Show that - a-1-b m- a-1-b n- b-1-a m - b-1-a n- a-b m-n

( a+1/b )^m× ( a 1/b )^n/ ( b+1/a )^m × ( b 1/a )^n= ( a/b )^m+n LHS= ( a+1/b )^m× ( a 1/b )^n/ ( b+1/a )^m × ( b 1/a )^n =( ab+1) ^m× ( ab 1 )^n/b^m.b^n/( ab+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Laws of Exponents for Real Numbers

Show that : a...

Question

Show that :

$\frac{{(a + \frac{1}{b})}^{m} \times {(a - \frac{1}{b})}^{n}}{{(b + \frac{1}{a})}^{m} \times {(b - \frac{1}{a})}^{n}} = {(\frac{a}{b})}^{m + n}$

Open in App

Solution

$\frac{{(a + \frac{1}{b})}^{m} \times {(a - \frac{1}{b})}^{n}}{{(b + \frac{1}{a})}^{m} \times {(b - \frac{1}{a})}^{n}} = {(\frac{a}{b})}^{m + n} L H S = \frac{{(a + \frac{1}{b})}^{m} \times {(a - \frac{1}{b})}^{n}}{{(b + \frac{1}{a})}^{m} \times {(b - \frac{1}{a})}^{n}} = \frac{\frac{(a b + 1)^{m} \times (a b - 1)^{n}}{b^{m} . b^{n}}}{\frac{(a b + 1)^{m}}{a^{m}} \times \frac{(a b - 1)^{n}}{a^{n}}} = \frac{(a b + 1)^{m} (a b - 1)^{n} \times a^{m} \times a^{n}}{b^{m} . b^{n} (a b + 1)^{m} (a b - 1)^{n}} = \frac{a^{m} . a^{n}}{b^{m} . b^{n}} = \frac{a^{m + n}}{b^{m + n}} = {(\frac{a}{b})}^{m + n} = R H S$

Suggest Corrections

Similar questions

Q. Show that:

\frac{{(a + \frac{1}{b})}^{m} \times {(a - \frac{1}{b})}^{n}}{{(b + \frac{1}{a})}^{m} \times {(b - \frac{1}{a})}^{n}} = {(\frac{a}{b})}^{m + n}

Q. Prove that

1.3 + 3.5 + 5.7 + . . . . + (2 n - 2 n + 1) = \frac{n (4 n^{2} + 6 n - 1)}{3}

Find a pattern for these sequences:

a = 1, 2, 3, 4...

b = 0, 2, 4, 6, 8...

c = 1, 3, 5, 7...

Q. Prove

1.3 + 3.5 + 5.7 + . . . + (2 n - 1) (2 n + 1) = \frac{n (4 n^{2} + 6 n - 1)}{3}

Q. If

a, a_{1}, a_{2}, . . . . . a_{2 n}, b

are in A.P.,

a, g_{1}, g_{2}, . . . . . g_{2 n}

are in G.P
and

a, h, b

are in H.P then:

\frac{a_{1} + a_{2 n}}{g_{1} g_{2 n}} + \frac{a_{1} + a_{2 n - 1}}{g_{1} g_{2 n - 1}} + . . . . . + \frac{a_{n} + a_{n + 1}}{g_{n} g_{n + 1}} =

Join BYJU'S Learning Program

Show that : (a+1b)m×(a−1b)n(b+1a)m×(b−1a)n=(ab)m+n

(a+1b)m×(a−1b)n(b+1a)m×(b−1a)n=(ab)m+nLHS=(a+1b)m×(a−1b)n(b+1a)m×(b−1a)n=(ab+1)m×(ab−1)nbm.bn(ab+1)mam×(ab−1)nan=(ab+1)m(ab−1)n×am×anbm.bn(ab+1)m(ab−1)n=am.anbm.bn=am+nbm+n=(ab)m+n=RHS

Show that :

$\frac{{(a + \frac{1}{b})}^{m} \times {(a - \frac{1}{b})}^{n}}{{(b + \frac{1}{a})}^{m} \times {(b - \frac{1}{a})}^{n}} = {(\frac{a}{b})}^{m + n}$