Keresés


Toplista

Toplista
  • betöltés...

Magántanár kereső

Ha szívesen korrepetálnál, hozd létre magántanár profilodat itt.
Ha diák vagy és korrepetálásra van szükséged, akkor regisztrálj be és írd meg itt, hogy milyen tantárgyban!

Matek

268
Köszönöm szépen!
Jelenleg 1 felhasználó nézi ezt a kérdést.
-1
Középiskola / Matematika

Válaszok

2
1,

`root()(16*9)` = `root()(144)` = 12

`root()(50*2)` = `root()(100)` = 10

`root()(24*6)` = `root()(144)` = 12

`root()(1000*0.1)` = `root()(100)` = 10

`root()(25/4)` = `5/2`

`root()(1/4)` = `1/2`

`root()(300/120)` = `root()(10/4)` = `root()(10)/2`

`root()(45/125)` = `root()(9/25)` = `3/5`

`root()(5^2)` = 5

`root()(2^6)` = `2^3`

`root()(4^3)` = `root()(2^6)` = `2^3`

`root()((-2)^4` = `root()(16)` = 4

`root()((-9)^3)` = `root()(-729)` `notin RR`

`root()(x^2)` = |x|

`root()(y^6)` = `|y^3|`

`root()(9a^2)` = `|3a|`

`root()(81y^6)` = `|9y^3|`

`root()((-a)^2)` = `|a|`

2,

`root()(18)+root()(50)-root()(98)` = `3*root()(2)+5*root()(2)-7*root()(2)` = `(3+5-7)*root()(2)` = `root()(2)`

`root()(27)+5*root()(3)+root()(108)` = `3*root()(3)+5*root()(3)+6*root()(3)` = `(3+5+6)*root()(3)` = `14*root()(3)`

`root()(28)+root()(7)-root()(63)` = `2*root()(7)+1*root()(7)-3*root()(7)` = `(2+1-3)*root()(7)` = `0*root()(7)` = 0

`root()(20)+root()(12)-root()(80)+root()(405)+root()(48)-root()(75)` = `2*root()(5)+2*root()(3)-4*root()(5)+9*root()(5)+4*root()(3)-5*root()(3)` =

= `(2-4+9)*root()(5)+(2+4-5)*root()(3)` = `7*root()(5)+root()(3)`

`root()(200)+root()(450)+3*root()(12)-8*root()(72)` = `10*root()(2)+15*root()(2)+6*root()(3)-48*root()(2)` = `(10+15-48)*root()(2)+6*root()(3)` = `6*root()(3)-23*root()(2)`

`(3*root()(2)+2*root()(8)-root()(27)+5*root()(3)))*(7*root()(2)-2*root()(3))` = `(7*root()(2)+2*root()(3))*(7*root()(2)-2*root()(3))` = `(7*root()(2))^2-(2*root()(3))^2` = `98-12` = 86

`(root()(45)+2*root()(5-root()(125)))*(3*root()(5)+2*root()(20))` = `(3+2-5)*root()(5)*...` az első tényező nulla, ezért ennek az értéke nulla.

`(3*root()(20)+2*root()(3))*(2*root()(45)-root()(12))` = `(6*root()(5)+2*root()(3))*(6*root()(5)-2*root()(3))` = `(6*root()(5))^2-(2*root()(3))^2` = `180-12` = 168

`(2*root()(3)+5*root()(27)-root()(147))*2*root()(3)` = `(10*root()(3))*2*root()(3)` = `10*2*root()(3)^2` = 60

`root()(5+root()(24))*root()(5-root()(24))` = `root()(2+3+2*root()(2*3))*root()(2+3-2*root()(2*3))` = `(root()(2)+root()(3))*(root()(3)-root()(2))` = `3-2` = 1

`(root()(7)+root()(2))*(root()(7)-root()(2))` = `(root()(7))^2-(root()(2))^2` = `7-2` = 5

`(root()(3)+root()(8)-2*root()(27))*(root()(3)+root()(2))` = `(root()(3)+2*root()(2)-6*root()(3))*(root()(3)+root()(2))` = `(2*root()(2)-5*root()(3))*(root()(3)+root()(2))` = `(2*2-5*3-5*root()(6)+2*root()(6))` = `7*root()(6)-11`

`(root()(27)+5*root()(3)+root()(108))*root()(3)` = `(3*root()(3)+5*root()(3)+6*root()(3))*root()(3)` = `14*root()(3)*root()(3)` = `14*3` = 42

`root()(6+2*root()(5))*root()(6-root()(20))` = `root()(1+5+2*root()(5))*root()(1+5-2*root()(5))` = `(1+root()(5))*(root()(5)-1))` = `5-1` = 4

Kb. A tévedés kockázatával :D
Módosítva: 3 éve
0

1.
`sqrt(16*9) = 4*3 = 12`

`sqrt(50*2) = sqrt 100 = 10`

`sqrt(24*6) = sqrt(36*4) = 6*2 = 12`

`sqrt(1000*"0,1") = sqrt 100 = 10`

`sqrt(25/4) = 5/2`

`sqrt(1/4) = 1/2`

`sqrt(300/120) = sqrt(5/2)`

`sqrt(45/125) = sqrt(9/25) = 3/5`

`sqrt(5^2) = 5`

`sqrt(2^6) = 2^3 = 8`

`sqrt(4^3) = sqrt(2^6) = 8`

`sqrt((-2)^4) = (-2)^2 = 4`

`sqrt((-9)^3) = sqrt((-(3)^2)^3) = sqrt(-3^6) = sqrt(-729)\ ""` nem értelmezett

`sqrt(x^2) = abs x`

`sqrt(y^6) = abs(y^3)`

`sqrt(9a^2) = 3 abs a`

`sqrt(81y^6) = 9 abs(y^3)`

`sqrt((-a)^2) = abs(-a) = abs a`

2.
`sqrt(18)+sqrt(50)-sqrt(98) = 3 sqrt(2)+5 sqrt(2)-7 sqrt(2) = sqrt(2)`

`sqrt(27)+5 sqrt(3)+sqrt(108) = 3 sqrt(3)+5 sqrt(3)+6 sqrt(3) = 14 sqrt(3)`

`sqrt(28)+sqrt(7)-sqrt(63) = 2 sqrt(7)+sqrt(7)-3 sqrt(7) = 0`

`sqrt(20)+sqrt(12)-sqrt(80)+sqrt(405)+sqrt(48)-sqrt(75) ``=`` 2 sqrt(5)+2 sqrt(3)-4 sqrt(5)+9 sqrt(5)+4 sqrt(3)-5 sqrt(3) = 7 sqrt(5)+sqrt(3)`

`sqrt(200)+sqrt(450)+3 sqrt(12)-8 sqrt(72) = 10 sqrt(2)+15 sqrt(2)+6 sqrt(3)-48 sqrt(2) = 6 sqrt(3)-23 sqrt(2)`

3.
`(3 sqrt 2+2 sqrt 8-sqrt 27+5 sqrt 3)*(7 sqrt 2-2 sqrt 3) ``=`` (3 sqrt 2+4 sqrt 2-3 sqrt 3+5 sqrt 3)(7 sqrt 2-2 sqrt 3) ``=`` (7 sqrt 2+2 sqrt 3)(7 sqrt 2-2 sqrt 3) ``=`` (7 sqrt 2)^2-(2 sqrt 3)^2 = 49*2-4*3 = 98-12 = 86`

`(sqrt 45+2 sqrt 5-sqrt 125)(3 sqrt 5+2 sqrt 20) ``=`` (3 sqrt 5+2 sqrt 5-5 sqrt 5)(3 sqrt 5+4 sqrt 5) ``=`` 0*(7 sqrt 5) = 0`

`(sqrt 7+sqrt 2)(sqrt 7-sqrt 2) = (sqrt 7)^2-(sqrt 2)^2 = 7-2 = 5`

`(3 sqrt 20+2 sqrt 3)*(2 sqrt 45-sqrt 12) ``=`` (6 sqrt 5+2 sqrt 3)(6 sqrt 5-2 sqrt 3) ``=`` (6 sqrt 5)^2-(2 sqrt 3)^2 = 36*5-4*3 = 180-12 = 168`

`(sqrt 3+sqrt 8-2 sqrt 27)(sqrt 3+sqrt 2) ``=`` (sqrt 3+2 sqrt 2-6 sqrt 3)(sqrt 3+sqrt 2) ``=`` (2 sqrt 2-5 sqrt 3)(sqrt 3+sqrt 2) ``=`` 2 sqrt 2*sqrt 3+2 sqrt 2*sqrt 2-5 sqrt 3*sqrt3-5 sqrt 3*sqrt2 ``=`` 2 sqrt 6+4-15-5 sqrt 6 = -11-3 sqrt 6`

`(2 sqrt 3+5 sqrt 27-sqrt 147)*2 sqrt 3 = (2 sqrt 3+15 sqrt 3-7 sqrt 3)*2 sqrt 3 = 10 sqrt 3*2 sqrt 3 = 20*3 = 60`

`(sqrt 27+5 sqrt 3+sqrt 108)*sqrt 3 = (3 sqrt 3+5 sqrt 3+6 sqrt 3)*sqrt 3 = 14 sqrt 3*sqrt 3 = 14*3 = 42`

`sqrt(5+sqrt 24)*sqrt(5-sqrt 24) = sqrt((5+sqrt 24)*(5-sqrt 24)) = sqrt(5^2-(sqrt 24)^2) = sqrt(25-24) = 1`

`sqrt(6+2 sqrt 5)*sqrt(6-sqrt 20) = sqrt((6+2 sqrt 5)*(6-2 sqrt 5)) = sqrt(6^2-(2 sqrt 5)^2) = sqrt(36-4*5) = sqrt 16 = 4`
Módosítva: 3 éve
0