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ne ban minh biet cau tra loi nhung lam the nao bam ngoac vuong
\(A=5+5^2+5^3+...+5^{2016}\)
\(5A=5^2+5^3+5^4+...+5^{2017}\)
\(\rightarrow5A-A=5^{2017}-5\)
\(4A=5^{2017}-5\)
\(\Rightarrow4A+5=5^{2017}-5+5\)
Mà \(4A+5=5^x\)
\(\Rightarrow5^x=5^{2017}\)
Vậy \(x=2017\)
a) \(3^{x+1}.15=135\)
\(\Rightarrow3^{x+1}=9\)
\(\Rightarrow3^{x+1}=3^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
b) \(x+2x+2^2x+....+2^{2016}x=2^{2017}-1\\ \Rightarrow x\left(2+2^2+...+2^{2016}\right)=2^{2017}-1\\ \Rightarrow x\left(2^{2017}-2\right)=2^{2017}-1\)
c) \(x\left(x-1\right)+\left(x-1\right)^2=0\\ \Rightarrow x\left(x-1\right)+\left(x-1\right)\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+\left(x-1\right)\right)=0\\ \Rightarrow\left(x-1\right)\left(2x-1\right)=0\\ \Rightarrow\begin{cases}x-1=0\\2x-1=0\end{cases}\)
d) \(2^2.2^5\le2^{x-5}\le2^{10}\\ \Rightarrow2^7\le2^{x-5}\le2^{10}\)
Số số hạng của biểu thức \(\left(20182018-122018\right):20000+1=1004\)
Tổng là \(\left(20182018+122018\right).1004:2\)
\(=20304036.1004:2=.......2\)
Suy ra chữ số tận cùng của tổng đố là chữ số 2
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
\(\left(3^{2016}\cdot11+3^{1018}\cdot50\right):\left(3^{2017}\cdot4^2\right)\)
\(=\left(3^{2016}\cdot11+3^{2016}\cdot3^2\cdot50\right):\left(3^{2017}\cdot4^2\right)\)
\(=\left(3^{2016}\cdot11+3^{2016}\cdot450\right):\left(3^{2017}\cdot16\right)\)
\(=\left[3^{2016}\cdot\left(11+450\right)\right]:\left(3^{2017}\cdot16\right)\)
\(=\left[3^{2016}\cdot461\right]:\left(3^{2017}\cdot16\right)\)
\(=\frac{3^{2016}\cdot461}{3^{2017}\cdot16}\)
\(=\frac{3^{2016}\cdot461}{3\cdot3^{2016}\cdot16}\)
\(=\frac{461}{3\cdot16}=\frac{461}{48}\)
Nghiệm?
\(\left(x-5\right)^{2016}+\left(x-5\right)^{2018}=0\\ \Rightarrow\left(x-5\right)^{2016}\left[1+\left(x-5\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\1+\left(x-5\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x\in\varnothing\end{matrix}\right.\)
Yêu cầu bài là gì bạn nhỉ?