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a) x+2x+...+50x =2550
x. [ 1+2+3+....+50]=2550
ta co :
so so hang cua day 1;2;3;4;...;50:
[50-1]:1+1=50
tong cua day tren la :
[50+1].50:2=1275
=> x.1275=2550
x=2550:1275
vay x=2
\(5^{x-1}=125\)
\(5^{x-1}=5^3\)
\(\Rightarrow x-1=3\)nên x = 4
707 - 3( x + 34 ) = 407
=> 3( x + 81 ) = 300
=> x + 81 = 100
=> x = 19
127 + 3( x + 17 ) = ( 47 : 45 )2
=> 3( x + 17 ) = ( 42 )2 - 127
=> 3( x + 17 ) = 256 - 127
=> 3( x + 17 ) = 129
=> x + 17 43
=> x = 26
x - 6 - ( 48 - 24. 2 : 6 - 3 ) = 100
x - 6 - ( 48 - 2 - 3 ) = 100
x - 6 - 43 = 100
x - 49 = 100
x = 149
[ ( 6x - 39 ) : 3 ] . 28 = 5628
( 6x - 39 ) : 3 = 201
6x - 39 = 603
6x = 642
x = 107
Chúc em hok tốt
Study well
\(707-3.(x+3^4)=407\)
\(\Rightarrow\)\(707-3.(x+81)=407\)
\(\Rightarrow\)\(3.(x+81)=707-407\)
\(\Rightarrow\)\(3.\left(x+81\right)=300\)
\(\Rightarrow\)\(\left(x+81\right)=300:3\)
\(\Rightarrow\)\(\left(x+81\right)=100\)
\(\Rightarrow\) \(x=100-81\)
Xin lỗi vì tôi có việc bận nên tí nữa sẽ làm tiếp
\(\Rightarrow\) \(x=19\)
\(A=2+2^2+2^3+2^4+2^5+...+2^{60}\)
\(=(2+2^2)+(2^3+2^4)+(2^5+2^6)+...+(2^{59}+2^{60})\)
\(=2(1+2)+2^3(1+2)+2^5(1+2)+...+2^{59}(1+2)\)
\(=2.3+2^3.3+2^5.3+...+2^{59}.3\)
\(=(2+2^3+2^5+...+2^{59}).3\)chia hết cho 3
Vậy \(A=2+2^2+2^3+2^4+2^5+...+2^{60}\)xhia hết cho 3
A = 2 + 22 + 23 + 24 + 25 + ..........+260
A = ( 2 + 22 + 23 ) + ..........+ ( 2 58+ 259 + 260 )
A = 2 ( 1+ 2 ) + 22 + ..............+ 258 ( 1 + 2 ) + 22
A = 2 x 3 + 4 + ................+ 258 x 3 + 4
Vì 3 chia hết cho 3 nên
A = 2 x 3 + 4 + ................+ 258 x 3 + 4 sẽ chia hết cho 3
1)\(2^3\cdot37-2^3\cdot63-10=2^3\left(37-63\right)-10=8\cdot-26-10\)=-218
2)\(2^3+2^2+2^4=2^2\left(1+2+4\right)=4\cdot7=28\)
3)\(5^3-5=5\left(5^2-1\right)=5\cdot24=120\)
4)\(3+3^2+3^4=3\left(1+3+3^3\right)=3\cdot13=39\)
5)\(x^{n+1}-x^n=x^n\left(x-1\right)\)
C = 4+42+43+....+4n
4.C = 42+43+44+....+4n+1
4.C - C = (42+43+44+....+4n+1) - ( 4+42+43+....+4n )
3.C = 4n+1 - 4
C = 4n+1 - 4/3
\(C=4+4^2+4^3+...+4^n\)
\(4C=4^2+4^3+4^4+...+4^{n+1}\)
\(4C-C=\left(4^2+4^3+...+4^{n+1}\right)-\left(4+4^2+4^3+...+4^n\right)\)
\(3C=4^{n+1}-4\)
\(C=\frac{4^{n+1}-4}{3}\)