hoa mắt chóng mặt

sao nhiều thế bạn

quá nhiều

a,

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

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

= \(\left(\left(a-\sqrt{2b}\right)\left(a+\sqrt{2b}\right)\right)\left(a^2+2b\right)\)

c, 

=\(4x^4+20x^2+25\)

=\(\left(2x^2\right)^2+2.2x^2.5+5^2\)

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

d,

=\(8x^6-27y^3\)

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

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

Câu b đề ghi ko rõ lắm

a: =xy(x^2-4xy^2+4y^4)

=xy(x-2y^2)^2

b:=(x^3-y)^2

c: =(a^2-b^2)(a^2+b^2)

=(a^2+b^2)(a-b)(a+b)

d: 64x^6-27y^6

=(4x^2-3y^2)(16x^4+12x^2y^2+9y^4)

e: =(2x)^3+(3y)^3

=(2x+3y)(4x^2-6xy+9y^2)

\(9x^4-12x^2y^3+16y^6=\left(3x^2+4y^3\right)^2\)

\(4x^4-16x^2y^3+16y^6=\left(2x^2-4y^3\right)^2\)

\(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

\(a^3+\frac{1}{27}=\left(a+\frac{1}{3}\right)\left(a^2-\frac{1}{3}a+\frac{1}{9}\right)\)

Câu a và câu b viết sai đề hết rồi!

a: =(6x)^2-(3x-2)^2

=(6x-3x+2)(6x+3x-2)

=(9x-2)(3x+2)

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

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

=8x(x^2+1)

e: =(4x)^2-2*4x*3y+(3y)^2

=(4x-3y)^2

f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)

\(=-\left(\dfrac{1}{2}x^2-2y^3\right)^2\)

g: =(4x)^3+1^3

=(4x+1)(16x^2-4x+1)

k: =x^3(27x^3-8)

=x^3(3x-2)(9x^2+6x+4)

l: =(x^3-y^3)(x^3+y^3)

=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2)

$36x^2-(3x-2)^2$

$=[6x-(3x-2)][6x+(3x-2)]$

$=(6x-3x+2)(6x+3x-2)$

$=(3x+2)(9x-2)$

$16(4x+5)^5-25(2x+2)^2$

Câu này không có dạng hằng đẳng thức hiệu hai bình phương vì số mũ của hai phần là $5$ và $2$.

Có thể đặt nhân tử chung $1$ nhưng không phân tích tiếp được bằng các phương pháp thông thường.

$(x-y+4)^2$

Đây đã là một tích:

$(x-y+4)^2=(x-y+4)(x-y+4)$

$(x+1)^4-(x-1)^4$

$=[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]$

$=[x^2+2x+1-x^2+2x-1][x^2+2x+1+x^2-2x+1]$

$=4x(2x^2+2)$

$=8x(x^2+1)$

$16x^2-24xy+9y^2$

$=(4x)^2-2\cdot4x\cdot3y+(3y)^2$

$=(4x-3y)^2$

$-\dfrac{x^4}{4}+2x^2y^3-4y^6$

$=-\dfrac14(x^4-8x^2y^3+16y^6)$

$=-\dfrac14(x^2-4y^3)^2$

$64x^3+1$

$=(4x)^3+1^3$

$=(4x+1)(16x^2-4x+1)$

$x^3y^6z^9-125$

$=(xy^2z^3)^3-5^3$

$=(xy^2z^3-5)(x^2y^4z^6+5xy^2z^3+25)$

$27x^6-8x^3$

$=x^3(27x^3-8)$

$=x^3[(3x)^3-2^3]$

$=x^3(3x-2)(9x^2+6x+4)$

$x^6-y^6$

$=(x^3-y^3)(x^3+y^3)$

$=(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)$

$27x^3-54x^2y+36xy^2-8y^3$

$=(3x)^3-3(3x)^2(2y)+3(3x)(2y)^2-(2y)^3$

$=(3x-2y)^3$

$y^9-9x^2y^6+27x^4y^3-27x^6$

$=(y^3)^3-3(y^3)^2(3x^2)+3(y^3)(3x^2)^2-(3x^2)^3$

$=(y^3-3x^2)^3$

\(a.6x^3-9x^2=3x^2\left(2x-3\right)\\ b.20x^2y-12x^3=4x^2\left(5y-3x\right)\)

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

b) \(20x^2y-12x^3=4x^2\left(5y-3x\right)\)

1: Ta có: \(x^{10}-4x^8+4x^6\)

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

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

2: Ta có: \(m^3+27\)

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

3: Ta có: \(x^3+8\)

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

4: Ta có: \(\frac{1}{27}+a^3\)

\(=\left(\frac{1}{3}+a\right)\left(\frac{1}{9}-\frac{a}{3}+a^2\right)\)

5: Ta có: \(8x^3+27y^3\)

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

6: Ta có: \(\frac{1}{8}x^3+8y^3\)

\(=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)

7: Ta có: \(8x^6-27y^3\)

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

8: Ta có: \(\frac{1}{8}x^3-8\)

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

9: Ta có: \(\frac{1}{64}x^6-125y^3\)

\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{4}x^2y+25y^2\right)\)

10: Ta có: \(\left(a+b\right)^3-c^3\)

\(=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)\cdot c+c^2\right]\)

\(=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)

11: Ta có: \(x^3-\left(y-1\right)^3\)

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

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

12: Ta có: \(x^6+1\)

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

1) \(x^{10}-4x^8+4x^6\)

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

2) \(m^3+27=m^3+3^3=\left(m+3\right)\left(m^2-3m+3^2\right)\)

3) \(x^3+8=x^3+2^3=\left(x+2\right)\left(x^2-2x+2^2\right)\)

4) \(\frac{1}{27}+a^3=\left(\frac{1}{3}\right)^3+a^3=\left(\frac{1}{3}+a\right)\left[\left(\frac{1}{3}\right)^2-\frac{1}{3}a+a^2\right]\)

5) \(8x^3+27y^3=\left(2x\right)^3+\left(3y\right)^3=\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

6) \(\frac{1}{8}x^3+8y^3=\left(\frac{1}{2}x\right)^3+\left(2y\right)^3=\left(\frac{1}{2}x+2y\right)\left[\left(\frac{1}{2}x\right)^2-\frac{1}{2}x.2y+\left(2y\right)^2\right]=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)

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

10) \(\left(a+b\right)^3-c^3=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]=\left(a+b-c\right)\left[\left(a^2+2ab+b^2\right)+ac+bc+c^2\right]=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)11) \(x^3-\left(y-1\right)^3=\left(x-y+1\right)\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]=\left(x-y+1\right)\left[x^2+xy-x+\left(y^2-2y+1\right)\right]=\left(x-y+1\right)\left(x^2+xy-x+y^2-2y+1\right)\)

P/s: Đăng ít thôi chớ bạn!

a) Ta có: \(x-2y+x^2-4y^2\)

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

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

b) Ta có: \(x^2+2xy+y^2-4x^2y^2\)

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

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

c) Ta có: \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

d) Ta có: \(x^3+3x^2+3x+1-8y^3\)

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

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

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