a: \(=x\left[49-x^2\left(2x+1\right)^2\right]\)

\(=x\left[49-\left(2x^2+x\right)^2\right]\)

\(=x\left[\left(7-2x^2-x\right)\left(7+2x^2+x\right)\right]\)

b: \(=5\left[25x^2-\left(y^2-4y+4\right)\right]\)

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

c: \(=1-4x^2-x\left(x^2-4\right)\)

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

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

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

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

e: =(x-9)(x+6)

4x(x+y)(x+y+z)(x+z)+(yz)^2

=(2x(x+y+z))(2(x+y)(x+z)+(yz)^2

=(2x^2+2xy+2xz)(2x^2+2xy+2xz+2yz)+(yz)^2

Đặt t=C

=(t-yz)(t+yz)-(yz)^2

=t^2-(yz)^2+(yz)^2=t^2=(2x^2+2xy+2xz+yz)^2

=

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Em hâm mộ JungKook

a) x²-x-y²-y=(x²-y²)-(x+y)=(x+y)(x-y)-(x+y)=(x+y)(x-y-1)

b)x²-2xy+y²-z²=(x-y)²-z²=(x-y-z)(x-y+z)

c)5x-5y+ax-ay=5(x-y)+a(x-y)=(x-y)(5+a)

d)a³-a²x-ay+xy=a²(a-x)-y(a-x)=(a-x)(a²-y)

e)4x²-y²+4x+1=[(2x)²+2.2x.1+1²]-y²=(2x+1)²-y²=(2x+1-y)(2x+1+y)

Chúc bạn học tốt!

Bạn học căn thức chưa ?

phan tich cac da thuc sau thanh nhan tu theo mau:

a)\(2x^3-x\)

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

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

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

b)\(5x^2\left(x-1\right)-15x\left(x-1\right)\)

\(=\left(5x^2-15x\right)\left(x-1\right)\)

\(=5x\left(x-3\right)\left(x-1\right)\)

d)\(3x\left(x-2y\right)+6y\left(2y-x\right)\)

\(=3x\left(x-2y\right)-6y\left(x-2y\right)\)

\(=\left(3x-6y\right)\left(x-2y\right)\)

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

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

Ta có (x^2 + y^2 )^3 + (z^2 – x^2 )^3 – (y^2 + z^2 )^3

= (x^2 + y^2 )^3 + (z^2 – x^2 )^3 + (-y^2 - z^2 )^3

Ta thấy x^2 + y^2 + z^2 – x^2 – y^2 – z^2 = 0

=> áp dụng nhận xét ta có: (x^2+y^2 )^3+ (z^2 -x^2 )^3 -y^2 -z^2 )^3

= 3(x^2 + y^2 ) (z^2 –x^2 ) (-y^2 – z^2 )

= 3(x^2+y^2 ) (x+z)(x-z)(y^2+z^2 )

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