rationalise the denominator of 1/7-2 root 3
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Do you want step by step explaination
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Step-by-step explanation:
1/(√7+√3-√2)
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
you can solve it further
=-(√7+√3+√2)2[4–(√21)]/20
=-(√7+√3+√2)(4–√21)/10
or further
=-4√7–4√3–4√2+√21*7+√21*3+√21*2/10
=-4√7–4√3–4√2+7√3+3√7+√42/10
solving similar terms
=[-√7+3√3–4√2+√42]/10
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