Bài 4:

a: \(216x^3+27y^3=27\left(8x^3+y^3\right)\)

\(=27\left[\left(2x\right)^3+y^3\right]\)

\(=27\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

b: \(64a^3-8=8\left(8a^3-1\right)\)

\(=8\left[\left(2a\right)^3-1^3\right]\)

\(=8\left(2a-1\right)\left(4a^2+2a+1\right)\)

c: \(x^3+8=x^3+2^3=\left(x+2\right)\left(x^2-2x+4\right)\)

d: \(27x^3-8y^3=\left(3x\right)^3-\left(2y\right)^3\)

\(=\left(3x-2y\right)\left[\left(3x\right)^2+3x\cdot2y+\left(2y\right)^2\right]\)

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

Bài 5:

a: \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)

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

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

\(=2y^2-10xy\)

b: \(\left(x-y\right)^3-3\left(x-y\right)^2\cdot x+3\left(x-y\right)\cdot x^2-x^3\)

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

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

c: \(\left(3x+3\right)^3-2\left(x+1\right)^3-\left(5x-1\right)^2\)

\(=27\left(x+1\right)^3-2\left(x+1\right)^3-\left(5x-1\right)^2\)

\(=25\left(x+1\right)^3-25x^2+10x-1\)

\(=25x^3+75x^2+75x+25-25x^2+10x-1\)

\(=25x^3+50x^2+85x+24\)

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

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

\(=\left(-3x+2\right)\left(4x^2-12x+9-2x^2+x+3+x^2+2x+1\right)+\left(3x-1\right)^2\)

\(=\left(-3x+2\right)\left(3x^2-9x+13\right)+\left(3x-1\right)^2\)

\(=-9x^3+27x^2-39x+6x^2-18x+26+9x^2-6x+1\)

\(=-9x^3+42x^2-63x+27\)

Đề yêu cầu cái gì thế? Em ơi!

????????

a: Ta có: \(\widehat{OMN}=\widehat{MQP}\)(hai góc đồng vị, MN//PQ)

\(\widehat{ONM}=\widehat{NPQ}\)(hai góc đồng vị, MN//PQ)

mà \(\widehat{MQP}=\widehat{NPQ}\)(MNPQ là hình thang cân)

nên \(\widehat{OMN}=\widehat{ONM}\)

=>ΔOMN cân tại O

b: Xét ΔMNQ và ΔNMP có

NM chung

NQ=MP

MQ=NP

Do đó: ΔMNQ=ΔNMP

c: H ở đâu vậy bạn?

\(x^2-10x-11=0\)

=>\(x^2-10x+25-36=0\)

=>\(\left(x-5\right)^2-6^2=0\)

=>(x-5-6)(x-5+6)=0

=>(x-11)(x+1)=0

=>\(\left[{}\begin{matrix}x-11=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-1\end{matrix}\right.\)

\(x^2\)\(-2.x.5+5^2\)\(-36\)\(=0\)

\(\Leftrightarrow\)\(\left(x-5\right)^2\)\(-36=0\)

\(\Leftrightarrow\left(x-5^{ }\right)^2\)\(=36\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-1\end{matrix}\right.\)

a: \(\left(3x+4y\right)^2+\left(4x-3y\right)^2\)

\(=9x^2+24xy+16y^2+16x^2-24xy+9y^2\)

\(=25x^2+25y^2\)

b: \(\left(x^2+6x+9\right)-\left(25x^2-40x+16\right)\)

\(=x^2+6x+9-25x^2+40x-16\)

\(=-24x^2+46x-7\)

a, ( 3x +4y)^2 + ( 4x-3y)^2

= ( 3x + 4y )^2 - ( 3y - 4x )^2 ( hằng đẳng thức số 2)

b, (x^2 +6x+9)-(25x^2-40x+16)

= (x^2 +3x +3x +9) - (25x^2 - 20x - 20x +16)

= [(x^2 + 3x) + (3x + 9 )] - [(25x^2 -20x)+(-20x+16)]

= [x(x+3)+3(x+3)] - [5x(5x-4)-4(5x-4)]

= (x+3)(x+3) - (5x-4)(5x-4)

= (x+3)^2 - (5x-4)^2 ( hằng đẳng thức số 2)

f(2)=0

=>\(a\cdot2^2+b\cdot2+c=0\)

=>4a+2b+c=0

=>c=-4a-2b

=>\(f\left(x\right)=ax^2+bx-4a-2b\)

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

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

\(=\left(x-2\right)\left(ax+2a+b\right)⋮x-2\)

`a, x^8 - 1`

`=(x^4)^2 - 1^2`

`= (x^4 - 1)(x^4 + 1)`

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

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

`b, x^10 - 1`

`= (x^5)^2-1^2`

`=(x^5-1)(x^5+1)`

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

g: n là số lẻ nên n=2k+1

Vì 5 là số nguyên tố nên \(n^5-n⋮5\)

\(n^5-n=n\left(n^4-1\right)=n\left(n^2-1\right)\left(n^2+1\right)\)

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

Vì n;n-1;n+1 là ba số nguyên liên tiếp

nên \(n\left(n-1\right)\left(n+1\right)⋮3!=6\)

=>\(n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮6\)

=>\(n^5-n⋮6\)

mà \(n^5-n⋮5;ƯCLN\left(5;6\right)=1\)

nên \(n^5-n⋮\left(5\cdot6\right)=30\)

\(n^5-n=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)\)

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

\(=\left(2k+1\right)\cdot2k\cdot\left(2k+2\right)\left(4k^2+4k+2\right)\)

\(=8k\left(k+1\right)\left(2k^2+2k+1\right)\left(2k+1\right)\)

Vì k;k+1 là hai số nguyên liên tiếp

nên k(k+1) chia hết cho 2

=>\(8k\left(k+1\right)⋮16\)

=>\(n^5-n⋮16\)

mà \(n^5-n⋮30\)

nên \(n^5-n⋮BCNN\left(30;16\right)\)

=>\(n^5-n⋮240\)

f: Tích của 5 số nguyên liên tiếp sẽ chia hết cho 5!

mà \(5!=1\cdot2\cdot3\cdot4\cdot5=120\)

nên tích của 5 số nguyên liên tiếp sẽ chia hết cho 120

e: \(n^3+3n^2+2n=n\left(n^2+3n+2\right)=n\left(n+1\right)\left(n+2\right)\)

Vì n;n+1;n+2 là ba số nguyên liên tiếp

nên \(n\left(n+1\right)\left(n+2\right)⋮3!=6\)

=>\(n^3+3n^2+2n⋮6\)