Bài 1:
a: \(\left(x+2y\right)^2=x^2+2\cdot x\cdot2y+\left(2y\right)^2=x^2+4xy+4y^2\)
b: \(\left(x-3y\right)\left(x+3y\right)=x^2-\left(3y\right)^2=x^2-9y^2\)
c: \(\left(5-x\right)^2=5^2-2\cdot5\cdot x+x^2=25-10x+x^2\)
d: \(\left(x-3\right)^3=x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=x^3-9x^2+27x-27\)
e: \(\left(2y+3\right)^3\)
\(=\left(2y\right)^3+3\cdot\left(2y\right)^2\cdot3+3\cdot2y\cdot3^2+3^3\)
\(=8y^3+36y^2+54y+27\)
f: \(\left(x-\frac12\right)^2=x^2-2\cdot x\cdot\frac12+\left(\frac12\right)^2=x^2-x+\frac14\)
Bài 2:
a: \(49x^2+14x+1=\left(7x\right)^2+2\cdot7x\cdot1+1^2=\left(7x+1\right)^2\)
b: \(\left(x-3y\right)^2+6\left(x-3y\right)+9\)
\(=\left(x-3y\right)^2+2\cdot\left(x-3y\right)\cdot3+3^2\)
\(=\left(x-3y+3\right)^2\)
c: \(x^2+x+\frac14=x^2+2\cdot x\cdot\frac12+\left(\frac12\right)^2=\left(x+\frac12\right)^2\)
Bài 3:
a: \(49x^2-4=0\)
=>\(49x^2=4\)
=>\(x^2=\frac{4}{49}\)
=>\(\left[\begin{array}{l}x=\frac27\\ x=-\frac27\end{array}\right.\)
b: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=11\)
=>\(x^2+6x+9-\left(x^2-4\right)=11\)
=>\(x^2+6x+9-x^2+4=11\)
=>6x+13=11
=>6x=-2
=>\(x=-\frac26=-\frac13\)
