Tìm x thuộc N biết
\(5^n+5^{n+2}=650\)
\(32^{-n}.16^n=1024\)
\(3^{-1}.3^n+5.3^{n-1}=162\)
Ai nhanh tik
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(2008^n=1\Rightarrow n=0\)
\(2^n=1024\Rightarrow2^n=2^{10}\Rightarrow n=10\)
tíc mình nha
\(a,2008^n=1=>n=0.\)
\(b,5^n+5^{n+2}=650\)
\(=>5^n\left(1+5^2\right)=650\)
\(=>5^n.26=650\)
\(=>5^n=650:26\)
tự tính tiếp nhé !!
\(c,2^n=1024=2^{10}=>n=10\)
\(d,3^{n-1}5.3^{n-1}=162\)mình nghĩ đề thiếu !
\(f\)) \(32^{-x}.16^x=1024\)
\(\left(2\right)^{-5x}.2^{4x}=2^{10}\)
\(\Leftrightarrow2^{4x-5x}=2^{10}\)
\(\Leftrightarrow2^{-x}=2^{10}\)
\(\Leftrightarrow-x=10\)
\(\Leftrightarrow x=-10\)
\(g\)) \(3^{x-1}.5+3^{x-1}=162\)
\(3^{x-1}.\left(5+1\right)=162\)
\(3^{x-1}.6=162\)
\(3^{x-1}=162:6\)
\(3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
\(h\)) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^6.\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(2x-1\right)^6.\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=1\\\left(2x-1\right)^2=\left(1,-1\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x-1=-1\\2x-1=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x=0\\2x=2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=0\\x=1\end{cases}}\)
\(i\)) \(5^x+5^{x+2}=650\)
\(5^x.\left(1+5^2\right)=650\)
\(5^x.26=650\)
\(5^x=650:26\)
\(5^x=25\)
\(\Leftrightarrow5^x=5^2\)
\(\Leftrightarrow x=2\)
7^6+7^5+7^4 chia hết cho 11
= 7^4.2^2+7^4.7+7^4
= 7^4.(2^2+7+1)
= 7^4. 11
Vì tích này có số 11 nên => chia hết cho 7
a) \(15+2^n=31\)
\(2^n=16\Rightarrow n=4\)
b) \(2.2^n+4.2^n=6.2^5\)
\(2^n\left(2+4\right)=6.2^5\)
\(2^n.6=6.2^5\Rightarrow n=5\)
c) \(32^n:16^n=1024\)
\(\left(2^5\right)^n:\left(2^4\right)^n=2^{10}\)
\(2^{5n}:2^{4n}=2^{10}\)
\(2^n=2^{10}\Rightarrow n=10\)
d) \(5^n+5^{n+2}=650\)
\(5^n+5^n.25=650\)
\(5^n\left(1+25\right)=650\)
\(5^n.26=650\)
\(5^n=25\Rightarrow n=2\)
e) \(3^n+5.3^{n+1}=432\)
\(3^n+5.3^n.3=432\)
\(3^n\left(1+15\right)=432\)
\(3^n.16=432\)
\(3^n=27\Rightarrow n=3\)
a)
\(\left(\frac{1}{3}\right)^n\cdot27^n=3^n\)
\(\Rightarrow\left(\frac{1}{3}\cdot27\right)^n=3^n\)
\(\Rightarrow9^n=3^n\)
\(\Rightarrow\left(3^2\right)^n=3^n\)
\(\Rightarrow3^{2n}=3^n\)
\(\Rightarrow2n=n\)
\(\Leftrightarrow n=0\)
Vậy \(n=0\)
d) Ta có:
\(6^{3-n}=216\)
\(\Rightarrow6^{3-n}=6^3\)
\(\Rightarrow3-n=3\)
\(\Rightarrow n=3-3\)
\(\Rightarrow n=0\)
Vậy \(n=0\)\(\text{ }\)
1. Ta có: \(x\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
=> \(x\left(6-x\right)^{2003}-\left(6-x\right)^{2003}=0\)
=> \(\left(6-x\right)^{2003}\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}\left(6-x\right)^{2003}=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}6-x=0\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=6\\x=1\end{cases}}\)
Bài 2. Ta có: (3x - 5)100 \(\ge\)0 \(\forall\)x
(2y + 1)100 \(\ge\)0 \(\forall\)y
=> (3x - 5)100 + (2y + 1)100 \(\ge\)0 \(\forall\)x;y
Dấu "=" xảy ra khi: \(\hept{\begin{cases}3x-5=0\\2y+1=0\end{cases}}\) => \(\hept{\begin{cases}3x=5\\2y=-1\end{cases}}\) => \(\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{1}{2}\end{cases}}\)
Vậy ...
\(A=1+3+3^2+3^3+...+3^{101}\)
\(3A=3+3^2+3^3+3^4+...+3^{101}\)
\(3A-A=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)
\(2A=3^{101}-1\)
\(A=\left(3^{101}-1\right):2\)
Thu gọn tổng sau:
A=1+3+32+33+...+3100
B= 2100-299-298-297-...-22-2
C= 3100-399+398-397-...+32-3+1
a, 5n+5n+2=650
=>5n+5n.52=650
=>5n(1+25)=650
=>5n.26=650
=>5n=25
=>5n=52
=>n=2
Vậy n=2