\(a,=x\left(x-2\right)^2\\ b,=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\\ c,=x^2\left(2x-1\right)-4\left(2x-1\right)=\left(x-2\right)\left(x+2\right)\left(2x-1\right)\\ d,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ e,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x\left[\left(x-2\right)^2-y^2\right]=x\left(x-y-2\right)\left(x+y-2\right)\\ g,=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\\ h,=x^3-x-2x+2=x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\\ =\left(x-1\right)\left(x^2+x-2\right)=\left(x-1\right)^2\left(x+2\right)\\ i,=3x^2+3x-10x-10=\left(x+1\right)\left(3x-10\right)\)

\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)

Dễ nhưng mà dài chết người 

Bài 1:

Ta có: \(x^2-2x+2=x^2-2x+1+1\)

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

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

Ta thấy rằng: \(\left(x-1\right)^2\ge0\) ( Với mọi \(x\in Z\) )

mà 1 > 0

=> \(\left(x-1\right)^2+1\ge0\)

<=> \(x^2-2x+1\ge0\)

Bài 3:

a) 53^2 + 47^2 + 94.53

= 53^2 + 47^2 + 2.47.53

= ( 53 + 47 )^2

= 100^2

= 10000

b) 50^2 - 49^2 + 48^2 - 47^2 + 2^2 - 1^2

= ( 50^2 - 49^2 ) + ( 48^2 - 47^2 ) + ( 2^2 - 1^2 )

= (50+49).(50-49) + (48+47).(48-47) + (2+1).(2-1)

= 50 + 49 + 48 + 47 + 2 + 1

= (49 + 1) + (48 + 2) + 50 + 47

= 50 + 50 + 50 + 47

= 197

Bài 1 : 

x²-2x+2>0 với mọi x

=x²-2.x.1/4+1/16+31/16

=(x-1/4)² + 31/16

Vì (x-1/4)² \(\ge\) 0 nên (x-1/4)² + 31/16 \(\ge\) 0 với mọi x (đfcm)

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

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

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

b, \(5x^3-5x^2y-10x^2+10xy\)

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

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

c, \(2x^2-5x=x\left(2x-5\right)\)

f, \(3x^2-7x-10=3x^2+3x^2-10x-10\)

\(=3x^2\left(x+1\right)-10\left(x+1\right)=\left(3x^2-10\right)\left(x+1\right)\)

d, \(x^3-3x^2+1-3x=x^3-3x^2-3x+1\)

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

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

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

e, \(3x^2-6xy+3y^2-12z^2\)

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

\(=3\left[\left(x-y\right)^2-4z^2\right]\)

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

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

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

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

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

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

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

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

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

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

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

b. \(5x^3-5x^2y-10x^2+10xy\)

\(=5\left[\left(x^3-x^2y\right)-\left(2x^2-2xy\right)\right]\)

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

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

c. \(2x^2-5x=x\left(2x-5\right)\)

d. \(x^3-3x^2+1-3x\)

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

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

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

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

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

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

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

e. \(3x^2-6xy+3y^2-12z^2\)

\(=3\left[x^2-2xy+y^2-4z^2\right]\)

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

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

f. \(3x^2-7x-10\)

\(=3x^2-7x-7-3\)

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

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

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

\(=\left(x+1\right)\left[3\left(x-1\right)-7\right]\)

\(=\left(x+1\right)\left(3x-8\right)\)

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

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

h. \(3x^2-3y^2-12x+12y\)

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

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

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

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

j. \(x^2-3x+2=x^2-x-2x+2\)

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

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

P/s: ( Có j sai ns nha nhiều số quá tui rối đầu )

mình cảm ơn 2 bạn rất nhìu

a, x^2-9+(x-3)^2 = (x-3)(x+3)+(x-3)^2=(x-3)(x+3+x-3)=2x(x-3)

b,có sai k ạ ! vì mình thấy tự nhiên có ẩn y ở đó , nếu đề bài 2 ẩn thì 1 trong 3 hạng tử chứa ẩn x kia phải có thêm 1 ẩn y

c,đề bài thiếu  ẩn ở hạng tử thứ nhất ạ !

b mình viết đúng rồi mà, c hạng tử 1 là x^3

câu c bạn dùng máy tính bấm nghiệm nhé , dùng mode =>5=> 4 nhé ! 

còn câu b mình chưa ra !

\(a.\left(8x^4-4x^3+x^2\right):2x^2=4x^2-2x+\frac{1}{2}\)

\(b.\left(2x^4-x^3+3x^2\right):\left(-\frac{1}{3x^2}\right)=-6x^6+3x^5-9x^4\)

\(c.\left(-18x^3y^5+12x^2y^2-6xy^3\right):6xy=-3x^2y^4+2xy-y^2\)

\(d.\left(\frac{3}{4x^3y^6}+\frac{6}{5x^4y^5}-\frac{9}{10x^5y}\right):-\frac{3}{5x^3y}=-\frac{5}{4y^5}-\frac{2}{xy^4}-\frac{3}{2x^2}\)