Trong SGK có chỉ đó b

Ko hiểu thì kb vs mik

mik chỉ thêm cho

Chỉ mình với nha

Ok

Mình không giảnh để trả lời câu hỏi này 

a) ( x 2  – 4x + 1)( x 2  – 2x + 3).     b) (3x – y – 1)(x – 7y – 1).

c, x⁴+6x³+11x²+6x+1
=x⁴+6x³+9x²+2x²+6x+1
=x⁴+9x²+1+6x³+2x²+6x
=(x²)²+(3x)²+1²+2.x².3x+2.x².1+2.3x.1       (1)
Áp dụng hằng đẳng thức (a+b+c)²=a²+b²+c²+2ab+2ac+2bc
=> (1)=(x²+3x+1)²

Câu a nhé bạn:
a,   3x²−22xy−4x+8y+7y²+1
=3x²-21xy-xy-3x-x+7y+y+7y²+1
=(3x²−21xy−3x)−(xy-7y²-y)−(x-7y-1)
=3x(x−7y−1)−y(x−7y−1)−(x−7y−1)
=(3x−y−1)(x−7y−1)

Cho mk hs bn Diệu Thúy để lm đc như câu a) thì có pp j k bn? Hay chỉ lag ngồi nhóm lại thôi hả bn? Trả lời nhanh dùm mk nha😘

1) \(x^2-6x+3\)

\(=x^2-6x+9-6\)

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

\(=\left(x-3+\sqrt{6}\right)\left(x-3-\sqrt{6}\right)\)

2) \(2m^2+10m+8\)

\(=2m^2+2m+8m+8\)

\(=2m\left(m+1\right)+8\left(m+1\right)\)

\(=\left(2m+8\right)\left(m+1\right)\)

\(=2\left(m+4\right)\left(m+1\right)\)

3) \(9x^2+6x-8\)

\(=\left(9x^2+6x+1\right)-9\)

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

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

4) \(x^3-5x^2-14x\)

\(=x\left(x^2-5x-14\right)\)

\(=x\left(x^2-2x+7x-14\right)\)

\(=x\left[x\left(x-2\right)+7\left(x-2\right)\right]\)

\(=x\left(x+7\right)\left(x-2\right)\)

1) x^2-6x+3

= (x^2-6x+9)-6

=(x-3)^2-6

=(x-3-căn 6)(x-3+căn 6)

2) =2(m^2+5m+4)

=2(m+1)(m+4)

3) =9x^2+6x+1-9

=(3x+1)^-9

=(3x-2)(3x+4)

4, x^3-5x^2-14x

=x(x-7)(x+2)

5, a^4+4a^2-5

=a^4+4a^2+4-9

=(a^2+2)^-9

=(a^2-1)(a^2+5)

6, x^3-7x-6

=(x-3)(x+1)(x+2).

5) \(a^4+4a^2-5\)

\(=a^4+5a^2-a^2-5\)

\(=a^2\left(a^2+5\right)-\left(a^2+5\right)\)

\(=\left(a^2-1\right)\left(a^2+5\right)\)

\(=\left(a+1\right)\left(a-1\right)\left(a^2+5\right)\)

6) \(x^3-7x-6\)

\(=x^3-x-6x-6\)

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

\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)

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

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

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

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

1)  \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)

2)  \(x^3-9x^2+6x+16\)

\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)

\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)

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

3)   \(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2-x-1\right)\)

4)  \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)

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

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

5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)

\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)

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

\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)

gửi phần này trước còn lại làm sau !!! tk mk nka !!!

nhiều thế

6) \(\left(x+y\right)^2-\left(x+y\right)-12\)\(=\left(x+y\right)^2-2\cdot\frac{1}{2}\left(x+y\right)+\frac{1}{4}-\frac{49}{4}\)

\(=\left(x+y-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)\(=\left(x+y-\frac{1}{2}-\frac{7}{2}\right)\left(x+y-\frac{1}{2}+\frac{7}{2}\right)\)

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

7)   \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)          (NHÂN x + 2 vs x +  5  và  x + 3 vs x + 4 )

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

ĐẶT   \(x^2+7x+11=y\)   ta được :  

\(\left(y+1\right)\left(y-1\right)-24=y^2-1-24\)

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

8)  \(4x^4-32x^2+1=4x^4+4x^2+1-36x^2\)

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

9) sai đề rùi bạn ơi ! đề đúng nè 

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

Ta thấy :  

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

Thay vào biểu thức bài cho ta được : 

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

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

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

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

bài ở trên câu 3 : kết luận là  \(\left(x-3\right)\left(x^2-x-6\right)\)bạn sửa lại giúp mk nka !!! Th@nk !!! Tk Mk vs