(2n+1)\(\inƯ\left(18\right)=\left\{\pm1;\pm2;\pm3;\pm6;\pm9;\pm18\right\}\)

ta có bảng sau

vậy....

bài 1

b.(x+y)²+2(x+y)(x-y)+(x-y)²

= [(x+y)+(x-y)]²

= (x+y-x+y)²

= (2y)²

= 4y²

bài 2

a. 2x²y+4xy+2y

=2y(x²+2x+1)

=2y(x+1)²

b.9x2+6xy-4z2+y2

= (9x²+6xy+y²)-4z²

= (3x+y)²-(2z)²

= (3x+y-2z)(3x+y+2z)

3x² - 2x - 1 = 0

=>3x2+x-3x-1=0

=>x(3x+1)-1(3x+1)=0

=>(x-1)(3x+1)=0

=> \(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=1\\3x=-1\Rightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)

vậy x=1 hoặc x=-\(\dfrac{1}{3}\)

a) 9x (3x-y) +3y (y-3x)

= 9x(3x-y)-3y(3x-y)

=(3x-y)(9x-3y)

= (3x-y)3(3x-y)

= 3(3x-y)²

b)x³ - 3x² - 9x + 27

=x³+3x²-6x²-18x+9x+27

= (x³+3x²)-(6x²+18x)+(9x+27)

= x²(x+3)-6x(x+3)+9(x+3)

= (x+3)(x²-6x+9)

=(x+3)(x-3)²

x²+x-6=0

=>x²+3x-2x-6=0

=>x(x+3)-2(x+3)=0

=> (x-2)(x+3)=0

=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

vậy ....

y3(x-1)-7y3+7xy3

=y³(x-1)+(7xy³-7y³)

= y³(x-1)+7y³(x-1)

=(x-1)(y³+7y³)

= (x-1)8y³

(3n+10) \(⋮\)(n+2)

vì\(\left(n+2\right)⋮\left(n+2\right)\)

=> \(3\left(n+2\right)⋮\left(n+2\right)\)

=> \(\left(3n+6\right)⋮\left(n+2\right)\)

=> \(\left(3n+10\right)-\left(3n+6\right)⋮\left(n+2\right)\)

=>\(\left(3n+10-3n-6\right)⋮\left(n+2\right)\)

=> \(4⋮\left(n+2\right)\)

=> \(n+2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

ta có bảng sau

vậy \(n\in\left\{-6;-4;-3;-1;0;2\right\}\)

(x+1)(x-2)-x²=0

=> x²-2x+x-2-x²=0

=> -x-2=0

=> -x=2

=> x=-2

(x-1)³-(x³-3x²-1)

=x³-3x²+3x-1-(x³-3x²-1)

= x³-3x²+3x-1-x³+3x²+1

= 3x

\(\left(2x+1\right)⋮\left(x+2\right)\)

vì \(\left(x+2\right)⋮\left(x+2\right)\)

=> \(2\left(x+2\right)⋮\left(x+2\right)\)

=> \(\left(2x+4\right)⋮\left(x+2\right)\)

=> \(\left(2x+1\right)-\left(2x+4\right)⋮\left(x+2\right)\)

=> \(\left(2x+1-2x-4\right)⋮\left(x+2\right)\)

=> \(-3⋮\left(x+2\right)\)

=> \(x+2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

ta có bảng sau

vậy ...

( a + 13 ) $⋮$ ( 2a + 1 )

vì \(\left(a+13\right)⋮\left(2a+1\right)\)

=> \(2\left(a+13\right)⋮\left(2a+1\right)\)

=> \(\left(2a+26\right)⋮\left(2a+1\right)\)

=>\(\left(2a+26\right)-\left(2a+1\right)⋮\left(2a+1\right)\)

=> \(\left(2a+26-2a-1\right)⋮\left(2a+1\right)\)

=>\(25⋮\left(2a+1\right)\)

=> \(2a+1\inƯ\left(25\right)=\left\{\pm1;\pm5;\pm25\right\}\)

vậy ....