| UHU-Startseite►Mathematik►Jahrgangsstufen►9►Weiterentwicklung der Zahlvorstellung►Umgehen mit einfachen Wurzeltermen► |
|
Lösungen dazu
b) `sqrt (81/1024) = sqrt (3^4/2^10) = 3^2/2^5 = 9/32` c) `sqrt sqrt 256 = sqrt sqrt (2^8) = sqrt (2^4) = 2^2 = 4` d) `sqrt (32/18) = sqrt (16/9) = 4/3` e) `sqrt 128 = sqrt (2^7) = sqrt (2^6*2) = 2^3*sqrt 2` f) `sqrt 162 = sqrt (81*2) = sqrt (3^4*2) = 3^2 * sqrt 2` g) `sqrt 125 = sqrt (5^3) = 5 sqrt 5` h) `sqrt 1000 = sqrt (10^3) = 10*sqrt 10` i) `sqrt (3^2/7^4) = 3/7^2` j) `sqrt (5^6/11^2) = 5^3/11` k) `sqrt (2^16/3^24) = 2^8/3^12` l) `sqrt (4^4/9^6) = 4^2/9^3` m) `sqrt (a^4/b^10) = a^2/b^5` n) `sqrt (c^30/d^12) = c^15/d^6` o) `sqrt ((4x^4)/(9y^6)) = (2 x^2)/(3 y^3)` p) `sqrt((0,01 z^4)/u^8) = (0,1 z^2)/u^4` q) `sqrt(a^5/b^10) = sqrt (a^4/b^10*a) = a^2/b^5*sqrt a` r) `sqrt(c^3/d^2) = c/d*sqrt c` s) `sqrt((3 x^4)/(9y^5)) = x^2 / (3y^2) * sqrt (3/y)` t) `sqrt((0,01 z^3)/u^5) = (0,1 z)/u^2 * sqrt( z/u)
b) `4 sqrt 5 = sqrt 16*5 = sqrt 80` c) `1/3 sqrt 9 = sqrt (1/9) * sqrt 9 = sqrt (1/9*9) = sqrt 1 = 1` d) `2/7 sqrt (49/8) = sqrt (4/49*49/8) = sqrt (1/2)` e) `0,1 sqrt 10 = sqrt(0,01*10) = sqrt 0,1` f) `0,5 sqrt 8 = sqrt (0,25*8) = sqrt 2` g) `1,5 sqrt 12 = sqrt (2,25*12) = sqrt 27` h) `4 sqrt (2,5) = sqrt (16*2,5) = sqrt 40` i) `a sqrt b = sqrt (a^2*b)` j) `2a sqrt (0,5 c) = sqrt (4 a^2*0,5 c) = sqrt(2 a^2 c)` k) `p/q sqrt q = sqrt (p^2/q^2*q ) = sqrt (p^2/q)` l) `m/n sqrt(n/m) = sqrt (m^2/n^2*n/m) = sqrt (m/n)`
b) `2/sqrt 8 = (2*sqrt 8)/(sqrt 8*sqrt 8) = (2*sqrt 8)/8 = 1/4 sqrt 8` c) `5/sqrt 15 = (5 sqrt 15)/15 = 1/3 sqrt 15` d) `2/sqrt 7 = (2 sqrt 7)/7 = 2/7 sqrt 7` e) `sqrt 5/sqrt 3 = (sqrt 5*sqrt 3)/3 = 1/3 sqrt 15` f) `sqrt 3/sqrt 5 = (sqrt 3*sqrt 5)/5 = 1/5 sqrt 15` g) `(6 sqrt 2)/sqrt 6 = (6 sqrt 2*sqrt 6)/6 = sqrt 12 = 2 sqrt 3` h) `(9 sqrt 10)/sqrt 8 = (9*sqrt 10*sqrt 8)/8 = (9 sqrt 80)/8 = 9/8*4*sqrt 5 = 9/2 sqrt 4` |