a Đề sai: )

b

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

c

\(4x^2-y^2+4x+1\\ =\left(2x\right)^2+2.2x.1+1-y^2\\ =\left(2x+1\right)^2-y^2\\ =\left(2x+1-y\right)\left(2x+1+y\right)\)

d

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

e

\(5x^2-10xy+5y^2-5z^2\\ =5\left(x^2-2xy+y^2-z^2\right)\\ =5\left[\left(x-y\right)^2-z^2\right]\\ =5\left(x-y-z\right)\left(x-y+z\right)\)

c: =(2x+1)^2-y^2

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

d: =x^2(x^2+2x+1)

=x^2(x+1)^2

e: =5(x^2-2xy+y^2-z^2)

=5[(x-y)^2-z^2]

=5(x-y-z)(x-y+z)

a)\(\left(3x-1\right)^2-16=\left(3x-1-16\right)\left(3x-1+16\right)\)

                                     \(=\left(3x-17\right)\left(3x+15\right)\)

c)\(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5+x-9\right)\left(2x+5-x+9\right)\)

                                                 \(=\left(x-4\right)\left(x+14\right)\)

      Aps dungj t/c a² - b² = ( a-b)(a+b)

a) ko bt làm

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

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

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

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

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

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 ~ 

1.a)\(x^2-ax+bx-ab=x\left(x-a\right)+b\left(x-a\right)=\left(x+b\right)\left(x-a\right)\)

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

c)\(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-4\right)\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)

2.a)\(2x^2-12x=-18=>2x^2-12x+18=0=>x^2-6x+9=0=>\left(x-3\right)^2=0=>x-3=0=>x=3\)b)\(\left(4x^2-4x+1\right)-x^2=0=>3x^2-3x-x+1=3x\left(x-1\right)-\left(x-1\right)=\left(3x-1\right)\left(x-1\right)=0\)

\(=>\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}=>\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)

a) 2x² - 12x = -18

<=> 2x² - 12x + 18 = 0

<=> 2(x² - 6x + 9) = 0

<=> 2(x² - 2.x.3 + 9) = 0

<=> 2(x - 3)² = 0

<=> x - 3 = 0

<=> x = 0 + 3

<=> x = 3

b) (4x² - 4x + 1) - x² = 0

<=> 4x² - 4x + 1 - x² = 0 

<=> 3x² - 4x + 1 = 0

<=> 3x² - x - 3x + 1 = 0

<=> x(3x - 1) - (3x - 1) = 0

<=> \(\orbr{\begin{cases}\left(3x-1\right)=0\\\left(x-1\right)=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)

bài 2 

a)\(2x^2-12x=-18\)

\(\Rightarrow2(x^2-6x+9)=0\) 

\(\Rightarrow\left(x-3\right)^2=0\) 

\(\Rightarrow x=3\) 

Vậy......

b) \(\left(4x^2-4x+1\right)-x^2=0\) 

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

\(\Rightarrow\left(3x-1\right)\left(x-1\right)=0\) 

\(\Rightarrow\orbr{\begin{cases}3x-1=0\Leftrightarrow x=\frac{1}{3}\\x-1=0\Leftrightarrow x=1\end{cases}}\) 

Vậy.....

\(A=4x^2+6x=2x\left(2x+3\right)\)

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

\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)

\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)

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

\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)

\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)

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

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

^^