Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : A = 3+32+33+...+32021
A = ( 3+32+33 )+ (34 + 35 + 36 )+ .... +( 32019 + 32020 + 32021)
A = 3. (1 + 3 + 32) + 34 . (1 + 3 + 32) + .... + 32019. (1 + 3 + 32)
A = 3 . 13 + 34 . 13 + ... + 32019 . 13
A = 13 . (3 + 34 + .... + 32019) chia hết cho 13.
Vậy tổng của A chia cho 13 có số dư là 0
S=1+32+34+36+.............................+398
9S=3+34+36+38+.........................+3100
=> 9S-S=3100-1
3100-1=(34)25-1
=(...1)25-1
=(.....1)-1
=(.....0) chia hết cho 10
Vậy S chia hết cho 10
a, \(S=1+3^2+3^4+3^6+...+3^{98}\)
\(\Rightarrow3^2S=3^2+3^4+3^6+3^8+...+3^{100}\)
\(\Rightarrow3^2S-S=\left(3^2+3^4+3^6+3^8+...+3^{100}\right)-\left(1+3^2+3^4+3^6+...+3^{98}\right)\)
\(\Rightarrow8S=3^{100}-1\)
\(\Rightarrow S=\frac{3^{100}-1}{8}\)
Vậy : \(S=\frac{3^{100}-1}{8}\)
b, \(S=1+3^2+3^4+3^6+...+3^{98}\)
\(S=\left(1+3^2\right)+\left(3^4+3^6\right)+...+\left(3^{96}+3^{98}\right)\)
\(S=\left(1+3^2\right)+3^4\left(1+3^2\right)+...+3^{96}\left(1+3^2\right)\)
\(S=1.10+3^4.10+...+3^{96}.10\)
\(S=\left(1+3^4+...+3^{96}\right).10\)
Vì : \(1+3^4+...+3^{96}\in N\Rightarrow S⋮10\)
Vậy : \(S⋮10\)
a) \(3-\left(\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3+\left(\dfrac{6}{7}\right)^0+\dfrac{1}{4}.\dfrac{1}{2}\)
\(=3+1+\dfrac{1}{8}=4+\dfrac{1}{8}\)
\(=\dfrac{33}{8}\)
b) \(64.2^3.\dfrac{1}{32^2}\)
\(=2^6.2^4.\dfrac{1}{32^2}=2^{10}.\dfrac{1}{32^2}\)
\(=\left(2^5\right)^2.\dfrac{1}{32^2}=32^2.\dfrac{1}{32^2}\)
= 1
c) \(\left(-2\right)^3+2^2+\left(-1\right)^{20}+\left(-2\right)^0\)
\(=-8+4+1+1\)
\(=-8+6=-2\)
d) \(2^3+3.\left(\dfrac{1}{2}\right)^0-\left(\dfrac{1}{2}\right)^2.4\left[\left(-2\right)^2:\dfrac{1}{2}\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+\left[4:\dfrac{1}{2}\right].8\)
\(=8+3-1+\left[4.2\right].8\)
\(=8+3-1+8.8\)
\(=10+64=74\)
Chúc bạn học tốt!!
a. \(6^2:4.3+2.5^2\)
= \(36:12+2.25\)
= \(3+50\)
=\(53\)
b. \(2.\left(5.4^2-18\right)\)
= \(2.\left(5.16-18\right)\)
= \(2.\left(80-18\right)\)
= \(2.62\)
= \(124\)
c. \(80:\left\{\left[\left(11-2\right).2\right]+2\right\}\)
\(=80:\left\{\left[9.2\right]+2\right\}\)
\(=80:\left\{18+2\right\}\)
\(=80:20\)
\(=4\)
Ta có: \(A=2+2^2+2^3+...+2^{100}\)
\(\Rightarrow A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(\Rightarrow A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{98}\left(1+2+2^2\right)\)
\(\Rightarrow A=2.7+2^4.7+...+2^{98}.7\)
\(\Rightarrow A=\left(2+2^4+...+2^{98}\right).7⋮7\)
\(\Rightarrow A⋮7\)
\(A=2+2^2+2^3+2^4+...+2^{100}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...\left(2^{99}+2^{100}\right)\)
\(=6+2^2.6+...+2^{98}.6⋮6\)
TL
=(2+22)+(23+24)+...(299+2100)
=6+26.6+...+298.6 chia hết cho 6
Hok tốt