\(=x\left(x-4\right)+5\left(x-4\right)=\left(x+5\right)\left(x-4\right)\)

x(x-4)+5(x-4)

(x+5)(x-4)

Phân tích thành đa nhân tử 

đc kết quả :

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

:)

Bài này ko thể phân tích theo kiểu lớp 8 được (chưa học căn thức)

\(2x^2-6x+1=\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\frac{3\sqrt{2}}{2}+\left(\frac{3\sqrt{2}}{2}\right)^2-\frac{7}{2}\)

\(=\left(\sqrt{2}x-\frac{3\sqrt{2}}{2}\right)^2-\left(\frac{\sqrt{14}}{2}\right)^2\)

\(=\left(\sqrt{2}x-\frac{3\sqrt{2}}{2}+\frac{\sqrt{14}}{2}\right)\left(\sqrt{2}x-\frac{3\sqrt{2}}{2}-\frac{\sqrt{14}}{2}\right)\)

\(=\left(\sqrt{2}x+\frac{\sqrt{14}-3\sqrt{2}}{2}\right)\left(\sqrt{2}x-\frac{\sqrt{14}+3\sqrt{2}}{2}\right)\)

\(2x^2-6x+1=2\left(x^2-3x+\frac{9}{4}-\frac{7}{4}\right)=2\left[\left(x-\frac{3}{2}\right)^2-\left(\frac{\sqrt{7}}{2}\right)^2\right]=2\left(x-\frac{3}{2}-\frac{\sqrt{7}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{7}}{2}\right)\)

\(=2\left(x-\frac{3+\sqrt{7}}{2}\right)\left(x-\frac{3-\sqrt{7}}{2}\right)\)

đề có sai ko bn?

\(6x^3+x^2+x+1=\left(6x^3+3x^2\right)+\left(-2x^2-x\right)+\left(2x+1\right)\)

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

K sai dau
giao an truong Tran dai nghia do

cái này dễ mà

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

= (2x-1)^3

(x² + 2.x.3 + 3² - 1).(x² + 2.x.4 + 16 - 1) - 24

=[(x+3)² - 1]. [(x+4)²-1] -24

=(x+3+1)(x+3-1)(x+4+1)(x+4-1) - 24

=(x+4)(x+2)(x+5)(x-3) - 24

(x²+6x+8)(x²+8x+15)-24

<=>(x²+4x+2x+8)(x²+5x+3x+15)-24

<=> [x(x+4)+2(x+4)][x(x+5)+3(x+5)]-24

<=> (x+4)(x+2)(x+5)(x+3)-24

<=> (x+4)(x+3)(x+2)(x+5)-24

<=>(x²+7x+12)(x²+7x+10)

đặt t=x²+7x+11 ta có:

(t-1)(t+1)-24

<=> t²-1-24

<=>t²-25

<=>(t-5)(t+5)

thay t=x²+7x+11 vào ta có:

(x²+7x+11-5)(x²+7x+11+5)

<=>(x²+7x+6)(x²+7x+16)