Usuń niewymierność z mianownika ułamka:
a) 10/√3+1
b) 1/√2-1
c) 4/√5+1
d) 2/√5-3
e) √2/√2-2
f) 2√5/2√5+4
g) 3√3/√3-3
h) 2√6/√6+2
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Rozwiązanie w załączniku
a) 10/√3+1 * √3-1/√3-1 = 10(√3-1)/3-1 = 10√3-10/2
b) 1/√2-1 * √2+1/√2+1 = √2+1/2-1 = √2+1/1 = √2+1
c) 4/√5+1 * √5-1/√5-1 = 4(√5-1)/5-1 = 4(√5-1)/4 = √5-1
d) 2/√5-3 * √5+3/√5+3 = 2(√5+3)/5-9 = 2(√5+3)/-4 = √5+3/-2
e) √2/√2-2 * √2+2/√2+2 = √2(√2+2)/2-4 = 2+2√2/-2
f) 2√5/2√5+4 * 2√5-4/2√5-4 = 2√5(2√5-4)/20-16 = 2√5(2√5-4)/4 = √5(2√5-4)/2 = 10-4√5/2
g) 3√3/√3-3 * √3+3/√3+3 = 3√3(√3+3)/3-9 = 3√3(√3+3)/-6 = √3(√3+3)/-2 = 3+3√3/-2
h) 2√6/√6+2 * √6-2/√6-2 = 2√6(√6-2)/6-4 = 2√6(√6-2)/2 = √6(√6-2) = 6-2√6