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a: =>y+3/10y=-1,3
=>13/10y=-1,3
hay y=-1
b: =>3/4y=1/2
hay y=2/3
c: \(\Leftrightarrow y\cdot\dfrac{19}{3}+16,75=-13,25\)
\(\Leftrightarrow y\cdot\dfrac{19}{3}=-30\)
hay y=-190
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\(a^2+b^2+1\ge ab+a+b\)
\(<=>2a^2+2b^2+2\geq 2ab+2a+2b\\<=>(a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1)\geq 0\\<=>(a-b)^2+(a-1)^2+(b-1)^2\geq 0\)
$\Rightarrow $ \(a^{2}+b^{2}\geq 2ab\) (1)
$\Rightarrow $ \(a^{2}+1\geq 2a\) (2)
$\Rightarrow $ \(b^{2}+1\geq 2b\) (3)
(1), (2) và (3)\(\Rightarrow a^{2}+b^{2}+1\geq ab+a+b\)
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2^1 + 2^2 + 2^3 +...+ 2^2010
= (2^1 + 2^2) + (2^3 + 2^4) + ... + (2^2009 + 2^2010)
= 2.(1 + 2) + 2^3.(1 + 2) + ... + 2^2009.(1 + 2) = 2.3 + 2^3.3 + ... + 2^2009.3 = 3.(2 + 2^3 + ... + 2^2009) => 2^1 + 2^2 + 2^3 +...+ 2^2010 chia hết cho 3 2^1 + 2^2 + 2^3 +...+ 2^2010 = (2^1 + 2^2 + 2^3) + ... + (2^2008 + 2^2009 + 2^2010) = 2.( 1 + 2 + 2^2) + ... + 2^2008.(1 + 2 + 2^2) = 2.7 + ... + 2^2008. 7 => 2^1 + 2^2 + 2^3 +...+ 2^2010 chia hết cho 7![](https://rs.olm.vn/images/avt/0.png?1311)
a, Nếu \(n=3k\left(k\in Z\right)\Rightarrow A=n^3-n=27k^3-3k⋮3\)
Nếu \(n=3k+1\left(k\in Z\right)\)
\(\Rightarrow A=n^3-n\)
\(=n\left(n-1\right)\left(n+1\right)\)
\(=\left(3k+1\right).3k.\left(3k+2\right)⋮3\)
Nếu \(n=3k+2\left(k\in Z\right)\)
\(\Rightarrow A=n^3-n\)
\(=n\left(n-1\right)\left(n+1\right)\)
\(=\left(3k+2\right)\left(n+1\right)\left(3k+3\right)⋮3\)
Vậy \(n^3-n⋮3\forall n\in Z\)
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Ta có:
`13^n-1(n in NN^**)`
`=(13-1)(13^{n-1}+........+1)`
`=12..... vdots 12`
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\(n.n+3n+6\)
\(=n^2+3n+6\)
Đặt cột dộc ta có :
n2 + 3n + 6 | n + 3
n2 + 3n | n
_________|
0 + 0 + 6
Để phép chia trên là phép chia hết thì :
\(6⋮n+3\Rightarrow n\inƯ\left(6\right)=\left\{1;-1;6;-6\right\}\)
+ ) n + 3 = 1
n = -2
+ ) n + 3 = -1
n = -4
+ ) n + 3 = 6
n = 3
+) n + 3 = -6
n = -9
Vậy \(n\in\left\{-9;3;-4;-2\right\}\)
n+13 \(⋮\) n+1
=>( n+1)+12 \(⋮\) n+1
=> 12 \(⋮\) n+1
=> n+1 \(\in\) Ư(12)
=> Ư(12) = { 1; 2; 3; 4; 6; 12}
=> n = { 0; 1; 2; 3; 5; 11}
Ta có: \(n+13⋮n+1\)
\(\Rightarrow\left(n+1\right)+12⋮n+1\)
\(\Rightarrow12⋮n+1\)
\(\Rightarrow n+1\in\left\{-1;1;-2;2;-3;3;-4;4;-6;6;-12;12\right\}\)
+) \(n+1=-1\Rightarrow n=-2\)
+) \(n+1=1\Rightarrow n=0\)
+) \(n+1=-2\Rightarrow n=-3\)
+) \(n+1=2\Rightarrow n=1\)
+) \(n+1=-3\Rightarrow n=-4\)
+) \(n+1=3\Rightarrow n=2\)
+) \(n+1=-4\Rightarrow n=-5\)
+) \(n+1=4\Rightarrow n=3\)
+) \(n+1=-6\Rightarrow n=-7\)
+) \(n+1=6\Rightarrow n=5\)
+) \(n+1=-12\Rightarrow n=-13\)
+) \(n+1=12\Rightarrow n=11\)
Vậy \(n\in\left\{-2;0;-3;1;-4;2;-5;3;-7;5;-13;11\right\}\)