1, \(a.\left(b+c\right)+3b+3c=a.\left(b+c\right)+3.\left(b+c\right)\)= \(\left(b+c\right).\left(3+a\right)\)

2, \(a.\left(m-n\right)+\left(m-n\right)=\left(m-n\right).\left(a+1\right)\)

3, \(7a^2-7ax-9a+9x=7a.\left(a-x\right)-9.\left(a-x\right)\)= \(\left(a-x\right).\left(7a-9\right)\)

4, \(4x+by+4y+bx=4.\left(x+y\right)+b.\left(x+y\right)\)= \(\left(x+y\right).\left(4+b\right)\)

5, \(ay-ax-2x+2y=a.\left(y-x\right)+2.\left(y-x\right)\)= \(\left(y-x\right).\left(a+2\right)\)

    Chúc bạn học tốt. Có gì không hiểu thì chat hỏi mik nhé. ^^

thank bn nhiều nha

1. a(b+c)+3b+3c=a(b+c)+3(b+c)=(b+c)(a+3)

2.a(m-n)+(m-n)=(m-n)(a+1)

3. 7a² -7ax-9a+9x=7a(a-x)-9(a-x)=(a-x)(7w-9

4. 4x+by+4y+bx=(4x+4y)+(by+bx)

5. ay-ax-2x+2y=(ay+2y)-(ax+2x)=y(a+2)-x(a+2)=(a+2)(y-x)

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

\(b,5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(5+a\right)\)

\(c,4x^2-25=\left(2x-5\right)\left(2x+5\right)\)

\(d,x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)

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

a ) 3x² + 2x

= x. ( 3x + 2 )

b ) 5x - 5y + ax - ay

= ( 5x + ax ) - ( 5y + ay )

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

= ( 5 + a ) ( x - y ) 

c ) 4x² - 25

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

d ) x² + 6x + 5

= x² + x + 5x + 5

= x.( x + 1 ) + 5.( x + 1 )

= ( x + 1 ) ( x + 5 )

e ) x² - y² + 2y - 1

= x² - ( y - 1 )²

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

f ) x³ - 3x + 2

= x³ + 2x² - 2x² - 4x + x + 2

= x² ( x + 2 ) - 2x ( x + 2 ) + ( x + 2 )

= ( x + 2 ) ( x² - 2x + 1 )

= ( x + 2 ) ( x - 1 )²

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

\(b)\)\(5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(a+5\right)\)

\(c)\)\(4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)

\(d)\)\(x^2+6x+5=\left(x^2+x\right)+\left(5x+5\right)=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)

\(e)\)\(x^2-y^2+2y-1\)

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

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

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

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

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

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

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

\(=\)\(\left(x-1\right)\left[x\left(x+1\right)-2\right]\)

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

Chúc bạn học tốt ~ 

a) Xem lại đề

b) x³ - 4x²y + 4xy² - 9x

= x(x² - 4xy + 4y² - 9)

= x[(x² - 4xy + 4y² - 3²]

= x[(x - 2y)² - 3²]

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

c) x³ - y³ + x - y

= (x³ - y³) + (x - y)

= (x - y)(x² + xy + y²) + (x - y)

= (x - y)(x² + xy + y² + 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

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

e) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

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

f) 3x² - 6xy + 3y² - 5x + 5y

= (3x² - 6xy + 3y²) - (5x - 5y)

= 3(x² - 2xy + y²) - 5(x - y)

= 3(x - y)² - 5(x - y)

= (x - y)[(3(x - y) - 5]

= (x - y)(3x - 3y - 5)

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

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

c) \(4x^2-4xy+y^2=\left(2x\right)^2-2.2x.y+y^2=\left(2x-y\right)^2\)

d) \(9x^3-9x^2y-4x+4y=9x^2\left(x-y\right)-4\left(x-y\right)=\left(9x^2-4\right)\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\)

e) \(x^3+2+3\left(x^3-2\right)=x^3+2+3x^3-6=4x^3-4=4\left(x^3-1\right)=4\left(x-1\right)\left(x^2+x+1\right)\)

\(a,\left(2a+3\right)x-\left(2a+3\right)y+\left(2a+3\right)\)

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

\(b,\left(4x-y\right)\left(a-1\right)-\left(y-4x\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)

​\(=\left(4x-y\right)\left(a-1\right)+\left(4x-y\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)​

\(=\left(4x-y\right)\left(a-1+b-1+1-c\right)\)

\(=\left(4x-y\right)\left(a+b-c-1\right)\)

\(c,x^k+1-x^k-1\)

\(=0?!?!\)

\(d,x^m+3-x^m+1\)

\(=4\)

\(e,3\left(x-y\right)^3-2\left(x-y\right)^2\)

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

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

\(f,81a^2+18a+1\)

\(=\left(9a\right)^2+2.9a+1\)

\(=\left(9a+1\right)^2\)

\(g,25a^2.b^2-16c^2\)

\(=\left(5ab\right)^2-\left(4c\right)^2\)

\(=\left(5ab+4c\right)\left(5ab-4c\right)\)

\(h,\left(a-b\right)^2-2\left(a-b\right)c+c^2\)

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

\(i,\left(ax+by\right)^2-\left(ax-by\right)^2\)

\(=\left(ax+by-ax+by\right)\left(ax+by+ax-by\right)\)

\(=2by.2ax\)

\(=4axby\)

\(\text{a) }\left(2a+3\right)x-\left(2a+3\right)y+\left(2a+3\right)\)

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

\(\text{b) }\left(4x-y\right)\left(a-1\right)-\left(y-4x\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)

\(=\left(4x-y\right)\left(a-1\right)+\left(4x-y\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)

\(=\left(4x-y\right)\left(a-1+b-1+1-c\right)\)

\(=\left(4x-y\right)\left(a+b-c-1\right)\)

\(\text{c) }x^k+1-x^k-1\)

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

\(=0\)

\(\text{d) }x^m+3-x^m+1\)

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

\(=4\)

\(\text{e) }3\left(x-y\right)^3-2\left(x-y\right)^2\)

\(=\left(x-y\right)^2\left[3\left(x-y\right)-2\right]\)

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

\(\text{f) }81a^2+18a+1\)

\(=\left(9a\right)^2+2.9a.1+1^2\)

\(=\left(9a+1\right)^2\)

\(\text{g) }25a^2b^2-16c^2\)

\(=\left(5ab\right)^2-\left(4c\right)^2\)

\(=\left(5ab+4c\right)\left(5ab-4c\right)\)

\(\text{h) }\left(a-b\right)^2-2.\left(a-b\right).c+c^2\)

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

\(\text{i) }\left(ax+by\right)^2-\left(ax-by\right)^2\)

\(=\left(ax+by+ax-by\right)\left(ax+by-ax+by\right)\)

\(=2by.2ax\)

\(=4byax\)

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

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

Bài 1.

a) -2x( -3x + 2 ) - ( x + 2 )²

= 6x² - 4x - ( x² + 4x + 4 )

= 6x² - 4x - x² - 4x - 4

= 5x² - 8x - 4

b) ( x + 2 )( x² - 2x + 4 ) - 2( x + 1 )( 1 - x )

= x³ + 8 + 2( x + 1 )( x - 1 )

= x³ + 8 + 2( x² - 1 )

= x³ + 8 + 2x² - 2

= x³ + 2x² + 6

c) ( 2x - 1 )² - 2( 4x² - 1 ) + ( 2x + 1 )²

= 4x² - 4x + 1 - 8x² + 2 + 4x² + 4x + 1

= 4

d) x² - 3x + xy - 3y

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

= ( x - 3 )( x + y )

Bài 2.

a) 4x² - 4xy + y² = ( 2x - y )²

b) 9x³ - 9x²y - 4x + 4y

= 9x²( x - y ) - 4( x - y )

= ( x - y )( 9x² - 4 )

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

c) x³ + 2 + 3( x³ - 2 )

= x³ + 2 + 3x³ - 6

= 4x³ - 4

= 4( x³ - 1 )

= 4( x - 1 )( x² + x + 1 )

Bài 3.

2( x - 2 ) = x² - 4x + 4

⇔ ( x - 2 )² - 2( x - 2 ) = 0

⇔ ( x - 2 )( x - 2 - 2 ) = 0

⇔ ( x - 2 )( x - 4 ) = 0

⇔ x = 2 hoặc x = 4

2/ (b⁴ - 4b² + 4) - 9a² = (b² - 2)²  - 9a² = (b² - 2 + 3a)(b² - 2 - 3a)

3/ (x² +1)(x² + x + 1)[x + (√13 - 7)/6][3x - (√13 + 7)/2]

bạn viết hẳn cách làm câu 3 cho mình đc k mình k hiểu lắm =)