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

b) \(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)

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

c) \(2x^3-12x^2+17x-2=2x^3-4x^2-8x^2+16x+x-2=2x^2\left(x-2\right)-8x\left(x-2\right)+\left(x-2\right).\)

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

d) \(2x^4+x^3-22x^2+15x+36=2x^4+2x^3-x^3-x^2-21x^2-21x+36x+36.\)

\(=2x^3\left(x+1\right)-x^2\left(x+1\right)-21x\left(x+1\right)+36\left(x+1\right)\)

\(=\left(x+1\right)\left(2x^3-x^2-21x+36\right)\)

a) \(8x\left(x-3\right)+x-3=0\)

\(\Rightarrow8x\left(x-3\right)+\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(8x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{8}\end{matrix}\right.\)

b) \(x^2+36=12x\)

\(\Rightarrow x^2-12x+36=0\)

\(\Rightarrow\left(x-6\right)^2=0\)

\(\Rightarrow x=6\)

Nếu bạn muốn có lời giải thì ít thôi @.@

Katherine Lilly Filbert nói rất đúng câu hỏi nhiều như vậy ai mà trả lời đc hết cơ chứ

\(x^4-2x^3-12x^2+12x+36=x^4+x^2+36-2x^3+12x-12x^2-x^2\)

\(=\left(x^2-x-6\right)^2-x^2=\left(x^2-6\right)\left(x^2-2x-6\right)\)

a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4 
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))

Bài 4:

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

\(5\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)

\(x^2-12x+36=\left(x-6\right)^2\)

Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn.

a.\(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

b.\(x^3-6x^2+12x-8=\left(x-2\right)^3\)

c.\(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)