B

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

Đặt \(x^2+3x+1,5=a\)

\(\Rightarrow A=\left(a-0,5\right)\left(a+0,5\right)-6\)

\(\Rightarrow A=a^2-0,25-6\)

\(\Rightarrow A=a^2-\frac{25}{4}\)

\(\Rightarrow A=\left(a-\frac{5}{2}\right)\left(a+\frac{5}{2}\right)\)

Thay \(a=x^2+3x+0,5\)vào A ta có :

\(A=\left(x^2+3x+0,5-\frac{5}{2}\right)\left(x^2+3x+0,5+\frac{5}{2}\right)\)

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

c, Đặt \(x^2+3x+2=a\)

Ta có : \(\left(a-1\right)a-6=a^2-a-6=\left(a^2-3a\right)+\left(2a-6\right)\)

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

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

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

Câu d làm tương tự .

Gợi ý : (x+3)(x+5) = x² + 8x + 15 

đặt bằng a rồi giải tiếp

d) Đặt \(B=\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)

\(B=\left(x^2+8x+7\right)\left(x^2+5x+3x+15\right)+15\)

\(B=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(a=x^2+8x+11\)

\(\Rightarrow B=\left(a-4\right)\left(a+4\right)+15\)

\(\Rightarrow B=a^2-16+15\)

\(\Rightarrow B=a^2-1\)

\(\Rightarrow B=\left(a-1\right)\left(a+1\right)\)

Thay \(a=x^2+8x+11\)vào B ta có :

\(B=\left(x^2+8x+11-1\right)\left(x^2+8x+11+1\right)\)

\(B=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

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

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

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

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

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

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

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

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

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

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

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

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

\(=(x-2y)(3x-1+5x)\)

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

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

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

5: \(\left(3x-y\right)^2-4y^2\)

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

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

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

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

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

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

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

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

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

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

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

đặt \(x^2+4x+8=a\)

=> \(A=a^2+3ax+2x^2=a^2+ax+2ax+2x^2=a\left(a+x\right)+2x\left(a+x\right)\)

          \(=\left(a+x\right)\left(a+2x\right)\)

b) ta có 

\(B=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

đặt \(x^2+8x+11=a\)

=> \(B=\left(a-4\right)\left(a+4\right)+15=a^2-16+15=a^2-1=\left(a-1\right)\left(a+1\right)\)

         \(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left(x^2+6x+2x+12\right)\)

         \(=\left(x^2+8x+10\right)\left[x\left(x+6\right)+2\left(x+6\right)\right]=\left(x^2+8x+10\right)\left(x+6\right)\left(x+2\right)\)

khó thế

Ta có (6x+5)²(3x+2)(x+1)-35

= (36x²+60x+25)(3x²+5x+2)-35 (1)

Đặt a=3x²+5x+2

=> 12a+1= 12(3x²+5x+2)+1 =36x²+60x+25

Thay a=3x²+5x+2 vào (1) ta được

(12a+1).a-35=12a²+a-35

= 12a²-20a+21a-35

= 4a(3a-5)+7(3a-5)

= (3a-5)(4a+7) (2)

Thay 3x²+5x+2=a vào (2) ta được

(9x²+15x+6-5)(12x²+20x+8+7)

= (9x²+15x+1)(12x²+20x+15)

Ta có: \(\left(6x+5\right)^2\left(3x+2\right)\left(x+1\right)-35\)

\(=\left(36x^2+60x+25\right)\left(3x^2+5x+2\right)-35\)(1)

Đặt \(3x^2+5x+2=y\)

\(\left(1\right)=\left(12y+1\right)y-35\)

\(=12y^2+y-35\)

\(=\left(3y-5\right)\left(4y+7\right)\)

\(=\left(9x^2+15x+1\right)\left(12x^2+20x+15\right)\)

b)(x²+x+1)(x²+x+2)-12

Đặt t=x²+x+1

t(t+1)-12=t²+t-12

=(t-3)(t+4)=(x²+x+1-3)(x²+x+1+4)

=(x²+x-2)(x²+x+5)

=(x-1)(x+2)(x²+x+5)

c)(x²+8x+7)(x²+8x+15)+15

Đặt t=x²+8x+7 

t(t+8)+15=t²+8t+15

=(t+3)(t+5)

=(x²+8x+7+3)(x²+8x+7+15)

=(x²+8x+10)(x²+8x+22)

d)(x+2)(x+3)(x+4)(x+5)-24

=(x²+7x+10)(x²+7x+12)-24

Đặt t=x²+7x+10

t(t+2)-24=(t-4)(t+6)

=(x²+7x+10-4)(x²+7x+10+6)

=(x²+7x+6)(x²+7x+16)

=(x+1)(x+6)(x²+7x+16)

a/ Đặt x² + 4x + 8 = a

Thì đa thức ban đầu thành

a² + 3ax + 2x² = (a² + 2ax + x²) + (ax + x²)

= (a + x)² + x(a + x) = (a + x)(a + 2x)

b/ Đặt x² + x + 1 = a thì đa thức ban đầu thành

a(a + 1) - 12 = a² + a - 12 = (a² - 3a) + (4a - 12) 

= (a - 3)(a + 4)

vài câu khó quá 

d)\(\left(x^2+y^2-z^2\right)^2-4x^2y^2\)

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

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

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

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

e)Đặt \(x^2+3x=a\)

Có: \(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)

\(=\left(a+1\right)\left(a-3\right)-5\)

\(=a^2-3a+a-3-5\)

\(=a^2-2a-8\)

\(=a^2+2x-4x-8\)

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

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

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

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

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

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

\(d,\left(x^2+y^2-z^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-z^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+y^2-z^2-2xy\right)\left(x^2+y^2-z^2+2xy\right)\)
\(=\left[\left(x^2-2xy+y^2\right)-z^2\right]\left[\left(x^2+2xy+y^2\right)-z^z\right]\)
\(=\left[\left(x-y\right)^2-z^2\right]\left[\left(x+y\right)^2-z^2\right]\)
\(=\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)\left(x+y+z\right)\)
\(e,\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\left(1\right)\)
\(\text{Đặt }x^2+3x+\frac{1-3}{2}=t\)
\(\text{hay }x^2+3x-2=t\left(2\right)\)
\(\left(1\right)\Leftrightarrow\left(t+3\right)\left(t-1\right)-5\)
\(\Rightarrow t^2-t+3t-3-5\)
\(=t^2+2t-8\)
\(=t^2-2t+4t-8\)
\(=t\left(t-2\right)+4\left(t-2\right)\)
\(=\left(t-2\right)\left(t+4\right)\left(3\right)\)
\(\text{Thay (2) vào (3),ta được:}\)
\(\left(x^2+3x-2-2\right)\left(x^2+3x-2+4\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x^2-x+4x-4\right)\left(x^2+x+2x+2\right)\)
\(=\left[x\left(x-1\right)+4\left(x-1\right)\right]\left[x\left(x+1\right)+2\left(x+1\right)\right]\)

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