1, (x-1)(x+2)(x+3)(x-6)+32x^2

= (x^2 - 7x + 6)(x^2 + 5x + 6) + 32x^2

đặt x^2 - x + 6 = a ta có

(a  - 6x)(a + 6x) + 32x^2

= a^2 - 36x^2 + 32x^2

= a^2 - 4x^2

= (a - 2x)(a + 2x)

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

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

2, (x+1)(x-4)(x+2)(x-8)+4x^2

= (x^2 + 7x - 8)(x^2 - 2x - 8) + 4x^2

đặt x^2 + 2,5x - 8 = a ta có

(a + 4,5x)(a - 4,5x)  + 4x^2 

= a^2 - 81/4x^2 + 4x^2

= a^2 - 65/4x^2

\(=\left(a-\sqrt{\frac{65}{4}}x\right)\left(a+\sqrt{\frac{65}{4}}x\right)=\left(x^2+\frac{5}{2}x-8+\sqrt{\frac{65}{4}}x\right)\left(x^2+\frac{5}{2}x-8-\sqrt{\frac{65}{4}x}\right)\) 

a, Ta có \(Q\left(x\right)=x+1=0\Leftrightarrow x=-1\)

Vậy P(x) chia hết cho Q(x) khi P(x) có nghiệm là -1 hay

\(3\left(-1\right)^3+2\left(-1\right)^2-5\left(-1\right)+m=0\Leftrightarrow m=-4\)

b.. ta có \(Q\left(x\right)=x^2-3x+2=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Vậy P(x) chia hết cho Q(x) khi P(x) có nghiệm là 1  và 2 hay

\(\hept{\begin{cases}2+a+b+3=0\\2.2^3+a.2^2+b.2+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=-5\\4a+2b=-19\end{cases}\Leftrightarrow}\hept{\begin{cases}a=-\frac{9}{2}\\b=-\frac{1}{2}\end{cases}}\)

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

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

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

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

\(a)\)

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

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

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

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

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

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

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

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

\(b)\)

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

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

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

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

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

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

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