Show that-i 11-x a-b-11-x b-a-1ii x a a-b x a a-b - x b b-a x b b-a a-b-1iii x 1 a-b 1 a-c - 1 b-c 1 b-a x 1 c-a 1 c-b-1iv x a 2-b 2 x a b a-b x b 2-c 2 x b c b-c x c 2-a 2 x a c a-c-x 2 a 3-b 3-c 3v x a-b a-b x b-c b-c x c-a c-a-1vi xa-a-11a-1aa-1-xvii a x-1 a y-1 x-y a y-2 a z-2 y-z a z-3 a x-3 z-x-1viii 3 a 3 b a-b 3 b 3 c b-c 3 c 3 a c-a-1

(i) 11+xa b+11+xb a =11+xaxb+11+xbxa=xbxb+xa+xaxa+xb=xb+xaxa+xb=1 (ii) xaa bxaa+b #247;xbb axbb+aa+b=xaa bxaa+b #215;xbb+axbb aa+b=xa2 abxa2+ab #215;xb2+ ...

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Byju's Answer

Standard IX

Mathematics

Visualisation of Addition of Polynomials

Show that:i 1...

Question

Show that:
(i) $\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1$

(ii) ${[\{\frac{x^{a (a - b)}}{x^{a (a + b)}}\} \div \{\frac{x^{b (b - a)}}{x^{b (b + a)}}\}]}^{a + b} = 1$

(iii) ${(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = 1$

(iv) ${(\frac{x^{a^{2} + b^{2}}}{x^{a b}})}^{a + b} {(\frac{x^{b^{2} + c^{2}}}{x^{b c}})}^{b + c} {(\frac{x^{c^{2} + a^{2}}}{x^{a c}})}^{a + c} = x^{2 (a^{3} + b^{3} + c^{3})}$

(v) ${(x^{a - b})}^{a + b} {(x^{b - c})}^{b + c} {(x^{c - a})}^{c + a} = 1$

(vi) ${\{{(x^{a - a^{- 1}})}^{\frac{1}{a - 1}}\}}^{\frac{a}{a + 1}} = x$

(vii) ${(\frac{a^{x + 1}}{a^{y + 1}})}^{x + y} {(\frac{a^{y + 2}}{a^{z + 2}})}^{y + z} {(\frac{a^{z + 3}}{a^{x + 3}})}^{z + x} = 1$

(viii) ${(\frac{3^{a}}{3^{b}})}^{a + b} {(\frac{3^{b}}{3^{c}})}^{b + c} {(\frac{3^{c}}{3^{a}})}^{c + a} = 1$

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Solution

(i)
$\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}}$
$= \frac{1}{1 + \frac{x^{a}}{x^{b}}} + \frac{1}{1 + \frac{x^{b}}{x^{a}}} = \frac{x^{b}}{x^{b} + x^{a}} + \frac{x^{a}}{x^{a} + x^{b}} = \frac{x^{b} + x^{a}}{x^{a} + x^{b}} = 1$

(ii)
${[\{\frac{x^{a (a - b)}}{x^{a (a + b)}}\} \div \{\frac{x^{b (b - a)}}{x^{b (b + a)}}\}]}^{a + b} = {[\{\frac{x^{a (a - b)}}{x^{a (a + b)}}\} \times \{\frac{x^{b (b + a)}}{x^{b (b - a)}}\}]}^{a + b} = {[\{\frac{x^{a^{2} - a b}}{x^{a^{2} + a b}}\} \times \{\frac{x^{b^{2} + a b}}{x^{b^{2} - a b}}\}]}^{a + b}$
$= {[\{x^{a^{2} - a b - a^{2} - a b}\} \times \{x^{b^{2} + a b - b^{2} + a b}\}]}^{a + b} = {[x^{- 2 a b} \times x^{2 a b}]}^{a + b} = {[x^{- 2 a b + 2 a b}]}^{a + b} = {[x^{0}]}^{a + b} = 1$

(iii)
${(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = {(x)}^{\frac{1}{a - b} \times \frac{1}{a - c}} {(x)}^{\frac{1}{b - c} \times}^{\frac{1}{b - a}} {(x)}^{\frac{1}{c - a} \times}^{\frac{1}{c - b}} = {(x)}^{\frac{1}{a - b} \times \frac{1}{a - c} + \frac{1}{b - c} \times \frac{1}{b - a} + \frac{1}{c - a} \times \frac{1}{c - b}} = {(x)}^{\frac{(b - c) - (a - c) + (a - b)}{(a - b) (a - c) (b - c)}} = x^{0} = 1$

(iv)
${(\frac{x^{a^{2} + b^{2}}}{x^{a b}})}^{a + b} {(\frac{x^{b^{2} + c^{2}}}{x^{b c}})}^{b + c} {(\frac{x^{c^{2} + a^{2}}}{x^{a c}})}^{a + c} = {(x^{a^{2} + b^{2} - a b})}^{a + b} {(x^{b^{2} + c^{2} - b c})}^{b + c} {(x^{c^{2} + a^{2} - a c})}^{a + c} = [x^{(a + b) (a^{2} + b^{2} - a b)}] [x^{(b + c) (b^{2} + c^{2} - b c)}] [x^{(a + c) (c^{2} + a^{2} - a c)}] = (x^{a^{3} + b^{3}}) (x^{b^{3} + c^{3}}) (x^{a^{3} + c^{3}}) = x^{2 (a^{3} + b^{3} + c^{3})}$

(v)
${(x^{a - b})}^{a + b} {(x^{b - c})}^{b + c} {(x^{c - a})}^{c + a} = [x^{(a - b) (a + b)}] [x^{(b - c) (b + c)}] [x^{(c - a) (c + a)}] = x^{(a^{2} - b^{2})} x^{(b^{2} - c^{2})} x^{(c^{2} - a^{2})} = x^{a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}} = x^{0} = 1$

(vi)
${\{{(x^{a - a^{- 1}})}^{\frac{1}{a - 1}}\}}^{\frac{a}{a + 1}} = \{{(x^{a - \frac{1}{a}})}^{\frac{1}{a - 1} \times \frac{a}{a + 1}}\} = {\{x^{\frac{a^{2} - 1}{a}}\}}^{\frac{a}{a^{2} - 1}} = x^{\frac{a^{2} - 1}{a} \times \frac{a}{a^{2} - 1}} = x^{1} = x$

(vii)
${(\frac{a^{x + 1}}{a^{y + 1}})}^{x + y} {(\frac{a^{y + 2}}{a^{z + 2}})}^{y + z} {(\frac{a^{z + 3}}{a^{x + 3}})}^{z + x} = {(a^{x + 1 - y - 1})}^{x + y} {(a^{y + 2 - z - 2})}^{y + z} {(a^{z + 3 - x - 3})}^{z + x} = {(a^{x - y})}^{x + y} {(a^{y - z})}^{y + z} {(a^{z - x})}^{z + x} = a^{x^{2} - y^{2} + y^{2} - z^{2} + z^{2} - x^{2}} = a^{0} = 1$

(viii)
${(\frac{3^{a}}{3^{b}})}^{a + b} {(\frac{3^{b}}{3^{c}})}^{b + c} {(\frac{3^{c}}{3^{a}})}^{c + a} = {(3^{a - b})}^{a + b} {(3^{b - c})}^{b + c} {(3^{c - a})}^{c + a} = 3^{a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}} = 3^{0} = 1$

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Similar questions

Q. Prove that

(i)

\sqrt{x^{- 1} y} \cdot \sqrt{y^{- 1} z} \cdot \sqrt{z^{- 1} x} = 1

.

(ii)

{(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} \cdot {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} \cdot {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = 1

(iii)

\frac{x^{a (b - c)}}{x^{b (a - c)}} \div {(\frac{x^{b}}{x^{a}})}^{c} = 1

(iv)

\frac{{(x^{a + b})}^{2} {(x^{b + c})}^{2} {(x^{c + a})}^{2}}{{(x^{a} x^{b} x^{c})}^{4}} = 1

Q. Prove that:

(i)

\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1

(ii)

\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1

Prove that:

(i) $\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1$

(ii) $\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1$

Q. If

y = \frac{1}{1 + x^{a - b} +^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} + \frac{1}{1 + x^{b - a} + x^{c - a}}

, then

\frac{d y}{d x}

is equal to
(a) 1
(b)

{(a + b + c)}^{x^{a + b + c - 1}}

\frac{1}{1 + x^{(b - a)} + x^{(c - a)}} + \frac{1}{1 + x^{(a - b)} + x^{(c - b)}} + \frac{1}{1 + x^{(b - c)} + x^{(a - c)}} = ?

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Show that: (i) 11+xa-b+11+xb-a=1 (ii) xaa-bxaa+b÷xbb-axbb+aa+b=1 (iii) x1a-b1a-cx1b-c1b-ax1c-a1c-b=1 (iv) xa2+b2xaba+bxb2+c2xbcb+cxc2+a2xaca+c=x2a3+b3+c3 (v) xa-ba+bxb-cb+cxc-ac+a=1 (vi) xa-a-11a-1aa+1=x (vii) ax+1ay+1x+yay+2az+2y+zaz+3ax+3z+x=1 (viii) 3a3ba+b3b3cb+c3c3ac+a=1