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Nguyễn Hoàng Sinh
Giới thiệu về bản thân
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Ta có \(A=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)
\(A=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{17}\left(1+2+2^2+2^3\right)\)
\(A=\left(1+2+2^2+2^3\right)\left(2+2^5+...+2^{17}\right)\)
\(A=15.\left(2+2^5+...+2^{17}\right)\)
Mà 15⋮5
⇒A⋮5 (đpcm)
Xin lỗi bạn bài đầu mình sai nhé, từ dòng 3 phải là:
\(A=\sqrt{70-\sqrt{3A}}\)
Phần sau tự giải giúp mình nha:))
Ta đặt A=\(\sqrt{70-\sqrt{3\sqrt{70-3\sqrt{70-...}}}}\)(A≥0)
Vì A là phép tính vô hạn tuần hoàn
⇒ A=\(\sqrt{70-3A}\)
⇔\(A^2=70-3A\)
⇔\(A^2+3A+\dfrac{9}{4}=70+\dfrac{9}{4}\)
⇔\(\left(A+\dfrac{3}{2}\right)^2=\dfrac{289}{4}\)
⇔\(A+\dfrac{3}{2}=\dfrac{\pm17}{2}\)
⇔\(A=7\)(Thỏa mãn) hoặc \(A=-10\)(Loại)
Vậy A=7 hay giá trị của phép tính trên là 7
Ta có n+2⋮n-1
=>(n-1) +3⋮ n-1
Mà n-1⋮n-1
=> 3⋮n-1 hay n-1ϵ Ư(3)={±1;±3}
Ta lại có n là số tự nhiên=> n≥0 => n-1≥-1
=> n-1 ϵ {±1;3}
=> n ϵ {0;2;4}
Ta có: aaa x b=(100a +10a+ a)x b =100ab +10ab +ab (1)
bbb x a=(100b+10b+b)x a=100ab+10ab+ab (2)
Từ (1) và (2)=> aaa x b = bbb x a