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12 + 22 + 33
= 1 + 4 + 27
= 5 + 27
= 32
( 24 + 32 ) : 2
= ( 16 + 9 ) : 2
= 25 : 2
= 12,5
~ Hok tốt ~
\(a,x-5⋮x+2\)
\(\Rightarrow x+2-7⋮x+2\)
\(\Rightarrow x+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
x + 2 = 1=> x = -1
x + 2 = -1 => x = -3
.... tương tự nhé ~
\(2x+3⋮x-5\)
\(\Rightarrow2x-10+7⋮x-5\)
\(\Rightarrow2\left(x-5\right)+7⋮x-5\)
\(\Rightarrow x-5\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
x - 5 = 1 => x = 6
....
5^6+5^7+5^8
=5^6.(1+5+5^2)
=5^6.31 chia hết cho 31
7^6+7^5-7^4
=7^4.(7^2+7-1)
=7^4.55 chia hết cho 11
BÀI 2:
a) \(5^6+5^7+5^8=5^6\left(1+5+5^2\right)=5^6.31\) \(⋮\)\(31\)
b) \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.55\)\(⋮\)\(11\)
c) \(2^3+2^4+2^5=2^3.\left(1+2+2^2\right)=2^3.7\)\(⋮\)\(7\)
d) mk chỉnh đề
\(1+2+2^2+2^3+...+2^{59}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{58}+2^{59}\right)\)
\(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{58}\left(1+2\right)\)
\(=\left(1+2\right)\left(1+2^2+...+2^{58}\right)\)
\(=3\left(1+2^2+...+2^{58}\right)\)\(⋮\)\(3\)
Ta có:
\(\hept{\begin{cases}72⋮x\\60⋮x\end{cases}\Rightarrow\hept{\begin{cases}x\inƯ\left(72\right)\\x\inƯ\left(60\right)\end{cases}\Rightarrow}x\inƯC\left(72,60\right)}\)
Ta có:ƯCLN(70;60)=12
=>ƯC(72;60)=Ư(12)={\(\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\)}
Vì x>6 nên x=12
72 / 60 = 1.2
1.2*10 = 12
=> x=12 (vi: 72 / 12 = 6; 60 / 12 = 5)
\(\text{a) Vì 35 ⋮ x}\)
\(\Rightarrow x\inƯ\left(35\right)\)
\(\RightarrowƯ\left(35\right)=\left\{1;5;7;35\right\}\)
\(\Rightarrow x\in\left\{1;5;7;35\right\}\)
\(\text{b) x - 1\inƯ(6)}\)
\(\RightarrowƯ\left(6\right)=\left\{1;2;3;6\right\}\)
Ta có bảng :
=> x thuộc { 2 ; 3 ; 4 ; 7 }
\(\text{c) 10 ⋮ ( 2x + 1 )}\)
\(\Rightarrow2x+1\inƯ\left(10\right)\)
\(\RightarrowƯ\left(10\right)=\left\{1;2;5;10\right\}\)
Ta có bảng :
=> x = 2
d) \(x⋮25,x< 100\)
\(\Rightarrow x\in B\left(25\right)\)
\(\Rightarrow B\left(25\right)=\left\{0;25;50;75;100;....\right\}\)
Mà x < 100
\(\Rightarrow x\in\left\{25;50;75\right\}\)
\(e)x+13⋮x+1\)
\(\Rightarrow x+1+12⋮x+1\)
\(\text{Vì x + 1 ⋮ x + 1 nên 12 ⋮ x + 1}\)
\(\Rightarrow x+1\inƯ\left(12\right)\)
Đến đây bn tự làm nốt nhé ...