đề như này thì phân tích kiểu gì bạn

a: 

Sửa đề: \(4x^2-4xy+y^2-25a^2+10a-1\)

\(=\left(2x-y\right)^2-\left(5a-1\right)^2\)

\(=\left(2x-y-5a+1\right)\left(2x-y+5a-1\right)\)

b: Sửa đề: \(ax^2+bx^2+2xy\left(a+b\right)+ay^2+by^2\) 

\(=ax^2+ay^2+2xya+bx^2+2xyb+by^2\)

\(=a\left(x+y\right)^2+b\left(x+y\right)^2\)

\(=\left(x+y\right)^2\left(a+b\right)\)

ax - ay + bx - by= (ax - ay) + (bx - by)

                    = a(x - y) + b(x - y)

                    = (x - y)(a + b)

x² - 2xy + y² - 1= (x² - 2xy + y²) - 1² 

                             = (x - y)²  - 1² 

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

9 - x² - 2xy - y² = 3² - ( x² + 2xy + y² )

                         = 3² - ( x + y)² 

                         = ( 3 - ( x + y)).(3 + ( x + y))

                         = (3 - x - y)(3 + x + y)

Bài 1 : 

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

b)  \(25-4x^2-4xy-y^2=5^2-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2=\left(5+2x+y\right)\left(5-2x-y\right)\)

c)  \(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z.\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)

d)   \(x^2-4xy+4y^2-z^2+4tz-4t^2=\left(x^2-4xy+4y^2\right)-\left(z^2-4tz+4t^2\right)\)

\(=\left(x-2y\right)^2-\left(z-2t\right)^2=\left(x-2y+z-2t\right).\left(x-2y-z+2t\right)\)

BÀi 2 : 

a)   \(ax^2+cx^2-ay+ay^2-cy+cy^2=\left(ax^2+cx^2\right)-\left(ay+cy\right)+\left(ay^2+cy^2\right)\)

\(=x^2.\left(a+c\right)-y\left(a+c\right)+y^2.\left(a+c\right)=\left(a+c\right).\left(x^2-y+y^2\right)\)

b)   \(ax^2+ay^2-bx^2-by^2+b-a=\left(ax^2-bx^2\right)+\left(ay^2-by^2\right)-\left(a-b\right)\)

\(=x^2.\left(a-b\right)+y^2.\left(a-b\right)-\left(a-b\right)=\left(a-b\right)\left(x^2+y^2-1\right)\)

c)  \(ac^2-ad-bc^2+cd+bd-c^3=\left(ac^2-ad\right)+\left(cd+bd\right)-\left(bc^2+c^3\right)\)

\(=-a.\left(d-c^2\right)+d.\left(b+c\right)-c^2.\left(b+c\right)=\left(b+c\right).\left(d-c^2\right)-a\left(d-c^2\right)\)

\(=\left(b+c-a\right)\left(d-c^2\right)\)

BÀi 3 : 

a)  \(x.\left(x-5\right)-4x+20=0\) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=4\end{cases}}}\)

b)  \(x.\left(x+6\right)-7x-42=0\)\(\Leftrightarrow x.\left(x+6\right)-7.\left(x+6\right)=0\) \(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\\x=7\end{cases}}}\)

c)   \(x^3-5x^2+x-5=0\) \(\Leftrightarrow x^2.\left(x-5\right)+\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)\)

\(\Leftrightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=-1\left(KTM\right)\\x=5\end{cases}}}\)

d)   \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\)  \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)

1)

\(x^2+4xy+4y^2-a^2+2ab-b^2\)

\(=(x^2+4xy+4y^2)-(a^2-2ab+b^2)\)

\(=(x+2y)^2-(a-b)^2\)

\(=(x+2y-a+b)(x+2y+a-b)\)

2)

\(m^2-6m+9-x^2+4xy-4y^2\)

\(=(m^2-6m+9)-(x^2-4xy+4y^2)\)

\(=(m-3)^2-(x-2y)^2\)

\(=[(m-3)-(x-2y)][(m-3)+(x-2y)]\)

\(=(m-3-x+2y)(m-3+x-2y)\)

3)

\(ax^2+bx^2+2axy+2bxy+ay^2+by^2\)

\(=a(x^2+y^2+2xy)+b(x^2+2xy+y^2)\)

\(=a(x+y)^2+b(x+y)^2\)

\(=(a+b)(x+y)^2\)

4)

\(ax^2+bx^2+6ax+6bx+9a+9b\)

\(=(ax^2+6ax+9a)+(bx^2+6bx+9b)\)

\(=a(x^2+6x+9)+b(x^2+6b+9)\)
\(=a(x+3)^2+b(x+3)^2=(a+b)(x+3)^2\)

5)

\(8a^2xy-18b^2xy\)

\(=2xy(a^2-9b^2)=2xy[a^2-(3b)^2]\)

\(=2xy(a-3b)(a+3b)\)

a) x^4 - x^3 - x + 1 

= x^3 ( x - 1 ) - ( x- 1 )

= ( x^3 - 1 )(x - 1)

= ( x- 1 )^2 (x^2 + x +  1 )

a)x⁴-x³-x+1

=x³(x-1)-(x-1)

=(x-1)(x³-1)

=(x-1)(x-1)(x²+x+1)

=(x-1)²(x²+x+1)

b)5x²-4x+20xy-8y

(sai đề)

e, x³ - 2x²  +x

= x(x² - 2x + 1)

= x(x-1)²

\(ax+bx+ay+by\)

\(=x\left(a+b\right)+y\left(a+b\right)\)

\(=\left(x+y\right)\left(a+b\right)\)

\(xy+1-x-y\)

\(=x\left(y-1\right)-\left(y-1\right)\)

\(=\left(x-1\right)\left(y-1\right)\)

\(x^2-y^2-5x-5y\)

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

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

a) \(x^3-2x^2+2x-1^3\)

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

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

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

b) \(x^2y+xy+x+1\)

\(=xy\left(x+1\right)+\left(x+1\right)\)

\(=\left(xy+1\right)\left(x+1\right)\)

c) \(ax+by+ay+bx\)

\(=a\left(x+y\right)+b\left(x+y\right)\)

\(=\left(a+b\right)\left(x+y\right)\)

d) \(x^2-\left(a+b\right)x+ab\)

\(=x^2-ax-bx+ab\)

\(=\left(x^2-ax\right)-\left(bx-ab\right)\)

\(=x\left(x-a\right)-b\left(x-a\right)\)

\(=\left(x-b\right)\left(x-a\right)\)

e) Ko biết làm

f) \(ax^2+ay-bx^2-by\)

\(=\left(ax^2+ay\right)-\left(bx^2+by\right)\)

\(=a\left(x^2+y\right)-b\left(x^2+y\right)\)

\(=\left(a-b\right)\left(x^2+y\right)\)

a, x³ - 2x² + 2x - 1³

= x³ - 2x² . 1+ 2x.1² - 1³

= (x - 3 )³

@Cauthuteamha Vietanh giúp mình