a, \(\frac{a}{5}=\frac{-3}{b}\)

=> ab = (-3).5

=> ab = -15

=> a và b thuộc Ư(-15) = {1;-1;3;-3;5;-5;15;-15}

Ta có bảng:

Vậy các cặp (a;b) là (1;-15) ; (-1;15) ; (3;-5) ; (-3;5) ; (5;-3) ; (-5;3) ; (15;-1) ; (-15;1)

b, \(\frac{a-1}{7}=\frac{-3}{b+3}\)

=> (a - 1)(b + 3) = (-3).7

=> (a - 1)(b + 3) = -21

=> a - 1 và b + 3 thuộc Ư(-21) = {1;-1;3;-3;7;-7;21;-21}

Ta có bảng:

Vậy các cặp (a;b) là (2;-24) ; (0;18) ; (4;-10) ; (-2;4) ; (8;-6) ; (-6;0) ; (22;-4) ; (-20;-2)

a, |3 - x| = 7

\(\Rightarrow\left\{\begin{matrix}3-x=7\\3-x=-7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=-4\\x=10\end{matrix}\right.\)

b, |x| < 4

=> x = {-3;-2;-1;0;1;2;3}

Vì \(\widehat{NAP}\) và \(\widehat{MAP}\) kề bù nên:

\(\widehat{NAP}=\widehat{MAN}-\widehat{MAP}=180^o-35^o=145^o\)

Vì tia AQ nằm giữa 2 tia AP và AN nên:

\(\widehat{PAQ}=\widehat{NAP}-\widehat{NAQ}=145^o-50^o=95^o\)

Vậy \(\widehat{PAQ}=95^o\)

\(2n+7⋮31\)

\(\Rightarrow2n+7+31⋮31\)

\(\Rightarrow2n+38⋮31\)

\(\Rightarrow2\left(n+19\right)⋮31\)

Vì \(2⋮̸31\) nên để \(2\left(n+19\right)⋮31\) thì \(n+19⋮31\)

\(\Rightarrow n+19\in B\left(31\right)\)

\(\Rightarrow n+19=31k\left(k\in N\right)\)

\(\Rightarrow n=31k-19\)

Vậy n có dạng 31k - 19

S = 1.2 + 2.3 + ... + 19.20

3S = 1.2.(3 - 0) + 2.3.(4 - 1) +...+ 19.20.(21 - 18)

3S = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 +...+ 19.20.21 - 18.19.20

3S = (1.2.3 + 2.3.4 +...+ 19.20.21) - (0.1.2 + 1.2.3 +...+ 18.19.20)

3S = 19.20.21 - 0.1.2

3S = 19.20.21

S = \(\frac{19.20.21}{3}\)

S = 2660

Vì \(\widehat{BOI}\) và \(\widehat{KOB}\) kề bù nên:

\(\widehat{BOI}+\widehat{KOB}=180^o\)

\(45^o+\widehat{KOB}=180^o\)

\(\widehat{KOB}=180^o-45^o\)

\(\widehat{KOB}=135^o\)

\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)

\(\Rightarrow\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2010}+1\right)-\left(\frac{x-3}{2009}+1\right)=\frac{x-4}{2008}+1\)

\(\Rightarrow\frac{x-1+2011}{2011}+\frac{x-2+2010}{2010}-\frac{x-3+2009}{2009}=\frac{x-4+2008}{2008}\)

\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}=\frac{x-2012}{2008}\)

\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)

\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

=> x - 2012 = 0

=> x = 2012

Vậy x = 2012

Ta có: \(\left\{\begin{matrix}\left|x\right|^3\ge0\\\left|x\right|^2\ge0\\\left|x\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x\right|^3-\left|x\right|^2-\left|x\right|\ge0\)

\(\Rightarrow7-\left|x\right|^3-\left|x\right|^2-\left|x\right|\ge7\)

\(\Rightarrow B\ge7\)

Dấu "=" xảy ra <=> x = 0

Vậy B_Max = 7 khi x = 0