find value of 2(√2+√6)/ 3√2+√3 (Hint: x= √2+√6;y=2+√3;find (x/y))
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2(√2+√6)=2√2+2√6
=√8+√24
3√2+√3=√18+√3
therefore, (√8+√24)/(√18+√3)
multiplying (√18-√3) both in denominator and nominator
=√8(√18-√3)+√24(√18-√3)/(√18)²-(√3)²
=√144-√24+√432-√72/18-3
=12-2√6+12√3-6√2/15
=4(3+3√3-2√2)/15
=4/15(3+3√3-2√2)
=√8+√24
3√2+√3=√18+√3
therefore, (√8+√24)/(√18+√3)
multiplying (√18-√3) both in denominator and nominator
=√8(√18-√3)+√24(√18-√3)/(√18)²-(√3)²
=√144-√24+√432-√72/18-3
=12-2√6+12√3-6√2/15
=4(3+3√3-2√2)/15
=4/15(3+3√3-2√2)
Answered by
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2(√2+√6)=2√2+2√6
=√8+√24
3√2+√3=√18+√3
therefore, (√8+√24)/(√18+√3)
multiplying (√18-√3) both in denominator and nominator
=√8(√18-√3)+√24(√18-√3)/(√18)²-(√3)²
=√144-√24+√432-√72/18-3
=12-2√6+12√3-6√2/15
=4(3+3√3-2√2)/15
=4/15(3+3√3-2√2)
=√8+√24
3√2+√3=√18+√3
therefore, (√8+√24)/(√18+√3)
multiplying (√18-√3) both in denominator and nominator
=√8(√18-√3)+√24(√18-√3)/(√18)²-(√3)²
=√144-√24+√432-√72/18-3
=12-2√6+12√3-6√2/15
=4(3+3√3-2√2)/15
=4/15(3+3√3-2√2)
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