\(A=\frac{4n+1}{2n+3}=\frac{2\left(2n+3\right)-5}{2n+3}=2-\frac{5}{2n+3}\)

Vậy để A nguyên thì 2n+3\(\in\)Ư(5)

Mà Ư(5)={1;-1;5;-5}

=>2n+3={1;-1;5;-5}

Ta có bảng sau

Vậy n={-1;-2;-4;1}

là con gái hay con trai vậy ? nhìn cái tên thì chẳng ai phân biệt được trai hay gái đâu.

a) \(219-7\left(x+1\right)=100\)

\(\Leftrightarrow219-7x-7=100\)

\(\Leftrightarrow-7x=100-219+7=-112\)

\(\Leftrightarrow x=16\)

b) \(\left(3x-6\right)\cdot3=3^4\)

\(\Leftrightarrow\left(x-2\right)\cdot3^2=3^4\)

\(\Leftrightarrow x-2=3^2\)

\(\Leftrightarrow x=9+2=11\)

\(3+\left(x\cdot4\right)-15=1\)

\(\Leftrightarrow4x=1+15-3=13\)

\(\Leftrightarrow x=\frac{13}{4}\)

\(3+\left(x\cdot4\right)-15=5\)

\(\Leftrightarrow4x=5+15-3=17\)

\(\Leftrightarrow x=\frac{17}{4}\)

4) \(3x^2\left(x+1\right)-2\left(x+1\right)=\left(x+1\right)\left(3x^2-2\right)\)

5) \(\left(a+b+c\right)^2-\left(ab+bc+ca\right)\left(a+b+c\right)+\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a+b+c-ab-bc-ca+1\right)\)

6) \(4x^2\left(x-2y\right)-20x\left(2y-x\right)=4x^2\left(x-2y\right)+20x\left(x-2y\right)=4x\left(x-2y\right)\left(x+5\right)\)

7) \(3x^2y^2\left(a-b+c\right)+2xy\left(b-a-c\right)=3x^2y^2\left(a-b+c\right)-2xy\left(a-b+c\right)\\ =xy\left(a-b+c\right)\left(3xy+2\right)\)

1) \(4x^3-14x^2=2x^2\left(2x-7\right)\)

2) \(5y^{10}+15y^6=5y^6\left(y^4+3\right)\)

3) \(9x^2y^2+15x^2y-21xy^2=3xy\left(3xy+5x-7y\right)\)