( x+2)^5 : (2x-1)^4 

(x+2)(x+2)(x+2)(x+2)(x+2) : (2x-1)(2x-1)(2x-1)(2x-1)

( x+2)( 1+1+1+1+1) : (2x-1)(1+1+1+1)

 (x+2) . 5 : (2x-1) . 4

 (x+2) . 5 : 2(x+2) -5 . 4

đề ghi thế làm sao giải

\(a)\)

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

\(=[\left(x+y\right)+\left(z+t\right)][\left(x+y\right)-\left(z-t\right)]\)

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

\(=x^2+2xy+y^2-z^2-2zt-t^2\)

\(\left(x+2y+3z+t\right)^3\)

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

\(=\left(4xy+6xz+2xt+x^2+4y^2+12yz+9z^2+4yt+6zt+t^2\right)\left(x+2y+3z+t\right)\)

\(=8y^3+12xy^2+36y^2z+12y^2t+6x^2y+54yz^2+36xyz+6yt^2+12xyt+36yzt+x^3+27z^3+27xz^2+9x^2z+t^3+3xt^2+9zt^2+3x^2t+18xzt+27z^2t\)

\(b)\)

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

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

\(=-z^2+2tz+y^2-2xy+x^2-t^2\)

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

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

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

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

a)  -4x^2   + 9y^2

=  9y^2   -  4x^2  

= (3y)^2   -  (2x)^2  

= ( 3y  - 2x )(3y + 2x)

b) ( x + 1 )^3  - ( 2-x)^3

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

= (2x - 1) ( x^2  - x + 7)

c)  8 + ( 4x - 3 )^3

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

= ( 2 + 4x - 3)( 4 - 8x + 6 + 16x^2  - 24x + 9)

= ( 4x - 1)( 16x^2 - 32x + 19)

d)  81 - (9 - x^2 )^2

= 9^2 -  (9 - x^2)^2

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

=  x^2( 18 - x^2)

 học tốt, mk ko tự tin với kết quả này lắm

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

b) \(y^2-13=\left(y-\sqrt{13}\right)\left(y+\sqrt{13}\right)\)

c) \(2x^2-4=2\left(x^2-2\right)=2\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)

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

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

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

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

f) \(a^6-b^6=\left(a^2\right)^3-\left(b^2\right)^3=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(a^4+a^2b^2+b^4\right)\)

27x³ - 27x² + 3x - 1 

= (3x)³ - 3.(3x)².1 + 3x.1² - 1 

= (3x - 1)³ 

ui sai hđt kìa b =))

(x+1)(x+2)(x+3)(x+4)-24

= (x^2 + 5x + 4)(x^2 + 5x + 6) - 24

đặt x^2 + 5x + 5 = a

ta có : (a - 1)(a + 1) - 24 = a^2 - 1 - 24

= a^2 - 25

= (a - 5)(a+5)

= (x^2 + 5x + 5 - 5)(x^2 + 5x + 5 + 5)

= (x^2 + 5x)(x^2 + 5x + 10)

= x(x + 5)(x^2 + 5x + 10)

x(x+1)(x+2)(x+3)+1

= (x^2 + 3x)(x^2 + 3x + 2) + 1

đặt x^2 + 3x + 1 = a

ta có : (a - 1)(a+1) + 1 = a^2 - 1 + 1 = a^2

= (x^2 + 3x + 1)^2

$\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24$

$=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24$

$=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24$

Đặt $t=x^2+5x+4$ , ta có
$t\left(t+2\right)-24$

$=t^2+2t-24$

$=\left(t^2+2t+1\right)-25$

$={\left(t+1\right)}^2-5^2$

$=\left(t+1-5\right)\left(t+1+5\right)$

$=\left(t-4\right)\left(t+6\right)$

$=\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)$

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

$x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1$

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

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

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

Đặt $x^2+3x=t\Rightarrow x^2+3x+2=t+2$

Do đó $\left(1\right)=t\left(t+2\right)+1=t^2+2t+1={\left(t+1\right)}^2$(*)

Vì $t=x^2+3x$ nên

$\left(\text{*}\right)={\left(x^2+3x+1\right)}^2$

 (2x⁴-8x²+8) : (4-2x²)

=  2(x⁴-4x²+4) : 2(2-x²)

= (x⁴-4x²+4) : (2-x²)

= (x²  - 2) : (2-x²)

=   - 1

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

\(\left(4-2x^2\right)=2\left(2-x^2\right)\Rightarrow\frac{2x^4+8x^2+8}{4-2x^2}=\frac{2\left(x^2+2\right)^2}{2\left(2-x^2\right)}=\frac{\left(x^2+2\right)^2}{2-x^2}\)

Nếu không sai đề thì tự phân tích rồi thực hiện phép chia đa thức

 a^2-2ab+1+2b-2a-3b^2  =  a^2 -2a . b+1 +2b-2a-3b^2 = -(b+a-1)(3b-a+1)

nha bạn chúc bạn học tốt 

cảm ơn bạn ♡•꧁༺༒✰ŞŦΔŘ✰༒༺꧁•♡✿(➻❥𝒢𝑜𝓁𝒹❃𝒮𝓉𝒶𝓇✤) ❀ nha  

a, x^3 + y^3  = (x + y)(x^2 - xy + y^2) = (x + y)[(x+y)^2 - 3xy]

mà x + y = 1 và xy = -1

=> x^3 + y^3 = 1(1^2 - 3(-1)) = 4

b, x^3 - y^3 = (x - y)(x^2 + xy + y^2) = (x-y)[(x-y)^2+3xy]

có x - y = 1 và xy = 6

=> 1(1^2 + 3.6) = 19