Bài 1:
a: \(\left(3x-\dfrac{1}{2}y\right)^2=\left(3x\right)^2-2\cdot3x\cdot\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\)
\(=9x^2-3xy+\dfrac{1}{4}y^2\)
b: \(\left(3x+2y\right)^2=\left(3x\right)^2+2\cdot3x\cdot2y+\left(2y\right)^2\)
\(=9x^2+12xy+4y^2\)
c: Sửa đề: \(8x^3+27y^3\)
\(=\left(2x\right)^3+\left(3y\right)^3\)
\(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
Bài 2:
a: \(4x^2+12x+9=0\)
=>\(\left(2x\right)^2+2\cdot2x\cdot3+3^2=0\)
=>\(\left(2x+3\right)^2=0\)
=>2x+3=0
=>2x=-3
=>\(x=-\dfrac{3}{2}\)
b: \(9x^2-16=0\)
=>(3x-4)(3x+4)=0
=>\(\left[{}\begin{matrix}3x-4=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
c: \(x^3-9x^2+27x-27=0\)
=>\(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=0\)
=>\(\left(x-3\right)^3=0\)
=>x-3=0
=>x=3
d: \(1-x+\dfrac{x^2}{3}-\dfrac{x^3}{27}=0\)
=>\(1^3-3\cdot1^2\cdot\dfrac{1}{3}x+3\cdot1\cdot\left(\dfrac{1}{3}x\right)^2-\left(\dfrac{1}{3}x\right)^3=0\)
=>\(\left(1-\dfrac{1}{3}x\right)^3=0\)
=>\(1-\dfrac{1}{3}x=0\)
=>x=3
