Math, asked by shakuntalamurmu57, 11 months ago

please find me this answer from Monday I have exams please please please please please please please please please please please please please please please please please please please please please​

Attachments:

Answers

Answered by TRISHNADEVI
3

 \huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \:   \: QUESTION \:  \: } \mid}}}}}

 \mathsf{ \:  \:Prove \:  \:  that, \:  \: }   \\  \\ \mathtt{  (a {}^{ \frac{1}{x - y} } ) {}^{ \frac{1}{x - z} }  \times (a {}^{ \frac{1}{y - z} } ) {}^{ \frac{1}{y - x} }  \times (a {}^{ \frac{1}{z - x} } ) {}^{\frac{1}{z - y}} = 1 }

 \huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \:   \: ANSWER \:  \: } \mid}}}}}

 \mathtt{L.H.S. =  (a {}^{ \frac{1}{x - y} } ) {}^{ \frac{1}{x - z} }  \times (a {}^{ \frac{1}{y - z} } ) {}^{ \frac{1}{y - x} }  \times (a {}^{ \frac{1}{z - x} } ) {}^{\frac{1}{z - y}} }  \\  \\  \mathtt{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: = (a) {}^{ \frac{1}{x - y} . \frac{1}{x - z} }  \times (a) {}^{ \frac{1}{y - z} . \frac{1}{y - x} }  \times (a) {}^{ \frac{1}{z - x} .\frac{1}{z - y} } }\\  \\ \mathtt{    \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: = (a) {}^{ \frac{1}{(x - y)(x - z)} }  \times (a)  {}^{ \frac{1}{(y - z)(y - x)} }  \times (a) {}^{ \frac{1}{(z - x)(z - y)} }}  \\  \\   \mathtt{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: =( a) {}^{\frac{1}{(x - y).(x - z)} +\frac{1}{(y - z).(y - x)} + \frac{1}{(z - x).(z - y)} }}  \\  \\ \mathtt{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:     =( a) {}^{\frac{1}{( - 1).(x - y).( z - x)} +\frac{1}{( - 1).(y - z).(x - y)} + \frac{1}{( - 1).(z - x).(y - z )} }} \\  \\    \mathtt{   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =( a) {}^{\frac{( - 1)}{(x - y).(z - x)} +\frac{( - 1)}{(y - z).(x - y)} + \frac{( - 1)}{(z - x).(y - z )} }}  \\  \\  \mathtt{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \: \: =  (a) {}^{ \frac{( - 1).(y - z) + ( - 1).(z - x) + ( - 1).(x - y)}{(x - y)(y - z)(z - x)} }}  \\  \\  \mathtt{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = (a) {}^{ \frac{ - y + z  - z + x - x + y}{(x - y)(y - z)(z - x)}}}   \\ \\   \mathtt{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: = (a)  {}^{ \frac{0}{(x - y)(y - z)(z - x)}} }  \\  \\  \mathtt{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = (a) {}^{0}}  \\  \\  \mathtt{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 1 \:  \: = R.H.S. }

FORMULA USED :-

 \mathtt{1. \: (p  {}^{ \frac{1}{m} } ) {}^{ \frac{1}{n} }  =  p {}^{ \frac{1}{mn} }  }\\  \\ \mathtt{2. \:  p {}^{m}   \times  p{}^{n}  = p{}^{m + n}}  \\  \\  \mathtt{3. \: p - q = ( - 1).(q - p)} \\  \\  \mathtt{4. \: p {}^{ \frac{0}{m} }  = p {}^{0}}  \\  \\  \mathtt{5. \:  p {}^{0 = 1} }

Similar questions