Question

a) √27+√243+√75-√108

c) √72-√98+√128

d)-3√98+2√162+5√147-√432

e) 0,2√75-5/4√48+2,(6)·√27


(Cu tot vu rezolvare pe caiet va rog)​

Answer 1

$\sqrt{27}=3\sqrt{3}$

$\sqrt{243} =9\sqrt{3}$

$\sqrt{75} =5\sqrt{3}$

$\sqrt{108} = 6\sqrt{3}$

a) √27+√243+√75-√108 = $3\sqrt{3} +9\sqrt{3}+5\sqrt{3}-6\sqrt{3}=11\sqrt{3}$

$\sqrt{72} = 6\sqrt{2}$

$\sqrt{98} = 7\sqrt{2}$

$\sqrt{128} = 8\sqrt{2}$

c) √72-√98+√128 = $6\sqrt{2} -7\sqrt{2} +8\sqrt{2} = 7\sqrt{2}$

$-3\sqrt{98} = -3*7\sqrt{2}=-21\sqrt{2}$

$2\sqrt{162} = 2*9\sqrt{2} = 18\sqrt{2}$

$5\sqrt{147} = 5*7\sqrt{3} =35\sqrt{3}$

$\sqrt{432} = \sqrt{3*144} =12\sqrt{3}$

d)-3√98+2√162+5√147-√432=$-21\sqrt{2} +18\sqrt{2} +35\sqrt{3} -12\sqrt{3} = -3\sqrt{2} +23\sqrt{3}$

$0,2\sqrt{75} = 0,2*5\sqrt{3} =\sqrt{3}$

$(5/4)*\sqrt{48}=1,25*\sqrt{4*3} =1,25*2\sqrt{3} =2,5\sqrt{3}$

$2,(6)*\sqrt{27} = (24/9)*3\sqrt{3} = (24/3)*\sqrt{3} = 8\sqrt{3}$

e) 0,2√75-5/4√48+2,(6)·√27 = $\sqrt{3} +2,5\sqrt{3}+8\sqrt{3} = 11,5\sqrt{3}$

sper să nu fi gresit la calcule.