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klio [65]
3 years ago
15

Help me with this attachment please?

Mathematics
2 answers:
Ilya [14]3 years ago
5 0

The three solutions are:

  • a = 1,\ b = 3. In fact, 1+333+333+333 = 1333
  • a = 2,\ b = 6. In fact, 2+666+666+666+666 = 2666
  • a = 3,\ b = 9. In fact, 3+999+999+999+999 = 3999

You can find the first one with a bit of trial and error. The second and third derive from the fact that if you multiply by 2 or 3 every term, the whole result will be multiplied by 2 or 3 as well

Volgvan3 years ago
4 0

<u>Answer</u>

1st solution: a = 1

                   b = 3

2nd solution: a = 2

                      b = 6


<u>Explanation</u>

1st solution

                  <em>         1</em>

<em>                       333</em>

<em>                       333</em>

<em>                       333</em>

<em>                   </em><em><u>  +333</u></em>

<em>               </em><em><u>       1333</u></em>

2nd solution

<em>         2</em>

<em>    666</em>

<em>    666</em>

<em>    666</em>

<em><u> + 666</u></em>

<em><u> 2666</u></em>

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(a-b)^4

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Substitute i = 2, we get

$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}

Substitute i = 3, we get

$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}

Substitute i = 4, we get

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$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}

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Hence the second term of the expansion is -4a^3b.

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