Bài 1: 

e: Ta có: \(x\left(y-x\right)^2-x^2+2xy-y^2\)

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

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

Bài 2: 

a: Ta có: \(M=m^2\left(m+n\right)-n^2m-n^3\)

\(=m^2\left(m+n\right)-n^2\left(m+n\right)\)

\(=\left(m+n\right)^2\cdot\left(m-n\right)\)

\(=\left(-2017+2017\right)^2\cdot\left(-2017-2017\right)\)

=0

\(\left(4x^2-7x-50\right)^2-16x^4-56x^3-49x^2\)

\(\text{Phân tích thành nhân tử}\)

\(\left(-4\right)\left(2x-5\right)\left(7x+25\right)\)

\(x^m+3.y-x^m+1.Y^3-x^3.y^m+1+xy^m+3\)

\(\text{Phân tích thành nhân tử}\)

\(-\left(x^3y^m-xy^m-y^3-3y-4\right)\)

Câu 3 ko hiểu >o<

hài bài khó quá mình cũng học lớp 8 nhưng kho lắm

2

\(M=2y-3x\sqrt{y}+x^2=y-2x\sqrt{y}+x^2+y-x\sqrt{y}\\ =\left(\sqrt{y}-x\right)^2+\sqrt{y}\left(\sqrt{y}-x\right)\\ =\left(\sqrt{y}-x\right)\left(\sqrt{y}-x+\sqrt{y}\right)\\ =\left(\sqrt{y}-x\right)\left(2\sqrt{y}-x\right)\)

b

\(y=\dfrac{18}{4+\sqrt{7}}=\dfrac{18\left(4-\sqrt{7}\right)}{16-7}=\dfrac{72-18\sqrt{7}}{9}=\dfrac{72}{9}-\dfrac{18\sqrt{7}}{9}=8-2\sqrt{7}\\ =7-2\sqrt{7}.1+1=\left(\sqrt{7}-1\right)^2\)

Thế x = 2 và y = \(\left(\sqrt{7}-1\right)^2\) vào M được:

\(M=2\left(\sqrt{7}-1\right)^2-3.2.\sqrt{\left(\sqrt{7}-1\right)^2}+2^2\\ =2\left(8-2\sqrt{7}\right)-6.\left(\sqrt{7}-1\right)+4\\ =16-4\sqrt{7}-6\sqrt{7}+6+4\\ =26-10\sqrt{7}\)

1:

a: =>2x-2căn x+3căn x-3-5=2x-4

=>căn x-8=-4

=>căn x=4

=>x=16

b: \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)-3\sqrt{x}\left(\sqrt{x}-2\right)=0\)

=>(căn x-2)(x-căn x+4)=0

=>căn x-2=0

=>x=4

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

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

2) \(4x(x-2y)-8y(2y-x)\)

\(=4x\left(x-2y\right)+8y\left(x-2y\right)\)

\(=\left(4x+8y\right)\left(x-2y\right)\)

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

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

\(=a^2\left(x-1\right)-b^2\left(x-1\right)\)

\(=\left(a^2-b^2\right)\left(x-1\right)\)

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

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

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

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

5) \(5x\left(x-y\right)^2+10y^2\left(y-x\right)^2\)

\(=5x\left(x-y\right)^2+10y^2\left(x-y\right)^2\)

\(=\left(5x+10y^2\right)\left(x-y\right)^2\)

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

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

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

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

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

7) \(x\left(m-a\right)^2-y\left(a-m\right)^2\)

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

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

8) \(6y^2\left(x-1\right)^2+9y\left(1-x\right)^2\)

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

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

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

#Ayumu

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

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

\(b,\left(4x-y\right)\left(a-1\right)-\left(y-4x\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)

​\(=\left(4x-y\right)\left(a-1\right)+\left(4x-y\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)​

\(=\left(4x-y\right)\left(a-1+b-1+1-c\right)\)

\(=\left(4x-y\right)\left(a+b-c-1\right)\)

\(c,x^k+1-x^k-1\)

\(=0?!?!\)

\(d,x^m+3-x^m+1\)

\(=4\)

\(e,3\left(x-y\right)^3-2\left(x-y\right)^2\)

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

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

\(f,81a^2+18a+1\)

\(=\left(9a\right)^2+2.9a+1\)

\(=\left(9a+1\right)^2\)

\(g,25a^2.b^2-16c^2\)

\(=\left(5ab\right)^2-\left(4c\right)^2\)

\(=\left(5ab+4c\right)\left(5ab-4c\right)\)

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

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

\(i,\left(ax+by\right)^2-\left(ax-by\right)^2\)

\(=\left(ax+by-ax+by\right)\left(ax+by+ax-by\right)\)

\(=2by.2ax\)

\(=4axby\)

\(\text{a) }\left(2a+3\right)x-\left(2a+3\right)y+\left(2a+3\right)\)

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

\(\text{b) }\left(4x-y\right)\left(a-1\right)-\left(y-4x\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)

\(=\left(4x-y\right)\left(a-1\right)+\left(4x-y\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)

\(=\left(4x-y\right)\left(a-1+b-1+1-c\right)\)

\(=\left(4x-y\right)\left(a+b-c-1\right)\)

\(\text{c) }x^k+1-x^k-1\)

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

\(=0\)

\(\text{d) }x^m+3-x^m+1\)

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

\(=4\)

\(\text{e) }3\left(x-y\right)^3-2\left(x-y\right)^2\)

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

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

\(\text{f) }81a^2+18a+1\)

\(=\left(9a\right)^2+2.9a.1+1^2\)

\(=\left(9a+1\right)^2\)

\(\text{g) }25a^2b^2-16c^2\)

\(=\left(5ab\right)^2-\left(4c\right)^2\)

\(=\left(5ab+4c\right)\left(5ab-4c\right)\)

\(\text{h) }\left(a-b\right)^2-2.\left(a-b\right).c+c^2\)

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

\(\text{i) }\left(ax+by\right)^2-\left(ax-by\right)^2\)

\(=\left(ax+by+ax-by\right)\left(ax+by-ax+by\right)\)

\(=2by.2ax\)

\(=4byax\)

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

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

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

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

Gọi f_{( x )} = x³ - x² - 11x + m

      g_{( x )}  = x - 3

Cho g_{( x )} = 0

\(\Rightarrow\)x - 3 = 0

\(\Rightarrow\)x      = 3

\(\Rightarrow\)f_{( 3 )} = 3³ - 3² - 11.3 + m

\(\Rightarrow\)f_{( 3 )} = - 15 + m

Để f_{( x )} \(⋮\)g_{( x )}

\(\Leftrightarrow\)- 15 + m = 0

\(\Rightarrow\)m              = - 15 

Vậy : m = - 15 thì M = x³ - x² - 11x + m \(⋮\)x - 3

c) \(\left(y^2+y+1\right)\left(y^2+y+2\right)-12\)(1)

Đặt \(y^2+y+1=t\)

\(\Rightarrow\left(1\right)=t\left(t+1\right)-12=t^2+t-12\)

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

Mà\(y^2+y+1=t\)(ẩn phụ) nên 

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

B1:

[(m+n)+(2m-3n)]^2

= (m+n)^2 + 2(m+n)(2m-3n) + (2m-3n)^2

= m^2 +2mn +n^2 + 4m^2 - 6mn + 4mn - 6n^2 + 4m^2 - 12mn + 9n^2

= 9m^2 - 12mn + 4n^2
 

B2,3

 bn lm theo hdt ( a +b + c) ^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc nha

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

b)(x+y+z)^2 đã thành nhân tử 

c)(2x-y+3z)^2 đã thành nhân tử, too

d)(x+y)(x-y)(2x+y).........

e)\(8x^6-y^6=\left(2x^2-y^2\right)\left(4x^4+2x^2y^2+y^4\right)\)

g)\(\left(m+n\right)^3-\left(2m+3n\right)^3\)

\(=-\left(m+2n\right)\left(7m^2+19mn+13n^2\right)\)