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

a) M = 0.                      b) N = 1820.

Bài 4: 

Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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

Bài 1: Ta có: \(A=1+4y-y^2=5-\left(y^2-4y+4\right)=5-\left(y-2\right)^2\le5\)

Dấu "=" xảy ra khi \(\left(y-2\right)^2=0\Rightarrow y-2=0\Rightarrow y=2\)

Vậy \(maxA=5\) khi \(y=2\)

Bài 2: Ta có: \(a^3+b^3+3ab=\left(a^3+3a^2b+3ab^2+b^3\right)-3a^2b-3ab^2+3ab\)

\(=\left(a+b\right)^3-3ab\left(a+b-1\right)=1-0=1\)

cảm ơn nhìu

a) \(3x^2-3y^2-12x+12y\)

\(=\left(3x^2-3y^2\right)-\left(12x-12y\right)\)

\(=3\left(x^2-y^2\right)-12\left(x-y\right)\)

\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)

\(=\left(x-y\right)\left(3x-3y-12\right)\)

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

b) \(4x^3+4xy^2+8x^2y-16x\)

\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)

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

c)    \(x^4-5x^2+4\)

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

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

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

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

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