\(3x^4-48=3\left(x^4-16\right)=3\left[\left(x^2\right)^2-4^2\right]\\ =3\left(x^2-4\right)\left(x^2+4\right)\\ =3\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)

x²-4x+x+4

=x²+x-4x+4

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

=(x+1)(x-4)

\(x^2-3x+4\)

\(=x^2+x-4x+4\)

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

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

\(=\left(x-4\right)\left(x+1\right)\).

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

nhớ cho mik nha

minh h cho 1  cai nhung o cho x^3-1 phai bang (x-1)(x^2+x+1)

x^5+2x^4+2x^3+2x^2+2x+1

=(x^5+x^4)+(x^4+x^3)+(x^3+x^2)+(x^2+x)+(x+1)

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

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

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

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

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

Câu a :

\(\dfrac{3x^2-12x+12}{x^4-8x}\)

\(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}\)

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

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

Câu b :

\(\dfrac{7x^2+14x+7}{3x^2+3x}\)

\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

\(x^8+3x^4+1=\left(x^8+\frac{2.3x^4}{2}+\frac{9}{4}\right)-\frac{5}{4}\)

\(=\left(x^4+\frac{3}{2}\right)^2-\frac{5}{4}=\left(x^4+\frac{3-\sqrt{5}}{2}\right)\left(x^4+\frac{3+\sqrt{5}}{2}\right)\)

Phân tích xấu lắm bạn ạ

Kết quá của t là: \(\left(x^4-\frac{-3+\sqrt{5}}{2}\right)\left(x^4+\frac{3+\sqrt{5}}{2}\right)\)
Ghê chưa