Question

Ma puteti ajuta cu teza 3?

Answer 1

1. 22/7 /11/5=22/7·5/11=10/7;
2. 112=2⁴·7
252=2²·3²·7
[112;252]=2⁴·3²·7=1008;
3. 24=2³·3
36=3²·2²
[24;36]=2³·3²=8·9=72;
Asadar avem 5/24=15/72>7/36=14/72 ⇔5/24>7/36;
4. 3¹·3²·.....3⁸=3¹⁺²⁺³⁺⁴⁺⁵⁺⁶⁺⁷⁺⁸=3³⁶=(3¹⁸)² ⇒3ˣ=3¹⁸ ⇒x=18;
5. 45 de grade;
6. Consideram urmatoarele variante
a. daca ΔABC-Δis unde AB=4cm si BC=6cm ⇒AC=4cm ⇒P=4+6+4=14cm(perimetrul triunghiului ABC);
b. daca ΔABC-Δis unde AB=4cm si BC=6cm ⇒AC=6cm ⇒P=4+6+6=16cm(perimetrul triunghiului ABC);
7. 90 de grade;
8. a+b+c+x/4=2,5 ⇔a+b+c+x=10 ⇒x=2,8;
9. 4cm;
10. a. 2x-6y+3=2(3y+7)-6y+3=6y+14-6y+3=17;
b. x+y>35 ⇔3y+7+y>35 ⇔4y+7>35 ⇔y>7 ⇒propozitie adevarata deoarece conform teoremei impartirii cu rest D=I·C+R ,I>R ⇔y>7. 
11. ΔABC≡ΔBAD deoarece [BC]≡[DA] respectiv <ABC≡<BAD ⇒[CA]≡[BD] si <ACD≡<BCD;