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Earmama's Math. Q: x, y 為實數, 且 [x+(x^2+1)^1/2][y+(y^2+1)^1/2] =1, please find: x+y =??
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青楓白霧
紫色2022
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青楓白霧 : 幹嘛自言自語.....
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earmama:...通通==>
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x+y =??= y+x
⎝( ̄(∞) ̄)⎠⊿
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x+y=0, 因為 [(x^2+1)^1/2 +x ] [(x^2+1)^1/2 -x] =1 ~~
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earmama: x= -y 帶入第2個 [(x^2+1)^1/2 +x ] [(x^2+1)^1/2 -x] =1
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earmama:[x+(x^2+1)^1/2] =1/[y+(y^2+1)^1/2] =[(y^2+1)^1/2] -y... (1) and [y+(y^2+1)^1/2] =1/[x+(x^2+1)^1/2] = =[(x^2+1)^1/2 -x....(2) , (1)+(2) 得 x+y= -(x+y) , x+y=0 [emo130]
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earmama: i need to write 4 times to teach a F..o..o..l
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earmama: 3 times only....
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