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Grade: 9
        
please give the solutions of these sums. please see the attachment. 
thanks
one year ago

Answers : (1)

Ram khilari Kushwah
28 Points
							
Q27:Let m=2x=3y=6-z
So m1/x=2-------(1)
m1/y=3-----------(2)
and m=6-z
m1/z=6-1=1/6-----(3)
 
On multiplication of uquarion (1).(2) and (3) we get
 
m1/x*m1/y*m1/z=2*3*1/6
m(1/x+1/y+1/z)=1=m0
Hence 1/x+1/y+1/z = 0
Hence proved
--------------------------------------------------------------------------------------------------
Q28:
Let us take first  7√3/(√10+√3
=7√3/(√10-√3)/(√10+√3)/(√10-√3)
=7*(√30-√3*√3)/(10-3)
So 7√3/(√10+√3=7/7(√30-3)=√30-3--------(1)
 
  2√5/(√6+√5) = 2√5*(√6-√5) / (√6+√5)(√6-√5)
=(2√30-2√5*√5) / (6-5)
2√5/(√6-√5)=(2√30+2*5)=2√30-10-----(2)
 
  3√2/(√15+3√2)=3√2(√15-3√2)  / (√15+3√2)(√15-3√2)
=(3√30-9*√2*√2)/(15-18)
=3/-3(√30 – 3*2)= -(√30-6)
32 / (√15+3√2) = 6-30---------------(3)
 
Hence the given expression using (1) (2) and (3)
 
=√30-3-(2√30-10)-(6-30)
=√30-3-2√30+10-6+30
=10-3-6
=1
Hence proved
 
 
 
one year ago
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