proszę rozwiąrzcie to
a) 3a√2(2/3√32-1/3√6+2√8)
b) (2a√12+a/3√27-√3)*√3
a) 3a√2(2/3√32-1/3√6+2√8)=3a√2(⅔·4√2-⅓√6+4√2)=3a√2(4⅔√2-⅓√6)=15a√4-6√2a=30a-6√2a=6a(5-√2)
b) (2a√12+a/3√27-√3)*√3=(2a·√4·√3+a/3√3·√9-√3)·√3=4√3a+a√3-√3)·√3=(5√3a-√3)·√3=15a-3=3(5a-1)
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
a) 3a√2(2/3√32-1/3√6+2√8)=3a√2(⅔·4√2-⅓√6+4√2)=3a√2(4⅔√2-⅓√6)=15a√4-6√2a=30a-6√2a=6a(5-√2)
b) (2a√12+a/3√27-√3)*√3=(2a·√4·√3+a/3√3·√9-√3)·√3=4√3a+a√3-√3)·√3=(5√3a-√3)·√3=15a-3=3(5a-1)