rationalise the denominator of 1 by7 ×1 by 4+ root 3
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Answer:
Step-by-step explanation:
multiply numerator and denominator by (√7+√3+√2)
=(√7+√3+√2)/[(√7+√3)-(√2)][(√7+√3)+(√2)]
=(√7+√3+√2)/(√7+√3)²-(√2)²
=(√7+√3+√2)/(√7)²+(√3)²+2(√7)(√3)-2
=(√7+√3+√2)/7+3+2(√21)-2
=(√7+√3+√2)/8+2(√21)
multiply numerator and denominator by 8–2(√21)
(√7+√3+√2)[8–2(√21)]/[8+2(√21)][8–2(√21)]
=(√7+√3+√2)[8–2(√21)]/[8]²-[2(√21)]²
=(√7+√3+√2)[8–2(√21)]/64–4*21
=(√7+√3+√2)[8–2(√21)]/64–84
=-(√7+√3+√2)[8–2(√21)]/20
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