(3x-1/3x)²-(3x+1/3x)(3x-1/3x)
Pls solve the question quickly...
rohitrajsingh:
2(1-9x^2)/9x^2
Yaa this one is right
Pls give the detailed answer
Answers
Answered by
3
Its of the form:
(a-b)^2 - (a+b)*(a-b)
= a^2 + b^2 - 2*a*b -(a^2 - b^2)
= a^2 + b^2 - 2*a*b -a^2 + b^2
= 2b^2 - 2*a*b = 2*b*[b - a]
ATQ,
a = 3x, b = 1/3x
2*b*[b-a] = 2*1/3x*[1/3x - 3x] = 2/3x*[(1- 9x^2)/3x] = 2*[1-9x^2]/9x^2
Ans: 2*[1-9x^2]/9x^2 OR 2*[(1-3x)*(1+3x)]/9x^2 OR " 2/9x^2 - 2 "
Hope it helps
(a-b)^2 - (a+b)*(a-b)
= a^2 + b^2 - 2*a*b -(a^2 - b^2)
= a^2 + b^2 - 2*a*b -a^2 + b^2
= 2b^2 - 2*a*b = 2*b*[b - a]
ATQ,
a = 3x, b = 1/3x
2*b*[b-a] = 2*1/3x*[1/3x - 3x] = 2/3x*[(1- 9x^2)/3x] = 2*[1-9x^2]/9x^2
Ans: 2*[1-9x^2]/9x^2 OR 2*[(1-3x)*(1+3x)]/9x^2 OR " 2/9x^2 - 2 "
Hope it helps
Answered by
5
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♦ EXPANDING EXPRESSIONS ♦
[tex]( 3x - \frac{1}{3x} )^2 - ( 3x + \frac{1}{3x} )( 3x - \frac{1}{3x} ) \\ \\ = ( 3x - \frac{1}{3x} )[ ( 3x - \frac{1}{3x} ) - ( 3x + \frac{1}{3x} ) ] \\ \\ = ( 3x - \frac{1}{3x} )[ - \frac{2}{3x} \ ] \\ \\ = \frac{2}{9x^2} - 2 = 2[ \ \frac{1 - 9x^2 }{9x^2} \ ][/tex]
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→ Simplified Fraction = 2[ ( 1 - 9x² ) / ( 9x² ) ]
♦ EXPANDING EXPRESSIONS ♦
[tex]( 3x - \frac{1}{3x} )^2 - ( 3x + \frac{1}{3x} )( 3x - \frac{1}{3x} ) \\ \\ = ( 3x - \frac{1}{3x} )[ ( 3x - \frac{1}{3x} ) - ( 3x + \frac{1}{3x} ) ] \\ \\ = ( 3x - \frac{1}{3x} )[ - \frac{2}{3x} \ ] \\ \\ = \frac{2}{9x^2} - 2 = 2[ \ \frac{1 - 9x^2 }{9x^2} \ ][/tex]
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→ Simplified Fraction = 2[ ( 1 - 9x² ) / ( 9x² ) ]
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