Advertisements
Advertisements
प्रश्न
If abc = 1, show that `1/(1+a+b^-1)+1/(1+b+c^-1)+1/(1+c+a^-1)=1`
उत्तर
Consider the left hand side:
`1/(1+a+b^-1)+1/(1+b+c^-1)+1/(1+c+a^-1)`
`=1/(1+a+1/b)+1/(1+b+1/c)+1/(1+c+1/a)`
`=1/((b+ab+1)/b)+1/((c+bc+1)/c)+1/((a+ac+1)/a)`
`=b/(b+ab+1)+c/(c+bc+1)+a/(a+ac+1)` ...........(1)
We know that abc = 1
`therefore c = 1/(ab)`
By substituting the value of c in equation (1), we get
`=b/(b+ab+1)+(1/(ab))/(1/(ab)+b(1/(ab))+1)+a/(a+a(1/(ab))+1)`
`=b/(b+ab+1)+(1/(ab))/(1/(ab)+b/(ab)+(ab)/(ab))+a/((ab)/b+1/b+b/b)`
`=b/(b+ab+1)+(1/(ab))/((1+b+ab)/(ab))+a/((ab+1+b)/(b))`
`=b/(b+ab+1)+(1/(ab)xxab)/(1+b+ab)+(axxb)/(ab+1+b)`
`=b/(b+ab+1)+1/(b+ab+1)+(ab)/(b+ab+1)`
`=(b+ab+1)/(b+ab+1)`
= 1
Therefore, LHS = RHS
Hence, proved
APPEARS IN
संबंधित प्रश्न
If a = 3 and b = -2, find the values of :
aa + bb
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
State the power law of exponents.
If 24 × 42 =16x, then find the value of x.
If a, m, n are positive ingegers, then \[\left\{ \sqrt[m]{\sqrt[n]{a}} \right\}^{mn}\] is equal to
(256)0.16 × (256)0.09
\[\frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^n - 2 \times 5^{n + 1}}\] is equal to
If x = \[\sqrt[3]{2 + \sqrt{3}}\] , then \[x^3 + \frac{1}{x^3} =\]
If \[\sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5}, \text { then a } =\]
Find:-
`32^(1/5)`