Question

va rog rezolvare corecta și explicativă!
a)√18(√147-√108+√300-√432)+√48(√50-√32+√128-√98)
b)√27(√32-√72+√162-√288)+√8(√108-√192+√243-√75)
c)3√150(6√98-8√50-7√108+10√48)+4√54(3√192-8√72-4√75+11√32)
d)5√24(4√192-3√432+3√243-5√75)-3√216(3√72-4√162+6√200-5√128)

Answer 1

a)√18(√147-√108+√300-√432)+√48(√50-√32+√128-√98)
Scoatem ce se poate de sub radical si asta se face numai dupa ce am descompus fiecare numar in factori primi.
√18=√(2·3²)=3√2
√147=√(3·7²)=7√3
√108=√(2²·3³)=6√3
√300=√(2²·3·5²)=10√3
√432=√(2⁴·3³)=12√3
√48=√(2⁴·3)=4√3
√50=√(2·5²)=5√2
√32=√(2⁵)=4√2
√128=√(2⁷)=8√2
√98=√(2·7²)=7√2
Acum ne-au ramas simple adunari, scaderi si inmultiri. Tinem cont ca ADUNARI SI SCADERI putem face doar intre termeni cu ACELASI RADICAL, INMULTIRILE se fac inmultind coeficientii radicalilor cat si radicalii intre ei !
3√2(7√3-6√3+10√3-12√3)+4√3(5√2-4√2+8√2-7√2)=
3√2(-√3)+4√3·2√2=
-3√6+8√6=
5√6

b)√27(√32-√72+√162-√288)+√8(√108-√192+√243-√75)
√27=√(3³)=3√3
√72=√(2³·3²)=6√2
√162=√(2·3⁴)=9√2
√288=√(2⁵·3²)=12√2
√192=√(2⁶·3)=8√3
√243=√(3⁵)=9√3
√75=√(3·5²)=5√3
3√3(4√2-6√2+9√2-12√2)+2√2(6√3-8√3+9√3-5√3)=
3√3(-5√2)+2√2·2√3=
-15√6+4√6=
-11√6

Restul incearca sa le faci si singur/a :) Succes !