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)

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 :

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

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

Đặt \(x^2-8x+4=t\)

Ta có:\(A=\left(t-2\right)\left(t+2\right)-5=t^2-4-5=t^2-9=\left(t-3\right)\left(t+3\right)\)

Hay \(A=\left(x^2-8x+1\right)\left(x^2-8x+7\right)\)

\(A=\left(x^2-8x+1\right)\left(x-7\right)\left(x-1\right)\)

Cái này nếu phân tích thành nhân tử tiếp thì phức tạp lắm,bạn để như thế này thì gọn hơn

Nếu phân tích ra thì được như thế này:

\(x^2-8x+1=-\left(x+\sqrt{10}+4\right)\left(x+\sqrt{10}-4\right)\)

Lời giải:

a. 

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

$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$

b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?

\(8x^3+36x^2y+54xy^2+27y^3\\ =\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\\ =\left(2x+3y\right)^3\\ =\left(2x+3y\right)\left(2x+3y\right)\left(2x+3y\right)\)

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

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

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

a: =(2x)^3+3*(2x)^2*3y+3*2x*(3y)^2+(3y)^3

=(2x+3y)^3

b: (x-y)^3-(x+y)^3

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

=-2y*[x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2]

=-2y(3x^2+y^2)

c: (x+1)^3+(x-1)^3

=(x+1+x-1)[(x+1)^2-(x+1)(x-1)+(x-1)^2]

=2x*[x^2+2x+1-x^2+1+x^2-2x+1]

=2x(x^2+3)

d: =(x-1-x-1)(x-1+x+1)

=2x*(-2)=-4x

\(8x^3+36x^2y+54xy^2+27y^3\)

\(=\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\)

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

\(------\)

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

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

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

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

\(------\)

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

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

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

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

\(------\)

\(\left(x-1\right)^2-\left(x+1\right)^2\)

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

\(=-2.2x=-4x\)

a. 12xy² - 8x²y = 4xy . (3y - 2x)

b. 3x + 3y - x² - xy = (3x + 3y) - (x² + xy) = 3 . (x + y) - x . (x + y) = (x + y)(3 - x)

GIUSP MIK VS MN ƠI

Bài 2:

a: =>2x^2-8+3x-2x^2=15

=>3x=23

=>x=23/3

b: \(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)

=>x=5 hoặc x=-3

c: =>x^2+7x-30=0

=>(x+10)(x-3)=0

=>x=3 hoặc x=-10

vài câu khó quá 

Đặt a=x²+3x+5 

ta có \(8a^2+7a-15\)

\(=8a^2-8a+15a-15=8a\left(a-1\right)+15\left(a-1\right)\)

\(=\left(8a+15\right)\left(a-1\right)\)

Trả lại biến

\(\left(8x^2+24x+40+15\right)\left(x^2+3x+5-1\right)\)

\(=\left(8x^2+24x+55\right)\left(x^2+3x+4\right)\)