Ta có:

\(\left(x^2-3\right)^2+16\)

\(=x^4-6x^2+9+16\)

\(=x^4-6x^2+25\)

\(=\left(x^4+10x^2+25\right)-16x^2\)

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

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

Mơn bạn nha <3

\(=x^4-6x^2+9+16\) 

\(=x^4-6x^2+25\) 

\(=\left(x^4+10x^2+25\right)-16x^2\) 

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

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

a: \(=4xy\left(1-5x^2y\right)\)

b: \(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)

c: \(=x\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(x+y\right)\)

d: \(=\left(x+2y\right)^2-36=\left(x+2y+6\right)\left(x+2y-6\right)\)

b: \(\left(x^2+4\right)^2-16x^2\)

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

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

c: \(x^5-x^4+x^3-x^2\)

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

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

Lời giải:

a. Bạn xem lại đề

b. \((x^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)

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

c.

\(x^5-x^4+x^3-x^2=x^4(x-1)+x^2(x-1)=(x^4+x^2)(x-1)\)

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

a) 7x.(-y)+2(y-x)2

=>-7xy+4y-4x

b)(x^2+4)-16x^2

=>x^2+4-16x^2

=>-15x^2+4

c)x^5-x^4+x^3-x^2

=>x^4(x-1)+x^2(x-1)

=>(x^4+x^2)(x-1)

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

\(x^2+8x+16=\left(x+4\right)^2\)

\(x^2+2x+1-16=\left(x+1\right)^2-4^2=\left(x+1-4\right).\left(x+1+4\right)=\left(x-3\right).\left(x+5\right)\)

\(x^2+2x+1-16=\left(x^2+2x+1\right)-4^2=\left(x+1\right)^2-4^2=\left(x+1-4\right)\left(x+1+4\right)=\left(x-3\right)\left(x+5\right)\)

\(x^2+2x+1-16=\left(x+5\right)\left(x-3\right)\)

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

2xy – x2 – y2 + 16 (Có 2xy ; x2 ; y2, ta liên tưởng đến HĐT (1) hoặc (2))

= 16 – (x2 – 2xy + y2)

= 42 – (x – y)2 (xuất hiện hằng đẳng thức (3))

= [4 – (x – y)][4 + (x - y)]

= (4 – x + y)(4 + x – y).

\(\left(x^2+4y^2-20\right)^2-16\left(xy-4\right)^2=\left(x^2+4y^2-20\right)^2-\left(4xy-16\right)^2=\left(x^2+4y^2-20-4xy+16\right)\left(x^2+4y^2-20+4xy-16\right)=\left[\left(x-2y\right)^2-4\right]\left[\left(x+2y\right)^2-36\right]=\left(x-2y-2\right)\left(x-2y+2\right)\left(x+2y-6\right)\left(x+2y+6\right)\)

\(\left(x^2+4y^2-20\right)^2-\left(4xy-16\right)^2\)

\(=\left(x^2+4y^2-20-4xy+16\right)\left(x^2+4y^2-20+4xy-16\right)\)

\(=\left[\left(x-2y\right)^2-4\right]\left[\left(x+2y\right)^2-36\right]\)

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