①(2 √2-1)(2√2+1)
=(2√2)²-1
=8-1
=7
②(2√x+√y)(√4x-√y)
=(2√x)²-(√y)²
=4x-y
③(3√2+2√3)(3√2-2√3)
=(3√2)²-(2√3)²
=18-12
=6
④(3-√3)(1+1/√3)
=1/3(3-√3)(3+√3)
=1/3(3²-√3²)
=1/3(9-3)
=2
⑤(1+√2)²(1-√2)²
=[1²-(√2)²]²
=(1-2)²
=1
①(2 √2-1)(2√2+1)
=(2√2)²-1
=8-1
=7
②(2√x+√y)(√4x-√y)
=(2√x)²-(√y)²
=4x-y
③(3√2+2√3)(3√2-2√3)
=(3√2)²-(2√3)²
=18-12
=6
④(3-√3)(1+1/√3)
=1/3(3-√3)(3+√3)
=1/3(3²-√3²)
=1/3(9-3)
=2
⑤(1+√2)²(1-√2)²
=[1²-(√2)²]²
=(1-2)²
=1