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ICE Princess25 [194]
3 years ago
8

Help me please fast ​

Mathematics
1 answer:
Natalija [7]3 years ago
3 0

Answer:

Step-by-step explanation:

\frac{5}{x-3}+\frac{2}{x+1}=3  \\\\\frac{5(x+1)+2(x-3)}{(x-3)(x+1)} =3\\\\5x+5+2x-6=3(x-3)(x+1)\\\\7x-1=3(x^2+x-3x-3)\\\\7x-1=3(x^2-2x-3)\\\\7x-1=3x^2-6x-9\\\\0=3x^2-6x-7x-9+1\\0=3x^2-13x-8

Hence proved!

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7 0
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I am a number greater than 40,000 and less than 60,000. My ones digit and tens digit are the same. My ten-thousands digit is 1 l
jarptica [38.1K]
I am a number greater than 40,000 and less than 60,000:

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4 ≤ n₁ < 6

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My ten thousands digit is 1 less than 3 times the sum of my ones digit and tens digit:

n₁ = 3*2n₄ - 1

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This means that:

n = 10,000*(6n₄-1) + 1,000n₂ + 100n₃ + 11n₄

n = 60,000n₄ - 10,000 + 1,000n₂ + 100n₃ + 11n₄

n = 60,011n₄ - 10,000 + 1,000n₂ + 100n₃

<span>My thousands digit is half my hundreds digit, and the sum of those two digits is 9:

n</span>₂ = 1/2 * n₃
<span>
n</span>₂ + n₃ = 9
<span>
Therefore:

n</span>₂ = 9 - n₃
<span>
Therefore:

9 - n</span>₃ = 1/2 * n₃
<span>
9 = 1/2 * n</span>₃ + n₃
<span>
9 = 1.5 * n</span>₃
<span>
Therefore:

n</span>₃ = 6
<span>
If n</span>₃=6, n₂=3.
<span>
This means that:

</span>n = 60,011n₄ - 10,000 + 1,000*3 + 100*6

n = 60,011n₄ - 10,000 + 3,000 + 600

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