How to prove this algebra question? [closed]
If $x=y^2+z^2$, $y=x^2+z^2,$ and $z=x^2+y^2$ then show that $$frac{x}{x+1}+frac{y}{y+1}+frac{z}{z+1}=1$$
algebra-precalculus
edited Dec 1 at 17:20
Sean Roberson
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asked Dec 1 at 17:07
Abhishek K.
164
closed as off-topic by Xander Henderson, darij grinberg, user10354138, Alexander Gruber♦ Dec 4 at 3:38
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If $x=y^2+z^2$, $y=x^2+z^2,$ and $z=x^2+y^2$ then show that $$frac{x}{x+1}+frac{y}{y+1}+frac{z}{z+1}=1$$
algebra-precalculus
edited Dec 1 at 17:20
Sean Roberson
6,40531327
asked Dec 1 at 17:07
Abhishek K.
164
closed as off-topic by Xander Henderson, darij grinberg, user10354138, Alexander Gruber♦ Dec 4 at 3:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, darij grinberg, user10354138, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
If $x=y^2+z^2$, $y=x^2+z^2,$ and $z=x^2+y^2$ then show that $$frac{x}{x+1}+frac{y}{y+1}+frac{z}{z+1}=1$$
algebra-precalculus
edited Dec 1 at 17:20
Sean Roberson
6,40531327
asked Dec 1 at 17:07
Abhishek K.
164
If $x=y^2+z^2$, $y=x^2+z^2,$ and $z=x^2+y^2$ then show that $$frac{x}{x+1}+frac{y}{y+1}+frac{z}{z+1}=1$$
algebra-precalculus
algebra-precalculus
edited Dec 1 at 17:20
Sean Roberson
6,40531327
asked Dec 1 at 17:07
Abhishek K.
164
edited Dec 1 at 17:20
Sean Roberson
6,40531327
asked Dec 1 at 17:07
Abhishek K.
164
edited Dec 1 at 17:20
Sean Roberson
6,40531327
edited Dec 1 at 17:20
Sean Roberson
6,40531327
edited Dec 1 at 17:20
Sean Roberson
6,40531327
6,40531327
asked Dec 1 at 17:07
Abhishek K.
164
asked Dec 1 at 17:07
Abhishek K.
164
asked Dec 1 at 17:07
Abhishek K.
164
164
closed as off-topic by Xander Henderson, darij grinberg, user10354138, Alexander Gruber♦ Dec 4 at 3:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, darij grinberg, user10354138, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Xander Henderson, darij grinberg, user10354138, Alexander Gruber♦ Dec 4 at 3:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, darij grinberg, user10354138, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
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2 Answers
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hint
$x-y=y^2-x^2=(y-x)(y+x)$. So either $x=y$ or $x+y+1=0$. Do the same with other pairs.
answered Dec 1 at 17:37
Anurag A
25.6k12249
add a comment |
$$x^2+y^2+z^2=x(x+1)impliesdfrac x{x+1}=?$$
answered Dec 1 at 17:50
lab bhattacharjee
222k15156274
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
hint
$x-y=y^2-x^2=(y-x)(y+x)$. So either $x=y$ or $x+y+1=0$. Do the same with other pairs.
answered Dec 1 at 17:37
Anurag A
25.6k12249
add a comment |
hint
$x-y=y^2-x^2=(y-x)(y+x)$. So either $x=y$ or $x+y+1=0$. Do the same with other pairs.
answered Dec 1 at 17:37
Anurag A
25.6k12249
add a comment |
hint
$x-y=y^2-x^2=(y-x)(y+x)$. So either $x=y$ or $x+y+1=0$. Do the same with other pairs.
answered Dec 1 at 17:37
Anurag A
25.6k12249
hint
$x-y=y^2-x^2=(y-x)(y+x)$. So either $x=y$ or $x+y+1=0$. Do the same with other pairs.
answered Dec 1 at 17:37
Anurag A
25.6k12249
answered Dec 1 at 17:37
Anurag A
25.6k12249
answered Dec 1 at 17:37
Anurag A
25.6k12249
answered Dec 1 at 17:37
Anurag A
25.6k12249
25.6k12249
add a comment |
add a comment |
$$x^2+y^2+z^2=x(x+1)impliesdfrac x{x+1}=?$$
answered Dec 1 at 17:50
lab bhattacharjee
222k15156274
add a comment |
$$x^2+y^2+z^2=x(x+1)impliesdfrac x{x+1}=?$$
answered Dec 1 at 17:50
lab bhattacharjee
222k15156274
add a comment |
$$x^2+y^2+z^2=x(x+1)impliesdfrac x{x+1}=?$$
answered Dec 1 at 17:50
lab bhattacharjee
222k15156274
$$x^2+y^2+z^2=x(x+1)impliesdfrac x{x+1}=?$$
answered Dec 1 at 17:50
lab bhattacharjee
222k15156274
answered Dec 1 at 17:50
lab bhattacharjee
222k15156274
answered Dec 1 at 17:50
lab bhattacharjee
222k15156274
answered Dec 1 at 17:50
lab bhattacharjee
222k15156274
222k15156274
add a comment |
add a comment |
