a/ \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\\3+2^{x+1}=24-\left[16-\left(4-1\right)\right]\)
\(3+2^{x+1}=24-\left(16-3\right)\\ 3+2^{x-1}=24-13\\ 3+2^{x-1}=11\\ 2^{x+1}=11-3\\ 2^{x-1}=8\)
\(2^{x-1}=2^3\\ \Rightarrow x-1=3\\x=3+1\\ x=4\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=205550\)
\(\left(x.100\right)+\left(1+2+3+....+100\right)=205550\)
Ta tính tổng \(1+2+3+...+100\\ \) trước
Số các số hạng: \(\left[\left(100-1\right):1+1\right]=100\)
Tổng :\(\left[\left(100+1\right).100:2\right]=5050\)
Thay số vào ta có được:
\(\left(x.100\right)+5050=205550\\ \\ x.100=205550-5050\\ \\x.100=20500\\ \\x=20500:100\\ \\\Rightarrow x=2005\)
\(\left|x-5\right|=18+2.\left(-8\right)\\\left|x-5\right|=18+\left(-16\right)\\\left|x-5\right|=2\: \)
\(\Rightarrow\left[\begin{array}{nghiempt}x-5=2\\\\x-5=\left(-2\right)\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=2+5\\\\x=\left(-2\right)+5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=7\\\\x=3\end{array}\right.}\)
=> x ϵ {7;3}
\(\left(3.x-2^4\right).7^5=2.7^6.\frac{1}{2009^0}\\ \left(3.x-16\right).7^5=2.7^6.\frac{1}{1}\)
\(\left(3.x-16\right).7^5=2.7^6.1\\\left(3.x-16\right).7^5=2.7^6 \)
\(3.x-16.=\frac{2.7^6}{7^5}\\3.x-16=2.7\\ 3.x-16=14\\ 3.x=14+16\\3.x=30\\x=30:3\\ x=10\)
ủa cái đó mk vào fx đánh ra rồi mà s h ns ko hgien65 ra z ta?? để mk hỏi thầy @phynit