Theo mình là đề bài sai.Giả sử nếu n = 2 thì biểu thức = 1.6-(-2).3 = 12 không chia hết cho 5

Theo mình phải là CHIA HẾT CHO 6

Câu này khá dễ bạn ạ

(n-1)(n+4)-(n-4)(n+1)

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

=6n luôn chia hết cho 6 với n thuộc Z ^_^

Ukm. mik lỡ nhập đề bài sai sorry bạn nha!!!

cảm ơn bạn nhìu

7(x - 3) - x(3 - x)

= (x - 3)(7 + x)

chỉ bt có v mà k bt có đúng k 

1 ) 7 ( x - 3 ) - x ( 3 - x ) 

= 7 ( x - 3 ) + x ( x - 3 )

= ( x - 3 ) ( 7 + x )

2 ) 4x² - 6x + 3 - 2x

= 4x² - 2x - 6x + 3

= 2x ( 2x - 1 ) - 3 ( 2x - 1 )

= ( 2x - 1 ) ( 2x - 3 )

3 ) ( 4 - x ) - 4x + x²

= ( 4 - x ) - x ( 4 - x )

= ( 4 - x ) (  1 - x )

4 ) x² - 2xy + y²

= ( x - y )²

1, \(7\left(x-3\right)-x\left(3-x\right)=7\left(x-3\right)+x\left(x-3\right)\)

\(=\left(x-3\right)\left(x+7\right)\)

2,\(4x^2-6x+\left(3-2x\right)=4x^2-8x+3\)

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

\(=\left(2x-3\right)\left(2x-1\right)\)

3,\(\left(4-x\right)-4x+x^2=x^2-5x+4\)

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

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

\(B=2x^2-6x+7\)

\(=2\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{2}+7\)

\(=2\left(x-\frac{3}{2}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)

Vậy \(MinB=\frac{5}{2}\Leftrightarrow x=\frac{3}{2}\)

\(C=\left(2x-5\right)^2-4\left(2x-5\right)\)

\(=\left(2x-5\right)\left(2x-5-4\right)=2x-5\)

\(=[\left(2x-5\right)^2-4\left(2x-5\right)+4]-4\)

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

\(=\left(2x-7\right)^2-4\ge-4\)

Vậy \(MinC=-4\Leftrightarrow x=\frac{7}{2}\)

(2x-5)^2 -4(2x-5)=(2x-5)^2 -4(2x-5)+4-4=(2x-7)^2 -4>=-4 suy ra C đạt gtnn là -4

d) \(x^2-y^2-2x+2y\)

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

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

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

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

\(4xy^2-12x^2y+8xy\)

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

\(3x^2-6xy+3y^2-12z^2\)

\(=3.\left(x^2-2xy+y^2-4z^2\right)\)

\(=3.\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

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

\(x^4y^4+4=\left[\left(x^2y^2\right)^2+2..x^2y^2.2+2^2\right]-\left(2xy\right)^2\)

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

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

Mina giúp Shino đây nè:3(lần lượt nhá)

Ta có:\(4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\frac{1}{2}\)

1/ f(x) = 4x² - 4x + 1

4x² - 4x + 1 = 0

=> 4x² + 2x + 2x + 1 = 0

=> 2x(2x + 1) + (2x + 1) = 0

=> (2x + 1)(2x + 1) = 0

=> (2x + 1)² = 0

=> 2x + 1 = 0

=> 2x = -1

=> x = -1/2

Vậy nghiệm của đa thức f(x) là x = -1/2

\(9x^2+6x+1=0\)

\(\Leftrightarrow\left(3x\right)^2+2\cdot3x\cdot1+1^2=0\)

\(\Leftrightarrow\left(3x+1\right)^2=0\)

\(\Leftrightarrow x=-\frac{1}{3}\)

Áp dụng 2 hằng đẳng thức:

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

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

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)

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

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

=x^3+2x^2-x-2

[X-1/2] [X+1/2] [4X-1]

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

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

1/2X²Y² [2X+Y] [2X-Y]

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

=\(2x^2y^2-\frac{1}{2}x^2y^4\)