Question

[Đức Nhật Huỳnh]

Tìm x biết :\(a\)) \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)\(b\)) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...............+\left(x+100\right)=205550\)\(c\)) \(\left|x-5\right|=18+2.\left(-8\right)\)\(d\)) \(\left(3.x-2^4\right).7^5=2.7^6.\frac{1}{2009^0}\)

Answer 1

[Phương Anh (NTMH)]

a/ \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\\3+2^{x+1}=24-\left[16-\left(4-1\right)\right]\)

\(3+2^{x+1}=24-\left(16-3\right)\\ 3+2^{x-1}=24-13\\ 3+2^{x-1}=11\\ 2^{x+1}=11-3\\ 2^{x-1}=8\)

\(2^{x-1}=2^3\\ \Rightarrow x-1=3\\x=3+1\\ x=4\)

Answer 2

[Phương Anh (NTMH)]

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=205550\)

\(\left(x.100\right)+\left(1+2+3+....+100\right)=205550\)

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:

\(\left(x.100\right)+5050=205550\\ \\ x.100=205550-5050\\ \\x.100=20500\\ \\x=20500:100\\ \\\Rightarrow x=2005\)