a) x =0 không là nghiệm

với x khác 0 ; chia cả 2 vế của pt cho x²

\(6x^2+5x-38+5.\frac{1}{x}+6.\frac{1}{x^2}=0\Leftrightarrow6\left(x^2+2+\frac{1}{x^2}\right)+5\left(x+\frac{1}{x}\right)-50=0\) 

\(\Leftrightarrow6\left(x+\frac{1}{x}\right)^2+5\left(x+\frac{1}{x}\right)-50=0\) đặt \(t=\left(x+\frac{1}{x}\right)\)=> 6t² +5t -50 =0 => t= -10/3 hoặc t =5/2

+x +1/x = -10/3 => 3x² +10x+3 =0  => x =-3 ; x =-1/3

+x+1/x = 5/2 => 2x² -5x +2 =0 => x=2; x =1/2

b) a⁴ +2a² + 1 - a² = ( a² +1)² -a² = (a² -a+1)(a²+a+1)

vao cau hoi tuong tu

a: Sửa đề: x^3-x^2+5x-5

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

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

b: x^3+4x^2+x-6

=x^3-x^2+5x^2-5x+6x-6

=(x-1)(x^2+5x+6)

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

c: \(=\left(x+2\right)^3+y^3\)

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

Phân tích thành đa nhân tử 

đc kết quả :

(2x+1)(3x+1)

:)

5x^2 + 5xy - x - y 

=5x.(x+y)-(x+y)

=(x+y)(5x-1)

7x - 6x^2 - 2

=-6x²+3x+4x-2

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

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

a)\(3x^2-11x+6=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(3x-2\right)\left(x-3\right)\)

b)\(8x^2+10x-3=8x^2+12x-2x-3=4x\left(2x+3\right)-\left(2x+3\right)=\left(4x-1\right)\left(2x+3\right)\)

c)\(8x^2-2x-1=8x^2+2x-4x-1=2x\left(4x+1\right)-\left(4x+1\right)=\left(2x-1\right)\left(4x+1\right)\)

\(y-2x\sqrt{y}-3x\sqrt{y}+6x^2=\sqrt{y}\left(\sqrt{y}-2x\right)-3x\left(\sqrt{y}-2x\right)=\left(\sqrt{y}-3x\right)\left(\sqrt{y}-2x\right)\)