Question

2√0,25=

√12 supra√3=

rationalizati fractia 6pe√5-√2=

aratati ca nr a=(√5+√2)² - √40 este rational

determinati x∈N incat x(√3-√2)=1pe√3+√2

demonstrati ca 3pe√3-2pe√2=1pe√3+√2

Answer 1

2√0,25=2√(25/100)=2x√25/√100=2x5/10=10/10=1

√12/√3=√(12/3)=√4=2

6/(√5-√2)=6(√5+√2)/[(√5-√2)(√5+√2)=
6(√5+√2)/(5-2)=
6(√5+√2)/3=2(√5+√2)

a=(√5+√2)² - √40 a=(5+√10+√10+2)-√(4x10)
a=7+2√10-2√10
a=7

x(√3-√2)=1/√3+√2
x=1/(√3+√2):(√3-√2)
x=1/(√3+√2)·1/(√3-√2)
x=1/(3-2)
x=1/1
x=1

3/√3-2/√2=1/√3+√2
3√3/3-2√2/2=(√3-√2)/[(√3+√2)(√3-√2)
√3-√2=(√3-√2)/(3-2)
√3-√2=√3-√2