \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
Ta có : \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n.\left(3^2+2\right)-2^n.\left(2^2+1\right)\)
\(=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10.\left(3^n-2^{n-1}\right)⋮10\) \(\left(dpcm\right)\)
Vậy : \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
\(.b.\) \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)
\(=3^n.30+2^n.12\)
\(=6\left(3^n.5+2^{n+1}\right)⋮6\) \(\left(dpcm\right)\)
Vậy : \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)
a)\(VT=3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(=3^n\cdot10-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot2\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot10\)
\(=10\cdot\left(3^n-2^{n-1}\right)⋮10\)
b)\(VT=3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=\left(3^{n+3}+3^{n+1}\right)+\left(2^{n+3}+2^{n+2}\right)\)
\(=3^{n+1}\left(3^2+1\right)+2^{n+2}\left(2+1\right)\)
\(=3^{n+1}\cdot10+2^{n+2}\cdot3\)
\(=3^n\cdot3\cdot2\cdot5+2^{n+1}\cdot2\cdot3\)
\(=3^n\cdot5\cdot6+2^{n+1}\cdot6\)
\(=6\cdot\left(3^n\cdot5+2^{n+1}\right)⋮6\)
Ta có
\(B=3+3^2+3^3+....+3^{2015}\)
\(3B=3^2+3^3+....+3^{2016}\)
\(\Rightarrow3B-B=\left(3^2+3^3+....+3^{2016}\right)-\left(3+3^2+....+3^{2015}\right)\)
\(\Rightarrow2B=3^{2016}-3\)
\(\Rightarrow2B+3=3^{2016}\)
Ta có:
\(B=3+3^2+...+3^{2015}\)
\(\Rightarrow3B=3^2+3^3+3^4+...+3^{2016}\)
\(\Rightarrow3B-B=\left(3^2+3^3+...+3^{2016}\right)-\left(3+3^2+...+3^{2016}\right)\)
\(\Rightarrow2B=3^{2016}-3\)
Thay 2B vào \(2B+3=3^n\) ta có:
\(3^{2016}-3+3=3^n\)
\(\Rightarrow3^{2016}=3^n\)
\(\Rightarrow n=2016\)
Vậy n = 2016