\(x^3-64=x^3-4^3\)

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

x³-64=x³-4³=(x-4)(x²+4x+16)

Ta có:\(x^3-64\)

\(=x^3-4^3\)

Áp dụng hằng đẳng thức:\(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)

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

( x-2 ) ^3 + 4^3

= hằng đẳng thức thứ 6 nha pạn

lam luon di bam dung cho

\(\left(x-2\right)^3+64\)

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

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

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

\(=\left(x+2\right)\left(x^2-8x+28\right)\)

Tham khảo nhé~

x⁸+64

=(x⁴)²+16x⁴+8²-16x⁴

=(x⁴+8)²-(4x²)²

=(x⁴+8-4x²)(x⁴+8+4x²)

Ta có

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)

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha

\(=6x^2+5x-3xy\)

\(=x\left(6x+5-3y\right)\)

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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