Mixing
Compute $frac{x}{y-z} +frac {y}{z-x}+frac {z}{x-y} $
$begingroup$
Compute $frac{x}{y-z} +frac {y}{z-x}+frac {z}{x-y} $ if $frac{x^2}{ (y-z)^2} +frac {y^2}{ (z-x)^2}+frac {z^2}{ (x-y)^2}=11. $
I have no idea how can I start.
algebra-precalculus summation factoring
edited Jan 6 at 20:21
Michael Rozenberg
99.3k1590189
asked Jan 6 at 17:39
rafarafa
598212
$endgroup$
add a comment |
$begingroup$
Compute $frac{x}{y-z} +frac {y}{z-x}+frac {z}{x-y} $ if $frac{x^2}{ (y-z)^2} +frac {y^2}{ (z-x)^2}+frac {z^2}{ (x-y)^2}=11. $
I have no idea how can I start.
algebra-precalculus summation factoring
edited Jan 6 at 20:21
Michael Rozenberg
99.3k1590189
asked Jan 6 at 17:39
rafarafa
598212
$endgroup$
add a comment |
$begingroup$
Compute $frac{x}{y-z} +frac {y}{z-x}+frac {z}{x-y} $ if $frac{x^2}{ (y-z)^2} +frac {y^2}{ (z-x)^2}+frac {z^2}{ (x-y)^2}=11. $
I have no idea how can I start.
algebra-precalculus summation factoring
edited Jan 6 at 20:21
Michael Rozenberg
99.3k1590189
asked Jan 6 at 17:39
rafarafa
598212
$endgroup$
Compute $frac{x}{y-z} +frac {y}{z-x}+frac {z}{x-y} $ if $frac{x^2}{ (y-z)^2} +frac {y^2}{ (z-x)^2}+frac {z^2}{ (x-y)^2}=11. $
I have no idea how can I start.
algebra-precalculus summation factoring
algebra-precalculus summation factoring
edited Jan 6 at 20:21
Michael Rozenberg
99.3k1590189
asked Jan 6 at 17:39
rafarafa
598212
edited Jan 6 at 20:21
Michael Rozenberg
99.3k1590189
asked Jan 6 at 17:39
rafarafa
598212
edited Jan 6 at 20:21
Michael Rozenberg
99.3k1590189
edited Jan 6 at 20:21
Michael Rozenberg
99.3k1590189
edited Jan 6 at 20:21
Michael Rozenberg
99.3k1590189
99.3k1590189
asked Jan 6 at 17:39
rafarafa
598212
asked Jan 6 at 17:39
rafarafa
598212
asked Jan 6 at 17:39
rafarafa
598212
598212
add a comment |
add a comment |
1 Answer
1
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$begingroup$
$$left(sum_{cyc}frac{x}{y-z}right)^2=sum_{cyc}frac{x^2}{(y-z)^2}+2sum_{cyc}frac{xy}{(y-z)(z-x)}=$$
$$=11+2sum_{cyc}frac{xy(x-y)}{(x-y)(y-z)(z-x)}=11+2frac{sumlimits_{cyc}(x^2y-x^2z)}{(x-y)(y-z)(z-x)}=$$
$$=11-frac{2(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=9.$$
answered Jan 6 at 19:33
Michael RozenbergMichael Rozenberg
99.3k1590189
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$$left(sum_{cyc}frac{x}{y-z}right)^2=sum_{cyc}frac{x^2}{(y-z)^2}+2sum_{cyc}frac{xy}{(y-z)(z-x)}=$$
$$=11+2sum_{cyc}frac{xy(x-y)}{(x-y)(y-z)(z-x)}=11+2frac{sumlimits_{cyc}(x^2y-x^2z)}{(x-y)(y-z)(z-x)}=$$
$$=11-frac{2(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=9.$$
answered Jan 6 at 19:33
Michael RozenbergMichael Rozenberg
99.3k1590189
$endgroup$
add a comment |
$begingroup$
$$left(sum_{cyc}frac{x}{y-z}right)^2=sum_{cyc}frac{x^2}{(y-z)^2}+2sum_{cyc}frac{xy}{(y-z)(z-x)}=$$
$$=11+2sum_{cyc}frac{xy(x-y)}{(x-y)(y-z)(z-x)}=11+2frac{sumlimits_{cyc}(x^2y-x^2z)}{(x-y)(y-z)(z-x)}=$$
$$=11-frac{2(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=9.$$
answered Jan 6 at 19:33
Michael RozenbergMichael Rozenberg
99.3k1590189
$endgroup$
add a comment |
$begingroup$
$$left(sum_{cyc}frac{x}{y-z}right)^2=sum_{cyc}frac{x^2}{(y-z)^2}+2sum_{cyc}frac{xy}{(y-z)(z-x)}=$$
$$=11+2sum_{cyc}frac{xy(x-y)}{(x-y)(y-z)(z-x)}=11+2frac{sumlimits_{cyc}(x^2y-x^2z)}{(x-y)(y-z)(z-x)}=$$
$$=11-frac{2(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=9.$$
answered Jan 6 at 19:33
Michael RozenbergMichael Rozenberg
99.3k1590189
$endgroup$
$$left(sum_{cyc}frac{x}{y-z}right)^2=sum_{cyc}frac{x^2}{(y-z)^2}+2sum_{cyc}frac{xy}{(y-z)(z-x)}=$$
$$=11+2sum_{cyc}frac{xy(x-y)}{(x-y)(y-z)(z-x)}=11+2frac{sumlimits_{cyc}(x^2y-x^2z)}{(x-y)(y-z)(z-x)}=$$
$$=11-frac{2(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=9.$$
answered Jan 6 at 19:33
Michael RozenbergMichael Rozenberg
99.3k1590189
answered Jan 6 at 19:33
Michael RozenbergMichael Rozenberg
99.3k1590189
answered Jan 6 at 19:33
Michael RozenbergMichael Rozenberg
99.3k1590189
answered Jan 6 at 19:33
Michael RozenbergMichael Rozenberg
99.3k1590189
99.3k1590189
add a comment |
add a comment |
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