Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(a^2+b^2+1\ge ab+a+b\)
\(<=>2a^2+2b^2+2\geq 2ab+2a+2b\\<=>(a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1)\geq 0\\<=>(a-b)^2+(a-1)^2+(b-1)^2\geq 0\)
$\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)\(\Rightarrow a^{2}+b^{2}+1\geq ab+a+b\)
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
\(n.n+3n+6\)
\(=n^2+3n+6\)
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 \(n\in\left\{-9;3;-4;-2\right\}\)
1: =>3n-12+17 chia hết cho n-4
=>\(n-4\in\left\{1;-1;17;-17\right\}\)
hay \(n\in\left\{5;3;21;-13\right\}\)
2: =>6n-2+9 chia hết cho 3n-1
=>\(3n-1\in\left\{1;-1;3;-3;9;-9\right\}\)
hay \(n\in\left\{\dfrac{2}{3};0;\dfrac{4}{3};-\dfrac{2}{3};\dfrac{10}{3};-\dfrac{8}{3}\right\}\)
4: =>2n+4-11 chia hết cho n+2
=>\(n+2\in\left\{1;-1;11;-11\right\}\)
hay \(n\in\left\{-1;-3;9;-13\right\}\)
5: =>3n-4 chia hết cho n-3
=>3n-9+5 chia hết cho n-3
=>\(n-3\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{4;2;8;-2\right\}\)
6: =>2n+2-7 chia hết cho n+1
=>\(n+1\in\left\{1;-1;7;-7\right\}\)
hay \(n\in\left\{0;-2;6;-8\right\}\)
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 \(n\in\left\{-2;0;-3;1;-4;2;-5;3;-7;5;-13;11\right\}\)