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

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

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

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

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

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

ĐẶT NHÂN TỬ CHUNG NHA!

1) ab - ac + ad = a( b- c +d ) 

2) ax - bx - cx + dx = x( a-b-c+d)

3) a.( b + c ) - d . ( b + c )= (b+c)(a-d)

4) ac - ad + bc - bd = a( c-d)  + b( c-d) = (a+b)(c-d)

5) ax + by + bx + ay= a( x+y) + b( x+y) = (a+b)(x+y)

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

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

1) ab + ac = a(b + c)

2) ab - ac + ad = a(b - c + d)

3) ax - bx - cx + dx = x(a - b - c + d)

4) a(b + c) - d(b + c) = (b + c)(a - d)

5) ac - ad + bc - bd = a(c - d) + b(c - d) = (c - d)(a + b)

6) ax + by + bx + ay = (ax + ay) + (bx + by) = a(x + y) + b(x + y) = (x + y)(a + b)

ab + ac = a ( b + c )

ab - ac + ad = a ( b - c + d )

ax - bx - cx + dx = x ( a - b - c + d )

1/ab+ac=a(b+c)

2/ab-ac+ad=a(b-c)+ad=a(b-c+d)

3/ax-bx-cx+dx=x(a-b-c)+xd=x(a-b-c+d)

4/a(b+c)-d(b+c)=(ab+ac)-(bd+cd)=b(a+d)-c(a+d)=a+d(b+c)

5/ac-ad+bc-bd=a(c-d)+b(c-d)=c-d(a+b)

6/ax+by+bx+ay=a(x+y)+b(x+y)=x+y(a+b)

1/ ab+ac=a(b+c)

2/ab-ac+ad=a(b-c+d)

3/ax-bx-cx+dx=x(a-b-c+d)

4/a(b+c)-d(b+c)=(b+c)(a-d)

5/ac-ad+bc-bd=a(c-d)+b(c-d)=(c-d)(a+b)

6/ax+by+bx+ay=a(x+y)+b(y+x)=(y+x)(a+b)

\(1,ab+ac\)

\(=a\left(b+c\right)\)

\(2,ab-ac+ad\)

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

\(3,ax-bx-cx+dx\)

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

\(4,a\left(b+c\right)-d\left(b+c\right)\)

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

\(5,ac-ad+bc-bd\)

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

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

\(6,ax+by+bx+ay\)

\(=\left(ax+bx\right)+\left(by+ay\right)\)

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

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

1? 

a(b+c)

a(b-c+d)

x(a-b-c+d)

(b+c)(a-d)

(c-d)(a+b)

(x+-y)(a+b)

1/ ab+ ac

=a(b+c)

2/ ab - ac + ad

=a(b-c+d)

3/ ax - bx - cx + dx

=x(a-b-c+d)

4/ a(b + c) - d(b + c)

=(a-d)(b+c)

5/ ac - ad + bc - bd

=a(c-d)+b(c-d)

=(a+b)(c-d)

6/ ax + by + bx + ay

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

=(x+y)(b+a)

A) ab + ac=a.(b+c)

B)ab - ac + ad=a.(b-c+d)

C) ã-bx-cx+dx=x.(a-b-c+d)

D)a(b+c) - d(b+c)=(b+c).(a-d)

E) ac-ad+bc-bd=a.(c-d)+b.(c-d)=(c-d).(a+b)

G) ax+by+bx+ay=x.(a+b)+y.(b+a)=(a+b).((x.y)

1) \(ab+ac=a\left(b+c\right)\)

2) \(ab-ac+ad=a\left(b-c+d\right)\)

3) \(ax-bx-cx+dx=x\left(a-b-c+d\right)\)

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

5) \(ac-ad+bc-bd=a\left(c-d\right)+b\left(c-d\right)=\left(a+d\right)\left(c-d\right)\)

6) \(ax+by+bx+ay=x\left(a+b\right)+y\left(a+b\right)=\left(x+y\right)\left(a+b\right)\)