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\(a,=\left(3x+\dfrac{y}{2}\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{y^2}{4}\right)\\ b,=\left(5x+3y\right)\left(25x^2+15xy+9y^2\right)\)
\(a,=\left(5x-1\right)^2\\ b,=\left(x+4\right)^2\\ c,=\left(4x+3y\right)^2\\ d,=\left(\dfrac{x}{4}+2y\right)^2\)
\(a,=\left(3x-11\right)\left(3x+11\right)\\ b,=\left(3x+1-x+2\right)\left(3x+1+x-2\right)\\ =\left(2x+3\right)\left(4x-1\right)\\ c,=\left(2x+1-8\right)\left(2x+1+8\right)=\left(2x-7\right)\left(2x+9\right)\)
a) \(\left(a+b\right)^3-\left(a-b\right)^3=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)=6a^2b+b^3=b\left(6a^2+b^2\right)\)
b) \(\left(x+y\right)^3+\left(x-y\right)^3=\left(x^3+3x^2y+3xy^2+y^3\right)+\left(x^3-3x^2y+3xy^2-y^3\right)=2x^3+6xy^2=2x\left(x^2+3y^2\right)\)
a) \(=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2b\left(3a^2+b^2\right)\)
b) \(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2x\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)
\(=2x\left(x^2+3y^2\right)\)
a: \(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3a^2b+b^3\)
\(=6a^2b+2b^3\)
\(=2b\left(3a^2+b^2\right)\)
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
\(a,=\left(3x+2y\right)^3\\ b,=\left(4-x\right)^3\\ c,=\left(\dfrac{1}{2}x-3y\right)^3\)