Question

Va rog frumos am nevoie de ajutor

Answer 1

1.

a) 3rad2+7rad2-10rad2= 0

b) 6rad3+(3rad3-4rad3)+3rad3= 8rad3

c) (3-3rad2)(1+rad2) = 3+3rad2-3rad2-6=3-6=-3

d) (rad3)²⁰¹⁸×(rad3)`²⁰¹⁶=(rad3)²⁰¹⁸`²⁰¹⁶=(rad3)²=3

e) (1/rad5)⁵:(rad5)³=(rad5)`⁵:(rad5)³=(rad5)`⁵`³= (rad5)`⁸=1/125

f)(-2rad7)⁷:(rad28)⁶=(-2rad7)⁷:(2rad7)⁶=-2rad7

2.

a) 1/rad11= rad11/11

b) 4/5rad28=4/10rad7=4rad7/70

c) 15/2rad60=15/4rad15=15rad15/4×15=rad15/4

d) -9/2rad18=-9/6rad2=-3rad2/4

3.

a) (rad18-rad8-rad2+rad50)²:50= (3rad2-2rad2-rad2+rad50)²:50=(rad50)²:50=50:50=1

b) [(2rad3)²]`³-(2rad3)`²×(2rad3)`⁴=(12)`³-1/12×1/12²= 1/12³-1/12³=0

c)

$( \frac{2 \sqrt{7 } - 3 \sqrt{5} }{ \sqrt{35} } )\times ( \frac{5}{ \sqrt{3} } + \frac{3}{ \sqrt{7}} - \frac{5 \sqrt{7} }{ \sqrt{21} } ) {}^{2} = ( \frac{14 \sqrt{5} - 15 \sqrt{7} }{35} ) \times ( \frac{5 }{ \sqrt{3} } + \frac{3}{ \sqrt{7} } - \frac{5 }{ \sqrt{3} } ) {}^{2} = ( \frac{14 \sqrt{5} - 15 \sqrt{7} }{35} ) \times ( \frac{3}{ \sqrt{7} } ) {}^{2} = ( \frac{14 \sqrt{5} - 15 \sqrt{7} }{35} ) \times \frac{9}{7} = \frac{126 \sqrt{5} - 135 \sqrt{7} }{245}$

Sper sa fie ok <3