Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3-2x+5\right)\)

\(=\left(2x+1\right)\left(8-2x\right)\)

\(=2\left(4-x\right)\left(2x+1\right)\)

b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

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

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

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

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

Bài 2: 

a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)

\(=\left(a-b\right)\left(2a-4b\right)\)

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

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

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

\(=\left(x-3y\right)\left(x-3y+4\right)\)

whatterfice

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

^^

Bài `1:`

`a)3x^3+6x^2=3x^2(x+2)`

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

Bài `2:`

`a)(2x-1)^2-25=0`

`<=>(2x-1-5)(2x-1+5)=0`

`<=>(2x-6)(2x+4)=0`

`<=>[(x=3),(x=-2):}`

`b)Q.(x^2+3x+1)=x^3+2x^2-2x-1`

`<=>Q=[x^3+2x^2-2x-1]/[x^2+3x+1]`

`<=>Q=[x^3-x^2+3x^2-3x+x-1]/[x^2+3x+1]`

`<=>Q=[(x-1)(x^2+3x+1)]/[x^2+3x+1]=x-1`

bn ko bik lm hay sao, hay là bn chỉ đăng đề lên thôi

sao nhìu... z p , đăq từq câu 1 thôy nha p

Ôi trời sao lắm thế ít thôi bạn nên tách ra mà bạn cần gấp lắm à

đúng rồi pn. giúp mik đc bài nào cũng đc

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

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

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

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

ỳtct7ct7c7c7t79tc9

1

a,\(\left(2x+1\right)\left(3x+1\right)-\left(6x-1\right)\left(x+1\right)\)

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

\(=6x^2+5x+1-6x^2-6x+x+1\)

\(=2\)

c,\(\left(a+1\right)\left(a^2-a+1\right)+\left(a+1\right)\left(a-1\right)\)

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

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

\(=a^3+a^2\)

2,

a,\(4ab+a^2-3a-12b\)

\(=\left(4ab-12b\right)+\left(a^2-3a\right)\)

\(=4b\left(a-3\right)+a\left(a-3\right)\)

\(=\left(4b+a\right)\left(a-3\right)\)

b,\(x^3+3x^2+3x+1-27y^3\)

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

\(=\left(x+1-3y\right)\left[\left(x+1\right)^2+\left(x+1\right).3y+\left(3y\right)^2\right]\)

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

4

a,\(2004^2-16\)

\(=2004^2-4^2\)

\(=\left(2004-4\right)\left(2004+4\right)\)

\(=2000.2008\)

\(=4016000\)

b,\(892^2+892.216+108^2\)

\(=\left(892+108\right)^2\)

\(=1000^2=1000000\)

c,\(10,2.9,8-9,8.0,2+10,2^2-10,2.0,2\)

\(=9,8\left(10,2-0,2\right)+10,2\left(10,2-0,2\right)\)

\(=9,8.10+10,2.10\)

\(=98+102\)

\(=200\)

d,\(36^2+26^2-52.36\)

=\(\left(36-26\right)^2\)

\(=10^2=100\)

3)\(A=-x^2+2x-3\)

\(\Leftrightarrow A=-x^2+2x-1-2\)

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

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

Vậy GTLN của A=-2 khi x=1