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Matek

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Jelenleg 1 felhasználó nézi ezt a kérdést.
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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
Módosítva: 3 hete
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 hete
0