x²+x-30

=x²-5x+6x-30

=x(x-5)+6(x-5)

=(x+6)(x-5)

a)

x²-4x+4-x²+5x=13

x+4=13

x=9

a)x(x^2-x+2)(x^2+x+2)

còn lại thì vào đây: https://coccoc.com/search/math

Đặt x²+10x+5=a.Ta có biểu thức là : a(a+8)+16=a²+8a+16=(a+4)²=(x²+10x+9)²

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

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

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

a) điều kiện xác định: x≠3 và x≠2

b) \(\dfrac{x^2-4}{\left(x-3\right)\left(x-2\right)}\)=\(\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x-2\right)}\)=\(\dfrac{x+2}{x-3}\)

Tại x=13 ta có \(\dfrac{13+2}{13-3}\)=\(\dfrac{3}{2}\)

 x⁴+2.x³-13.x²-14x+24

=x³.(x+2)-13x²+12x-26x+24

=x³.(x+2)-x.(13x-12)-2.(13x-12)

=x³.(x+2)-(13x-12)(x+2)

=(x+2)(x³-13x+12)

=(x+2)(x³-x-12x+12)

=(x+2)[x.(x²-1)-12.(x-1)]

=(x+2)[x.(x-1)(x+1)-12.(x-1)]

=(x+2)(x-1)[x.(x+1)-12]

=(x+2)(x-1)(x²+x-12)

=(x+2)(x-1)(x²-3x+4x-12)

=(x+2)(x-1)[x.(x-3)+4.(x-3)]

=(x+2)(x-1)(x-3)(x+4)

\(x^4+8x^3+28x^2+48x-13\)

\(=x^4+4x^3+13x^2+4x^3+16x^2+52x-x^2-4x-13\)

\(=x^2\left(x^2+4x+13\right)+4x\left(x^2+4x+13\right)-\left(x^2+4x+13\right)\)

\(=\left(x^2+4x-1\right)\left(x^2+4x+13\right)\)

C1

a) -7x(3x-2)=-21x^2+14x

b) 87^2+26.87+13^2=87^2+2.13.87+13^2=(87+13)^2=100^2

C2

a) (x-5)(x+5)

b)3x(x+5)-2(x+5)=(3x-2)(x+5)=0

\(\Rightarrow\left[\begin{array}{nghiempt}3x-2=0\\x+5=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{3}\\x=-5\end{array}\right.\)

Vậy S={-5;2/3}

C3:

a)3x^3-2x^2+2=(x+1)(3x^2-5x-5)-3

b) Để A chia hết cho B=> x+1\(\inƯ\left(-3\right)\)

\(\Rightarrow\begin{cases}x+1=3\\x+1=-3\\x+1=1\\x+1=-1\end{cases}\)\(\Rightarrow\begin{cases}x=2\\x=-4\\x=0\\x=-2\end{cases}\)