\(4n+1⋮\left(n+1\right)\\ =>4\left(n+1\right)-3⋮\left(n+1\right)\\ =>3⋮\left(n+1\right)\\ =>n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Với `n` là số tự nhiên `=>n+1\ge 1;n+1\in NN*`

Do đó, nên : `n+1\in {1;3}=>n\in {0;2}` (TM)

Vậy `n\in {0;2}` thì `4n+1` chia hết cho `n+1`

`(a):(2x-5)^{3}-6(x-2)(x^{2}+2x+4)-x(4x^{2}-3)`

`=(2x)^{3}-3.(2x)^{2}.5+3.2x.5^{2}-5^{3}-6(x^{3}-2^{3})-4x^{3}+3x`

`=8x^{3}-60x^{2}+150x-125-6x^{3}+48-4x^{3}+3x`

`=-2x^{3}-60x^{2}+153x-77`

`(b):(x^{2}+3y)^{2}-4x^{2}(y+x^{2})(x^{2}-y)`

`=(x^{2})^{2}+2.x^{2}.3y+(3y)^{2}-4x^{2}(x^{4}-y^{2})`

`=x^{4}+6x^{2}y+9y^{2}-4x^{6}+4x^{2}y^{2}`

`\sqrt{x^{2}+3x}-2=x`

`<=>\sqrt{x^{2}+3x}=x+2\    (ĐK:x+2\ge0 <=> x\ge -2)`

`<=>x^{2}+3x=(x+2)^{2}`

`<=>x^{2}+4x+4=x^{2}+3x`

`<=>x+4=0`

`<=>x=-4\   (KTMDK)`

Vậy pt vô nghiệm

`\sqrt{x^{2}+3x}-2=x`

`<=>\sqrt{x^{2}+3x}=x+2\    (ĐK:x+2\ge0<=>x\ge -2)`

`<=>x^{2}+3x=x+2`

`<=>x^{2}+2x-2=0`

`<=>(x+1)^{2}-3=0`

`=>x+1=\sqrt{3}` hoặc `x+1=-\sqrt{3}`

`<=>x=\sqrt{3}-1\ (TMDK)` hoặc `x=-1-\sqrt{3}\   (KTMDK)`

Vậy `S={\sqrt{3}-1}`

\(3n+17⋮\left(n+2\right)=>3\left(n+2\right)+11⋮\left(n+2\right)\\ =>11⋮\left(n+2\right)\\ =>n+2\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

Với `n` là số tự nhiên `=>n+2\ge 2;n+2\in NN`

`=>n+2=11`

`=>n=9` (TM)

Vậy `n=9` thì `3n+17` chia hết cho `n+2`

`A=xy+3xy+5xy+...+2021xy`

`=(1+3+5+...+2021).xy`

Xét tổng : `1+3+5+...+2021`

Dãy trên có : `(2021-1):2+1=1011` (số hạng)

Tổng dãy trên :   `(2021+1).1011:2=1022121`

Do vậy, nên `A=1022121.xy`

`(9-2x)^{3}=125=5^{3}`

`=>9-2x=5`

`=>2x=9-5=4`

`=>x=2`

`A=x^{3}+x^{2}z+y^{2}z-xyz+y^{3}`

`=(x^{3}+y^{3})+(x^{2}z+y^{2}z-xyz)`

`=(x+y)(x^{2}-xy+y^{2})+z(x^{2}+y^{2}-xy)`

`=(x^{2}-xy+y^{2})(x+y+z)`

`=(x^{2}-xy+y^{2}).0`

`=0` (DPCM)

\(\sqrt{3x-2}=x-1< =>\left\{{}\begin{matrix}x-1\ge0\\3x-2=\left(x-1\right)^2\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\ge1\\x^2-2x+1=3x-2\end{matrix}\right.< =>\left\{{}\begin{matrix}x\ge1\\x^2-5x+3=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x=\dfrac{5+\sqrt{13}}{2}\\x=\dfrac{5-\sqrt{13}}{2}\end{matrix}\right.\end{matrix}\right.< =>x=\dfrac{5+\sqrt{13}}{2}\)

`(g):(6)/(13)+(-14)/(39)`

`=(18)/(39)-(14)/(39)`

`=(4)/(39)`

`(h):(-3)/(21)+(6)/(42)`

`=(-1)/(7)+(1)/(7)`

`=0`

`(k):(7)/(21)+(9)/(-36)`

`=(1)/(3)-(1)/(4)`

`=(4-3)/(12)`

`=1/12`