Question

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Descompuneti in factori: (poza)
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Answer 1

Răspuns:

$1 + \frac{3}{5} a + \frac{9}{25} {a}^{2} + \frac{27}{125} {a}^{3} = 1 + \frac{3}{5} a + ( \frac{3}{5} ) ^{2} \times {a}^{2} + (\frac{3}{5} ) ^{3} \times {a}^{3} = \\ = 1 + \frac{3a}{5} + {( \frac{3a}{5} })^{2} +( \frac{3a}{5} ) ^{3} = \\ = \frac{3a}{5} (1 \div \frac{3a}{5} + 1 + \frac{3a}{5} + \frac{3 {a}^{2} }{5} ) = \\ = \frac{3a}{5} ( \frac{5}{3a} + 1 + \frac{3a + 3 {a}^{2} }{5} ) = \\ = \frac{3a}{5} ( \frac{5}{3a} + \frac{5}{5} + \frac{3a + 3 {a}^{2} }{5} ) \\ = \frac{3a}{5} ( \frac{5}{3a} + \frac{5 + 3a + 3 {a}^{2} }{5} ) = \frac{3a}{5} ( \frac{25}{15a} + \frac{(3 {a}^{2} + 3a + 5) \times 3a}{15a} ) = \\ = \frac{3a}{5} ( \frac{25}{15 {a} } + \frac{9 {a}^{3} + 9 {a}^{2} + 15a }{15a} ) \\ = \frac{3a}{5} ( \frac{9 {a}^{3} + 9 {a}^{2} + 15a + 25 }{15a} ) = \\ = \frac{3a}{5} ( \frac{3a(3 {a}^{2} + 3a + 5) + 25 }{15a} )$

Answer 2

 

$\displaystyle\bf\\1+\frac{3}{5}a+\frac{9}{25}a^2+\frac{27}{125}a^3=\\\\\\=1+\frac{3}{5}a+\frac{3^2}{5^2}a^2+\frac{3^3}{5^3}a^3=\\\\\\=\Bigg(1+\frac{3}{5}a\Bigg)+\Bigg(\frac{3}{5}a\Bigg)^2+\Bigg(\frac{3}{5}a\Bigg)^3=\\\\\\=\Bigg(1+\frac{3}{5}a\Bigg)+\Bigg(\frac{3}{5}a\Bigg)^2\Bigg(1+\frac{3}{5}a\Bigg)= \\\\\\=\Bigg(1+\frac{3}{5}a\Bigg)\left(1+\Bigg(\frac{3}{5}a\Bigg)^2\right)=\\\\\\=\Bigg(1+\frac{3}{5}a\Bigg)\Bigg(1+\frac{3^2}{5^2}a^2\Bigg)$