Question

Va rog mult, dau 50 de pcte

Answer 1

Ai raspuns pe foaie.
a) 3^101/3^101•(1+3)=1/4
b) 7^10•(1+2•7+3•7^2)/(12•7^10)=162/12=27/2
c)(80•81)/3•(1+2+3+...+80)=(80•81)/3•(80•81)/2=(80•81) • 2/3•(80•81)= 2/3
d) [5•(20•21)/2] : [4•(20•21)/2]=5•(20•21)/2 • 2/4•(20•21) = 5/4

Answer 2

$\dfrac{ {3}^{101} }{ {3}^{101} + {3}^{102} } ^{ \div {3}^{101} } = \dfrac{1}{1 + 3} = \dfrac{1}{4}$




$\dfrac{ {7}^{10} + 2 \times {7}^{11} + 3 \times {7}^{12} }{12 \times {7}^{10} } ^{ \div {7}^{10} } = \dfrac{1 + 2 \times 7 + 3 \times {7}^{2} }{12 \times 1} = \dfrac{1 + 14 + 3 \times49 }{12} = \dfrac{1 + 14 + 147}{12} = \dfrac{162}{12} ^{ \div 2} = \dfrac{81}{6}^{ \div 3} = \dfrac{27}{2}$




$\dfrac{2 + 4 + 6 + ... + 160}{3 + 6 + 9 + ... + 240} = \dfrac{2 \times (1 + 2 + 3 + ... + 80}{3 \times (1 + 2 + 3 + ... + 80} ^{1 + 2 + 3 +. .. + 80} = \dfrac{2 \times 1}{3 \times 1} = \dfrac{2}{3}$




$\dfrac{5 + 10 + 15 + ... + 100}{4 + 8 + 12 + ... + 80} = \dfrac{5 \times (1 + 2 + 3 + ... + 20)}{4 \times (1 + 2 + 3 + ... + 20)}^{ \div 1 + 2 + 3 + ... + 20} = \dfrac{5 \times 1}{4 \times 1} = \dfrac{5}{4}$