\(x^2+2xy+y^2-xz-yz\)

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

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

mk chỉnh lại đề

\(x^2-2xy+y^2-z^2+2zt+t^2\)

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

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

mk chỉnh lại đề:

\(ax^2+cx^2-ay+ay^2-cy+cy^2\)

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

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

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

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

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

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

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

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

trả lời giùm mình với

a) 8x² + 4xy - 2ax - ay = (8x² + 4xy) - (2ax + ay) = 4x(2x + y) - a(2x + y) = (4x - a)(2x + y)

b) 2xy - x² - y² = 16 - (-2xy + x² + y²) = 4² - (x - y)² = (4 - x + y)(4 + x - y)

c) x² - y² - 2yz - z² = x² - (y² + 2yz + z²) = z² - (y + z)² = (z - y - z)(z + y + z)

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

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

3 ) \(5x-5y+ax-ay=5.\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(5+a\right)\)

4 ) \(a^3-a^2x-ay+xy=a^2.\left(a-x\right)-y.\left(a-x\right)=\left(a-x\right)\left(a^2-y\right)\)

5 ) \(xy.\left(x+y\right)+yz.\left(y+z\right)+xz.\left(x+z\right)+2xyz\)

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

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

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

\(=\left(x+y\right)\left(xy+yz+z^2+xz\right)=\left(x+y\right)\left[\left(xy+xz\right)+\left(yz+z^2\right)\right]\)

\(=\left(x+y\right)\left[x.\left(y+z\right)+z.\left(y+z\right)\right]=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

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

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

a) 2x^2 -  2xy - 5x +5y

= (2x^2 - 2xy ) - ( 5x- 5 y)

=2x(x-y) - 5(x-y)

=(x- y). (2x- 5)

b)8x²  +4xy-2ax-ay

=(8x² +4xy) -(2ax+ay)

=4x(2x+y)-a(2x+y)

=(2x+y).(4x-a)

c)=x(x² -4x +4)

=x(x-2)²

d)=16- (x² -2xy +y^2)

=4^2-(x-y)^2

=(4-x+y).(4+x-y)

các câu còn lại tg tự

chúc bn hok tốt

\(b,=\left(x+y\right)-x\left(x+y\right)=\left(1-x\right)\left(x+y\right)\\ c,=x\left(x-y\right)-7\left(x-y\right)=\left(x-7\right)\left(x-y\right)\\ d,=x^2\left(a+c\right)-y\left(a+c\right)+y^2\left(a+c\right)\\ =\left(x^2-y+y^2\right)\left(a+c\right)\)

Câu a sai đề, không phân tích được

a: \(xy+xz-5x-5y=\left(x+y\right)\left(z-5\right)\)

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

c: \(x^2-xy-7x+7y=\left(x-y\right)\left(x-7\right)\)

Đa thức này không phân tích được thành nhân tử. Bạn coi lại đề.

Đề là gì vậy bạn

de la phan h da thuc do ban

1 : 8x²+4xy-2ax -ay=4x(2x+y)-a(2x+y)=(2x+y)(4x-a)

2,3 tương tự