Răspuns :

Schiyucaidid

a) [tex]3\sqrt{2} <5\sqrt{2}[/tex]

b) [tex]-6\sqrt{3} <4\sqrt{7}[/tex]

c)[tex]5\sqrt{3} <7\sqrt{2}[/tex]

d)[tex]2\sqrt{3} < 7\sqrt{2} \\[/tex]

e)[tex]2\sqrt{3} = 3\sqrt{2}[/tex]

f)[tex]2\sqrt{3}- 3\sqrt{2}=0[/tex]

g)[tex]0<2\sqrt{5}-4[/tex]

h)[tex]5\sqrt{2}-7>0[/tex]\

i)[tex]-3\sqrt{3} +5<-5[/tex]

Vadancorneliuid

Răspuns:

Explicație pas cu pas:

a)

3√5=√(5·9)=√45

5√2=√(2·25)=√50

=> 3√5<5√2

b)

-6√3=-√(3·36)=-√108

-4√7=-√(7·16)=-√112

=> -6√3>-4√7

c)

5√3=√(3·25)=√75

7√2=√(2·49)=√98

=> 5√3<7√2

d)

2√3=√(3·4)=√12

7√2=√(2·49)=√98

=> 2√3<7√2

e)

2√3=√(3·4)=√12

3√2=√(2·9)=√18

=> 2√3<3√2

f)

2√3=√(3·4)=√12

3√2=√(2·9)=√18

=> 2√3<3√2 => 2√3-3√2<0

g)

2√5=√(5·4)=√20

4=√16

=> 4<2√5 => 2√5-4>0

h)

5√2=√(2·25)=√50

7=√49

=> 5√2>7 => 5√2-7>0

i)

3√3=√(3·9)=√27

10=√100

=> 3√3<10 => 0<10-3√3 => 0<5+5-3√3 => -5<5-3√3 => -3√3+5>-5

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