Question

Cum se face 4 si 5? Please help

Answer 1

Trebuie să știm câteva pătrate perfecte:

4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, ...

4a)

[tex]\it \sqrt{18} +\sqrt8-\sqrt{32} = \sqrt{9\cdot2}+\sqrt{4\cdot2}-\sqrt{16\cdot2}= \\\;\\ = 3\sqrt2+2\sqrt2-4\sqrt2=5\sqrt2-4\sqrt2=\sqrt2[/tex]

d)

[tex]\it \sqrt{24}-\sqrt{54}-6\sqrt6 =\sqrt{4\cdot6}-\sqrt{9\cdot6}-6\sqrt6= \\\;\\ =2\sqrt6-3\sqrt6-6\sqrt6=2\sqrt6-9\sqrt6 =-7\sqrt6[/tex]

5a)

[tex]\it (\sqrt{18}+\sqrt{288}) - (\sqrt{98}+\sqrt{200})=(\sqrt{9\cdot2}+\sqrt{144\cdot2}) -\\\;\\ -(\sqrt{49\cdot2}+\sqrt{100\cdot2})= (3\sqrt2+12\sqrt2) - (7\sqrt2+10\sqrt2)= \\\;\\ =15\sqrt2- 17\sqrt2 =-2\sqrt2[/tex]

b)


[tex]\it 4\sqrt{18}+5\sqrt{108}-2\sqrt{72} =4\sqrt{9\cdot2}+5\sqrt{36\cdot3}-2\sqrt{36\cdot2}= \\\;\\ 4\cdot3\sqrt2 + 5\cdot6\sqrt3-2\cdot6\sqrt2 =12\sqrt2+30\sqrt3-12\sqrt2=30\sqrt3[/tex]

c)

[tex]\it 3\sqrt{162} -5\sqrt{12}-4\sqrt{75}+2\sqrt{50} = 3\sqrt{81\cdot2}-5\sqrt{4\cdot3}-4\sqrt{25\cdot3} + \\\;\\ 2\sqrt{25\cdot2} = 3\cdot9\sqrt2-5\cdot2\sqrt3-4\cdot5\sqrt3+2\cdot5\sqrt2=27\sqrt2- \\\;\\ -10\sqrt3-20\sqrt3+10\sqrt2= 37\sqrt2-30\sqrt3[/tex]

Answer 2

V - inseamna radical pt ca nu am simbolul 

4. a) 3V2 + 2V2 - 4V2 = 5V2 - 4V2 = V2
b) 3V5 + 5V5 + 2V5 = 8V5+2V5+10V5
c) V3 + 3V3 - 2V3 + 4V3 = 2V3 + 4V3 = 6V3
d) 2V6 - 3V6 - 6V6 = -V6 - 6V6 = -7V6

5. a) (3V2 + 12V2) - (7V2 + 10V2) = 15V2 - 17V2 = -2V2 
b) 4*3V2 + 5*6V3 - 2*6V2 = 12V2 + 30V3 - 12V2 = 30V3 
c) 3*9V2 - 5*2V3 - 4*5V3 + 2*5V2 = 27V2 - 10V3 - 20V3 + 10V2 =37V2 -30V3 
d) -4*6V6 + 3*4V6 - 3V6 + 2*10V6 = -24V6 + 12V6 - 3V6 + 20V6 = -12V6 -3V6 + 20V6 = -15V6 + 20V6 = 5V6