Question

Bài 1: tìm số nguyên n để giá tri của 2n\(^2\) + 3n +3 chia hết cho giá tri của 2n-1

bài 2 phân tích đa thức thành nhân tử

a, 3x\(^2\)-3y\(^2\)+4x-4y

b, 12x\(^2\)-3xy+8x-2y

c, x\(^3\)+x\(^2\)y-x\(^2\)z-xyz

d, xy+y-2x-2

e, x\(^3\)-3x\(^2\)+3x-9

giúp mk nha mn

Answer 1

1 ) Ta có : \(2n^2+3n+3\)

\(=2n^2-n+4n-2+5\)

\(=n\left(2n-1\right)+2\left(2n-1\right)+5\)

\(=\left(n+2\right)\left(2n-1\right)+5\)

Để \(2n^2+3n+3⋮2n-1\)

\(\Leftrightarrow\left(n+2\right)\left(2n-1\right)+5⋮2n-1\)

\(\Leftrightarrow5⋮2n-1\)

Do \(n\in Z\Rightarrow2n-1\in Z\)

\(\Rightarrow2n-1\in\left\{1;-1;5;-5\right\}\)

\(\Rightarrow2n\in\left\{2;0;6;-4\right\}\)

\(\Rightarrow n\in\left\{1;0;3;-2\right\}\)

Vậy ...

Bài 2 :

a ) \(3x^2-3y^2+4x-4y=3\left(x^2-y^2\right)+4\left(x-y\right)=3\left(x-y\right)\left(x+y\right)+4\left(x-y\right)\)

\(=\left(x-y\right)\left[3\left(x+y\right)+4\right]=\left(x-y\right)\left(3x+3y+4\right)\)

b ) \(12x^2-3xy+8x-2y\)

\(=3x\left(4x-y\right)+2\left(4x-y\right)\)

\(=\left(3x+2\right)\left(4x-y\right)\)

c ) \(x^3+x^2y-x^2z-xyz\)

\(=x^2\left(x+y\right)-xz\left(x+y\right)\)

\(=\left(x^2-xz\right)\left(x+y\right)\)

\(=x\left(x-z\right)\left(x+y\right)\)

d ) \(xy+y-2x-2\)

\(=y\left(x+1\right)-2\left(x+1\right)\)

\(=\left(y-2\right)\left(x+1\right)\)

e ) \(x^3-3x^2+3x-9\)

\(=x^2\left(x-3\right)+3\left(x-3\right)\)

\(=\left(x^2+3\right)\left(x-3\right)\)

Answer 2

1.\(\left(2n^2+3n+3\right):\left(2n-1\right)=n+2\) dư 5 (đoạn này bạn tự chia nha)

Muốn \(2n^2+3n+3\)\(⋮2n-1\) thì \(5⋮2n-1\)

\(\Rightarrow2n-1\inƯ\left(5\right)\)

\(\Rightarrow2n-1=\left\{-5;-1;1;5\right\}\)

\(\Rightarrow2n=\left\{-4;0;2;6\right\}\)

\(\Rightarrow n=\left\{-2;0;1;3\right\}\)

bài 2 nhờ Nguyễn Thanh Hằng hay Mysterious Person giải nha

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