a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

= 

c)4x+4y-x^2(x+y)

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

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

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

d) y^2+y-x^2+x

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

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

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

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

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

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

f)9x^2-25y^2-6x+10y

= 

ko biết làm

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a,\(xy+3x-7y-21\)

\(=x\left(y+3\right)-7\left(y+3\right)\)

\(=\left(y+3\right)\left(x-7\right)\)

\(b,2xy-15-6x+5y\)

\(=\left(2xy-6x\right)+\left(-15+5y\right)\)

\(=2x\left(y-3\right)-5\left(3-y\right)\)

\(=2x\left(y-3\right)+5\left(y-3\right)\)

\(=\left(y-3\right)\left(2x+5\right)\)

Bài 4 :

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

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

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

d)

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

e) Trùng câu d

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

Bài 5:

a) \(x^3-x^2-x+1=0\)

\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy ...

b) Sửa đề : \(\left(2x-3\right)^2-\left(4x^2-9\right)=0\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3-2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(-6\right)=0\)\

\(\Leftrightarrow2x-3=6\)

\(\Leftrightarrow x=\frac{9}{2}\)

vậy........

c) \(x^4+2x^3-6x-9=0\)

\(\Leftrightarrow\left(x^4-9\right)+\left(2x^3-6x\right)=0\)

\(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)+2x\left(x^2-3\right)=0\)

\(\Leftrightarrow\left(x^2-3\right)\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow x^2-3=0\Leftrightarrow x^2=3\Leftrightarrow x=\pm\sqrt{3}\)

Vậy

d) \(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Vậy ........

dài wa,lm xong chắc đến năm sau

1)A=3(x-1)^2-(x+1)^2+2(x-3)(x+3)-(2x+3)^2-(5-20x)

=3(x^2-2x+1)-(x^2+2x+1)+2(x^2-9)-(4x^2+12x+9)-(5-20x)

=3x^2-6x+3-x^2-2x-1+2x^2-18-4x^2-12x-9-5+20x

=-30

2)B=5x(x-7)(x+7)-x(2x-1)^2-(x^3+4x^2-246x)-175

=5x(x^2-49)-x(4x^2-4x+1)-x^3-4x^2+246x-175

=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175

=-175

cn lại lm tg tự nha bn