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

\(=x^3-3x^2+3x-1-x\left(x^2-6x+9\right)+1\)

\(=x^3-3x^2+3x-1-x^3+6x^2-9x+1=3x^2-6x\)

Rút gọn biểu thức: (x-1)³-x(x-3)²+1

=  x³ - 3x² + 3x - 1 - x ( x² - 6x + 9 ) + 1

 = x³ - 3x² + 3x - 1 - x³ + 6x² - 9x + 1

 =          3x² - 6x     

=          3x ( x - 2 )

1.  (x-1)(x-3)(x-5)(x-7)-20=0
<=> (x-1)(x-7)(x-3)(x-5)-20=0
<=> (x^2-8x+7)(x^2-8x+15)-20=0
Đặt x^2-8x+7=a => x^2-8x+15= a+8
=> a(a+8)-20=0
<=> a^2+8a-20=0
<=>(a^2+8a+16)-36=0
<=> (a+4)^2=36
=> {a+4=6a+4=−6{a+4=6a+4=−6
<=>{a=2a=−10{a=2a=−10
*a=2 => x^2-8x+7=2
<=> x^2-8x+5=0
<=>(x^2-8x+16)-11=0
<=>(x-4)^2=11
<=>x-4=√11
<=> x=√11 +4
*a=-10 => x^2-8x+7=-10
<=> x^2-8x+17=0
<=> (x^2-8x+16)+1=0
<=> (x-4)^2=-1 (PT vô nghiệm)
Vậy pt có nghiệm x=√11 +4

mk chỉ biết vậy thôi

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

Đặt \(x^2+x=t\)

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

Theo cách đặt \(\left(x^2+x-3\right)\left(x^2+x+1\right)\)

\(6y^2\left(x-1\right)+9y\left(x-1\right)\)

\(=\left(6y^2+9y\right)\left(x-1\right)=3y\left(2y+3\right)\left(x-1\right)\)

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

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