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# 10th

Posted in Math, asked by anisha, 6 years ago. 1662 hits.

## 0

(1/a)+(1/b)+(1/x)=1/a+b+x

we will expand the L.H.S and do cross multiplication herr

=> (bx+ax+ab)/abx=1/a+b+x

=>( bx+ax+ab)(a+b+x)=abx

=> (bx+ax+ab)a+(bx+ax+ab)b+(bx+ax+ab)x=abx

=>abx+a^2x+a^2b+b^2x+abx+ab^2+bx^2+abx+ax^2=abx

we will take the R.H.S into L.H.S then

abx+a^2x+a^2b+b^2x+abx+ab^2+bx^2+abx+ax^2-abx=0

now we will rewrite the equation

= x^2(a+b)+x(a^2+ab+ab+ab+b^2-ab)+a^2b+ab^2=0

= x^2(a+b)+x(a^2+2ab+b^2)+ab(a+b)=0

=x^2(a+b)+x(a+b)^2+ab(a+b)=0

in all the above terms,(a+b) we have common,so

(a+b)(x^2+(a+b)x+ab)=0

now, it will come ,x^2+(a+b)x+ab=0

now again we rewrite it, x^2+ax+bx+ab=0

x(x+a)+b(x+a)=0

(x+a)(x+b)=0

x+a=0,x+b=0

x=-a,x=-b

we can solve this type problems in many ways.i think its useful and simple way to understand the students

Venkat Nagendra Thati - 6 years ago
Ask Venkat Nagendra Thati for further help.
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