Bài 7: Phân tích đa thức thành nhân tử bằng phương pháp dùng hằng đẳng thức

a) A = x² - 2x + 1 - y² + 2x - 1 

       = (x² - 2x + 1)-( y²-2x+1) 

       = (x-1)²-(y-1)²

       = (x-1-y+1)(x-1+y-1)
b) A = x² - 4x + 4 - y² - 6y - 9

        = (x² - 4x + 4)-(y²+6y+9)

        = (x-2)²-(y+3)²

        = (x-2-y-3)(x-2+y+3)
c) A = 4x² - 4x + 1 - y² - 8y - 16

       = (4x² - 4x + 1) - (y²+8y+16)

       = (2x-1)²-(y+4)²

       = (2x-1-y-4)(2x-1+y+4)

d) A = x² - 2xy + y² - z² + 2zt - t²

       =(x² - 2xy + y²)-(z²- 2zt + t²)

      = (x-y)²-(z-t)²

       =(x-y-z+t)(z-y+z-t)

câu d mik có sửa lại đề vì mik thấy đề hơi sai

a) A =

= x² - y² + 2x - 2x + 1 - 1

= x² - y²  = (x-y) (x+y)

b) A=

= (x-2)² - (y+3)² = (x-y-5) (x+y+1)

c) A=

= (2x-1)² - (y+4)²

= (2x+y+3) (2x-y-5)

d) đề có thể sai

a)\(A=3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)b) \(A=\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)

c) \(A=x^2+y^2+2xy+yz+zx=\left(x+y\right)^2+z\left(x+y\right)=\left(x+y\right)\left(x+y+z\right)\)

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

\(4x^2-4x+1-\left(y^2+8y+16\right)=\left(2x-1\right)^2-\left(y+4\right)^2\)

\(=\left(2x-1-y-4\right)\left(2x-1+y+4\right)=\left(2x-y-5\right)\left(2x+y+3\right)\)

Bài a thì bạn thử xem lại đề nhé.

(x+y)^2 - 5y^2

x^2- 4y^2 + 4xy

= x^2 + 4xy - 4y^2

=x^2 + 2x2y - (2y)^2

= ( x - 2y )^2

 x² - 4y² + 4xy = x² + 4xy - 4y² = - (x² - 4xy + 4y²) = - ((x² - 2x.2y + (2y)²)

= - ( x - 2y)²

\(a-\sqrt{a}-2=\left(\sqrt{a}+1\right)\left(\sqrt{a}-2\right)\)

\(a-\sqrt{a}-2=\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)\)

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

e) \(a^3b-ab^3\\ =ab\left(a^2-b^2\right)\\ =ab\left(a-b\right)\left(a+b\right)\)

f) \(a^2+2a+1-b^2\\ =\left(a^2+2a+1\right)-b^2\\ =\left(a+1\right)^2-b^2\\ =\left(a+1-b\right)\left(a+1+b\right)\)

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

e: \(a^3b-ab^3=ab\left(a-b\right)\left(a+b\right)\)

f: \(a^2+2a+1-b^2=\left(a+1-b\right)\left(a+1+b\right)\)

\(n\left(3n+1\right)^2-16n\\ =n\left(9n^2+6n+1\right)-16n\\ =9n^3+6n^2+n-16n\\ =9n^3+6n^2-15n\\ =9n^3-9n^2+15n^2-15n\\ =\left(9n^3-9n^2\right)+\left(15n^2-15n\right)\\ =9n^2\left(n-1\right)+15n\left(n-1\right)\)

\(=\left(9n^2+15n\right)\left(n-1\right)\\ =3n\left(n+5\right)\left(n-1\right)⋮3\)

đến dây mik chx bt lm, bn tự chứng minh chia hết cho 2 nhé

Đặt \(N=n\left(3n+1\right)^2-16n\)

\(=n\left(9n^2+6n+1\right)-4n-12n\)

\(=n\left(9n^2+6n-3\right)-12n\)

\(=3n\left(n+1\right)\left(3n-1\right)-12n\)

Ta có: n và n+1 là 2 số nguyên liên tiếp nên tích luôn chia hết cho 2

\(\Rightarrow3n\left(n+1\right)\left(3n-1\right)⋮6\)

Hiển nhiên \(12n⋮6\)

\(\Rightarrow3n\left(n+1\right)\left(3n-1\right)-12n⋮6\)

Hay N chia hết cho 6 với mọi n nguyên

Ta có: \(n\left(3n+1\right)^2-16n\)

\(=n\left[\left(3n+1\right)^2-16\right]\)

\(=n\left(3n+1-4\right)\left(3n+1+4\right)\)

\(=n\left(3n-3\right)\left(3n+5\right)\)

\(=3n\left(n-1\right)\left(3n+5\right)⋮6\)

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

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

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

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

Ta có: \(A=x^3+y^3+z^3-3xyz\)

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

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

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

\(4abcd+\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(a^2c^2+b^2d^2+2abcd\right)+\left(a^2d^2+b^2c^2+2abcd\right)\)

\(=\left(ac+bd\right)^2+\left(ad+bc\right)^2\)

\(2cd\left(a^2+b^2\right)+2ab\left(c^2+d^2\right)=\left(2a^2cd+2abc^2\right)+\left(2b^2cd+2abd^2\right)\)

\(=2ac\left(ad+bc\right)+2bd\left(bc+ad\right)=2\left(ac+bd\right)\left(bc+ad\right)\)

Do đó:

\(M=\left[\left(ac+bd\right)^2+\left(ad+bc\right)^2-2\left(ac+bd\right)\left(ad+bc\right)\right]\left[\left(ac+bd\right)^2+\left(ad+bc\right)^2+2\left(ac+bd\right)\left(ad+bc\right)\right]\)

\(=\left(ac+bd-ad-bc\right)^2\left(ac+bd+ad+bc\right)^2\)

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