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x^3-64=x^3-4^3=(x-4)^3=x^3-3x^2.4+3x.4^2-4^3=x^3-12x^2+48x-64
\(x^3-64x\)
\(=x\left(x^2-64\right)=x\left(x^2-8^2\right)\)
\(=x\left(x-8\right)\left(x+8\right)\)
a) = (x+1-x+1)(x2+2x+1+x2-1+x2-2x+1)- 6(x2-1)
= 2( 3x2+1)- 6(x2-1)
= 2( 3x2+1-3x2+3)
=2. 4
=8
\(=3x^2\left(x^2-1\right)+\left(x^8-3x^4+3x^2-1\right)-\left(x^8-1\right)\)
\(=3x^4-3x^2+x^8-3x^4+3x^2+1-x^8+1\)
\(=2\)
=2 nha ban
(con cach lam ban nhan dang thuc len rui rut gon lai)
\(\left(2x^m+7y^n\right)^2=4x^{2m}+28x^my^n+49y^{2n}\)
\(\left[\left(x-3\right)-z\right]^2=\left(x-3\right)^2-2\left(x-3\right)z+z^2=x^2-6x+9-2xz+6z+z^2\)
\(\left(4a^2-3b^2\right)\left(3b^2+4a^2\right)=\left(4a^2\right)^2-\left(3b^2\right)^2=16a^4-9b^4\)
Tham khảo nhé~
a) \(\left(m-\dfrac{1}{4}\right)^3=\left(m^3-3m^2.\dfrac{1}{4}+3m\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)^3\right)\\ =\left(m^3-\dfrac{3}{4}m^2+\dfrac{3}{16}m-\dfrac{1}{64}\right)\)
b)\(\left(\dfrac{2}{3}-n\right)^3=\left(\dfrac{2}{3}\right)^3-3\left(\dfrac{2}{3}\right)^2n+3.\dfrac{2}{3}n^2-n^3\\ =\dfrac{8}{27}-\dfrac{4}{3}n+2n^2-n^3\)
c)\(m^3-125=m^3-5^3=\left(m-5\right)\left(m^2+5m+25\right)\)
d)\(m^3+\dfrac{1}{64}=m^3+\left(\dfrac{1}{4}\right)^3=\left(m+\dfrac{1}{4}\right)\left(m^2-\dfrac{1}{4}m+\dfrac{1}{16}\right)\)