X²+X-3X-3=15

X²-2X-3=15

X²-2X=18

X.X-2X=18

X(X-2)=18

TH1

X=18

TH2

X-2=18

X=20

Vậy Xϵ{18;20}

   (\(x\) - 3).(\(x\) + 1) = 15

   \(x^2\) - 3\(x\) + \(x\) - 3 = 15

  \(x^2\) - 2\(x\) - 3 + 4 =  15 + 4

   \(x^2\) - 2\(x\)  + 1    = 19

  (\(x\) - 1)²            = 19

 \(\left[{}\begin{matrix}x-1=-\sqrt{19}\\x-1=\sqrt{19}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1-\sqrt{19}\\x=1+\sqrt{19}\end{matrix}\right.\)

(x-3)(x+1)=15

x-3=15 hoặc x+1=15

x=15+3 hoặc x=15-1

x=18 hoặc x=14

có gì sai sót mong mọi người chiếu cố

(x-3)(x+1)=15

=>\(x^2+x-3x-3=15\)

=>\(x^2-2x-18=0\)

=>\(x^2-2x+1-19=0\)

=>\(\left(x-1\right)^2=19\)

=>\(\left[{}\begin{matrix}x-1=\sqrt{19}\\x-1=-\sqrt{19}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{19}+1\\x=-\sqrt{19}+1\end{matrix}\right.\)

56 + 29

-10 < x ≤ 13

⇒ x ∈ {-9; -8; -7; ...; 7; 8; 9; 10; 11; 12; 13}

Tổng của chúng là:

-9 + (-8) + (-7) + ... + 7 + 8 + 9 + 10 + 11 + 12 + 13

= 10 + 11 + 12 + 13

= 46

Chọn D

Đáp án D.46

12n + 1 = 12n + 18 - 17

= 6(2n + 3) - 17

Để (12n + 1) ⋮ (2n + 3) thì 17 ⋮ (2n + 3)

⇒ 2n + 3 ∈ Ư(17) = {-17; -1; 1; 17}

⇒ 2n ∈ {-20; -4; -2; 14}

⇒ n ∈ {-10; -2; -1; 7}

Mà n là số tự nhiên

n = 7

b: \(B=16^5+2^{15}\)

\(=\left(2^4\right)^5+2^{15}\)

\(=2^{20}+2^{15}\)

\(=2^{15}\left(2^5+1\right)=2^{15}\cdot33⋮33\)

c: \(45⋮9;99⋮9;180⋮9\)

Do đó: \(45+99+180⋮9\)

=>\(C⋮9\)

d: \(D=2+2^2+2^3+...+2^{60}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{58}\right)⋮7\)

\(D=2+2^2+2^3+...+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{57}\right)\)

=>D chia hết cho cả 3 và 5

a: A=1+2-3-4+5+6-7-8+...+97+98-99-100

=(1+2-3-4)+(5+6-7-8)+...+(97+98-99-100)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)(25 số -4)

\(=-4\cdot25=-100\)

b: \(B=1+3+3^2+...+3^{11}\)

\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{10}+3^{11}\right)\)

\(=\left(1+3\right)+3^2\left(1+3\right)+...+3^{10}\left(1+3\right)\)

\(=4\left(1+3^2+...+3^{10}\right)⋮4\)

\(\dfrac{9}{14}+\dfrac{8}{21}=\dfrac{27}{42}+\dfrac{16}{42}=\dfrac{43}{42}\)

\(\dfrac{13}{15}-\dfrac{7}{12}=\dfrac{52}{60}-\dfrac{35}{60}=\dfrac{17}{60}\)

\(\dfrac{9}{14}+\dfrac{8}{21}=\dfrac{27}{42}+\dfrac{16}{42}=\dfrac{43}{42}\)

\(\dfrac{13}{15}-\dfrac{7}{12}=\dfrac{52}{60}-\dfrac{35}{60}=\dfrac{17}{60}\)

#hn

Ta có: \(2^2+2^3+2^4+2^5\)

\(=\left(2^2+2^3\right)+\left(2^4+2^5\right)\)

\(=12+2^2.\left(2^2+2^3\right)\)

\(=12+2^2.12\)

\(=12.\left(1+2^2\right)\)

Vì \(12⋮3\) nên \(12.\left(1+2^2\right)⋮3\)

Vậy \(2^2+2^3+2^4+2^5⋮3\)

CM: Tổng \(2^2+2^3+2^4+2^5⋮3\)

= \(2^2.\left(1+2\right)\) \(+2^4.\left(1+2\right)\)

= \(2^2.3+2^4.3\)

= \(3.\left(2^2+2^4\right)\)\(⋮3\)

Ta có: 

\(x-6⋮3\)

\(\Rightarrow x⋮3\) và \(x\in B\left(3\right)\)

Xét:

\(12⋮3\) ; \(19⋮̸3\) ; \(45⋮3\) ; \(70⋮̸3\)

⇒ \(x\in\left\{12;45\right\}\)