If a - xm - n - y1- b - xn - 1 - ym and c - x1 - m - yn-then prove that am - n - bn - 1 - c1 - m -1

⇒ a^m n· b^n 1· c^1 m = 1 As given a = x^m+n· y^1; #160; #160; b =x^n + 1· y^m #160;and #160;c = x^1 + m· y^n ⇒ ( x^m ...

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Byju's Answer

Standard XII

Mathematics

Sigma n2

If a = xm +...

Question

If $a = x^{m + n} \cdot y^{1}; b = x^{n + 1} \cdot y^{m} a n d c = x^{1 + m} \cdot y^{n}$ ,
then prove that $a^{m - n} \cdot b^{n - 1} \cdot c^{1 - m} = 1$

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Solution

$\Rightarrow a^{m - n} \cdot b^{n - 1} \cdot c^{1 - m} = 1$
As given $a = x^{m + n} \cdot y^{1}; b = x^{n + 1} \cdot y^{m} a n d c = x^{1 + m} \cdot y^{n}$
$\Rightarrow {(x^{m + n} . y^{1})}^{m - n} \cdot {(x^{n + 1} . y^{m})}^{n - 1} \cdot {(x^{1 + m} . y^{n})}^{1 - m} = 1$
$∵ {(a^{m})}^{n} = a^{m n}$
So $\Rightarrow x^{m^{2} - n^{2}} . y^{m - n} . x^{n^{2} - 1} . y^{m n - m} . x^{1 - m^{2}} . y^{n - n m}$
$∵ a^{m} . a^{n} = a^{M + n}$
So $\Rightarrow x^{m^{2} - n^{2} + n^{2} - 1 + 1 - m^{2}} \cdot y^{m - n + m n - m + n - m n}$
$\Rightarrow x^{0} \cdot y^{0}$
$∵ a^{0} = 1$
$\Rightarrow 1 \times 1 = 1$
$L H S = R H S$

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Similar questions

State true or false.

a = x^{m + n} . y^{1}; b = x^{n + 1} . y^{m}

and

c = x^{1 + m} . y^{n}

, then

a^{m - n} . b^{n - 1} . c^{1 - m} = 1

Q. If

a = x^{m + n} . y^{l}; b = x^{n + l} . y^{m}

and

c = x^{l + m} . y^{n}

prove that :

a^{m - n} . b^{n - l} . c^{l - m} = 1

Q. (i) If

a = x^{m + n} y^{l}, b = x^{n + l} y^{m} and c = x^{l + m} y^{n}

, prove that

a^{m - n} b^{n - l} c^{l - m} = 1

.

(ii) If

x = a^{m + n}, y = a^{n + l} and z = a^{l + m}

, prove that

x^{m} y^{n} z^{l} = x^{n} y^{l} z^{m}

Simplify: $(i) {(\frac{x^{a + b}}{x^{c}})}^{a - b} {(\frac{x^{b + c}}{x^{a}})}^{b - c} {(\frac{x^{c + a}}{x^{b}})}^{c - a} (i i) \sqrt[l m]{\frac{x^{l}}{x^{m}}} \times \sqrt[m n]{\frac{x^{m}}{x^{n}}} \times \sqrt[n l]{\frac{x^{n}}{x^{l}}}$

Q. Prove that the coefficient of

x^{m}

in the expansion of

(1 + x)^{m} + 2 (1 + x)^{n - 1} + 3 (1 + x)^{n - 2} + (n - m + 1) (1 + x)^{m}

where

0 \geq m \geq n i s^{n + 2} C_{m + 2}

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If a=xm+n⋅y1;b=xn+1⋅ymandc=x1+m⋅yn,then prove that am−n⋅bn−1⋅c1−m=1

⇒am−n⋅bn−1⋅c1−m=1As given a=xm+n⋅y1;b=xn+1⋅ym and c=x1+m⋅yn⇒(xm+n.y1)m−n⋅(xn+1.ym)n−1⋅(x1+m.yn)1−m=1∵(am)n=amnSo⇒xm2−n2.ym−n.xn2−1.ymn−m.x1−m2.yn−nm∵am.an=aM+nSo⇒xm2−n2+n2−1+1−m2⋅ym−n+mn−m+n−mn⇒x0⋅y0∵a0=1⇒1×1=1LHS=RHS

If $a = x^{m + n} \cdot y^{1}; b = x^{n + 1} \cdot y^{m} a n d c = x^{1 + m} \cdot y^{n}$ ,
then prove that $a^{m - n} \cdot b^{n - 1} \cdot c^{1 - m} = 1$