Bài 8: Phân tích đa thức thành nhân tử bằng phương pháp nhóm các hạng tử

\(5x^2y^4-10x^4y^2+5x^2y^2=5x^2y^2\cdot\left(y^2-2x^2+1\right)\)

5x²y⁴ - 10x⁴y² + 5x²y²
= 5x²y². y² - 5x²y² . x² + 5x²y².1
=5x²y².(y²-x²+1)

\(1,3ax^2+3bx^2+ax+bx+5a+5b\)

\(=3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)=\left(a+b\right)\left(3x^2+x+5\right)\)

\(2,ax-bx-2cx-2a+2b+4c=x\left(a-b-2c\right)-2\left(a-b-2c\right)=\left(x-2\right)\left(a-b-2c\right)\)

\(a,3x^3-x^2-21x+7\)

\(=x^2\left(3x-1\right)-7\left(3x-1\right)\)

\(=\left(3x-1\right)\left(x^2-7\right)\)

\(b,x^3-4x^2+8x-8\)

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

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

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

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

\(c,x^3-5x^2-5x+1\)

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

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

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

\(\left(a+b\right)^3+\left(b+c\right)^3+\left(a+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(=2a^3-6abc+2b^3+2c^3\)

Khỏi ghi lại đề:

\(=a^3+3a^2b+3ab^2+b^3+b^3+3b^2c+3bc^2+c^3+c^3+3c^2a+3a^2c+a^3-3.\left(2abc+a^2b+ac^2+a^2c+b^2c+ab^2+bc^2\right)\)

\(=2a^3+2b^3+2c^3-6abc\)

ta có : (5n + 2)² - 4 = ((5n)² + 2.2.5n + 2²) - 4 = (5n)² + 20n + 4 - 4
= 25n² + 20n = 5n(5n + 4)

\(\Rightarrow\) (5n + 2)² - 4 = 5n(5n + 4)\(⋮\)5 \(\Rightarrow\) (5n + 2)² - 4 chia hết cho 5 với mọi n thuộc Z (đpcm)

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

\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+3^2\right)+x\left(2+x\right)\left(2-x\right)-1=0\)\(\Leftrightarrow x^3-27+x\left(4-x^2\right)-1=0\)

\(\Leftrightarrow x^3-27+4x-x^3=0\)

\(\Leftrightarrow4x=27\Rightarrow x=\dfrac{27}{4}\)

a , \(8x^3-27=\left(2x\right)^3-3^3=\left(2x-3\right)\left(4x^2+6x+9\right)\)

b , \(-x^4y^2-16-8x^2y=-\left[\left(x^2y\right)^2+4.x^2y+4^2\right]=-\left[x^2y+4\right]^2\)

c , \(2xy-x^2-y^2+16=-\left[\left(x^2-2xy+y^2\right)-16\right]=-\left[\left(x-y\right)^2-4^2\right]=-\left[\left(x-y-4\right)\left(x-y+4\right)\right]\)

\(a,8x^3-27=\left(2x\right)^3-3^3=\left(2x-3\right)\left(4x^2+6x+9\right)\)\(b,-x^4y^2-16-8x^2y=-\left(x^4y^2+8x^2y+16\right)=-\left(x^2y+4\right)^2\)\(c,2xy-x^2-y^2+16=16-\left(x^2-2xy+y^2\right)=4^2-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)

a) \(8x^3-27=\left(2x-3\right)\left(4x^2+6x+9\right)\)

b) \(-x^4y^2-16-8x^2y\)

\(=-\left(x^4y^2+8x^2y+16\right)\)

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

c) \(2xy-x^2-y^2+16\)

\(=-\left(x^2-2xy+y^2-16\right)\)

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

\(n^3-n⋮3\)

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

Ta có: \(n\left(n+1\right)\left(n-1\right)\) là tích ba số nguyên liên tiếp

\(\Rightarrow n\left(n+1\right)\left(n-1\right)⋮3\)

\(\Rightarrow n^3-n⋮3\)

Ta có: \(n^3-n=n\left(n^2-1\right)=n\left(n+1\right)\left(n-1\right)\)

Ta có tích của ba số nguyên liên tiếp luôn chia hết cho 3. Vậy suy ra:

\(\left(n^3-n\right)\)chia hết cho 3 (đpcm)

a) x^2+ 2x + 1

= x^2+2.x.1+1^2

=(x+1)^2
b) x^2 + 6x + 9

= x^2+2.x.3+3^2

=(x+3)^2
c) x^2 - 6x +9

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

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

=x^2-2.x.1+1^2

=(x-1)^2
e) 4x^2 + 4xy + y^2

=(2x)^2+2.2x.y+y^2

=(2x+y)^2

mọi người ơi bằng phương pháp dùng hằng thức đáng nhớ nha

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

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

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

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

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