Question

Efectua e simplifica as seguintes expressões algébricas:
(as expressões estão na foto anexada)
Me ajudem por favor​

Answer 1

Resposta:

Se temos apenas que simplificar, fica mais fácil. Vamos lá:

a) $\frac{3}{2a^2}-\frac{3}{2a}-\frac{4}{5}$

$MMC \; de \; 2a^2,2a,5 = 10a^2\\\\\rightarrow \frac{3}{2a^2}=\frac{3\cdot \:5}{2a^2\cdot \:5}=\frac{15}{10a^2}\\\\\rightarrow -\frac{3}{2a}=-\frac{3\cdot \:5a}{2a\cdot \:5a}=-\frac{15a}{10a^2}\\\\\rightarrow -\frac{4}{5}=\frac{4\cdot \:2a^2}{5\cdot \:2a^2}=\frac{8a^2}{10a^2}\\\\Logo, \; =\frac{15}{10a^2}-\frac{15a}{10a^2}-\frac{8a^2}{10a^2} =\frac{15-15a-8a^2}{10a^2}$

b) $\frac{1}{x^2+2x-1}-\frac{1}{x^2-1}$

$\mathrm{Fatorando}\:\;x^2-1= \left(x+1\right)\left(x-1\right)\\\\=\frac{1}{x^2+2x-1}-\frac{1}{\left(x+1\right)\left(x-1\right)}\\\\MMC \; de \; x^2+2x-1,\:\left(x+1\right)\left(x-1\right) = \left(x+1\right)\left(x-1\right)\left(x^2+2x-1\right)$$\rightarrow \frac{1}{x^2+2x-1}=\frac{1\cdot \left(x+1\right)\left(x-1\right)}{\left(x^2+2x-1\right)\left(x+1\right)\left(x-1\right)}=\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)\left(x^2+2x-1\right)}\\\\\rightarrow -\frac{1}{\left(x+1\right)\left(x-1\right)}=-\frac{1\cdot \left(x^2+2x-1\right)}{\left(x+1\right)\left(x-1\right)\left(x^2+2x-1\right)}=-\frac{x^2+2x-1}{\left(x+1\right)\left(x-1\right)\left(x^2+2x-1\right)}$$= \frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)\left(x^2+2x-1\right)}-\frac{x^2+2x-1}{\left(x+1\right)\left(x-1\right)\left(x^2+2x-1\right)} =\frac{\left(x+1\right)\left(x-1\right)-\left(x^2+2x-1\right)}{\left(x+1\right)\left(x-1\right)\left(x^2+2x-1\right)}\\\\\rightarrow \left(x+1\right)\left(x-1\right)-\left(x^2+2x-1\right) = x^2-1-\left(x^2+2x-1\right) = x^2-1-x^2-2x+1 = -2x\\\\=\frac{-2x}{\left(x+1\right)\left(x-1\right)\left(x^2+2x-1\right)}$

c) $\frac{x-2}{x^2-4}+\frac{x-2}{x}$

$\mathrm{Fatorando}\;\:x^2-4= \left(x+2\right)\left(x-2\right)\\\\=\frac{x-2}{\left(x+2\right)\left(x-2\right)} +\frac{x-2}{x}\\\\=\frac{1}{x+2}+\frac{x-2}{x}\\\\MMC \; de \; x+2,\:x = x\left(x+2\right)\\\\\rightarrow \frac{1}{x+2}=\frac{1\cdot \:x}{\left(x+2\right)x}=\frac{x}{x\left(x+2\right)}\\\\\rightarrow \frac{x-2}{x}=\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}\\\\=\frac{x}{x\left(x+2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}$

$=\frac{x+\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}$

d) $\frac{5x+10}{x^2-1} / \frac{3x+6}{x+1}$

$=\frac{\left(5x+10\right)\left(x+1\right)}{\left(x^2-1\right)\left(3x+6\right)}=\frac{5\left(x+2\right)\left(x+1\right)}{\left(x^2-1\right)\left(3x+6\right)}=\frac{5\left(x+2\right)\left(x+1\right)}{\left(x^2-1\right)\cdot \:3\left(x+2\right)}\\\\=\frac{5\left(x+1\right)}{3\left(x^2-1\right)} =\frac{5\left(x+1\right)}{3\left(x+1\right)\left(x-1\right)}\\\\=\frac{5}{3\left(x-1\right)}$

e) $\frac{12ax-9bx}{a^2-9} / \frac{4a-3b}{a^2x-9x}$

$=\frac{\left(12ax-9bx\right)\left(a^2x-9x\right)}{\left(a^2-9\right)\left(4a-3b\right)}=\frac{3x^2\left(4a-3b\right)\left(a^2-9\right)}{\left(a^2-9\right)\left(4a-3b\right)}\\\\=\frac{3x^2\left(a^2-9\right)}{a^2-9}\\\\=3x^2$

f) $\frac{x^2-y^2}{x^2-2xy+y^2} / \frac{x^3+y^3}{x^3-x^2y}$

$=\frac{\left(x^2-y^2\right)\left(x^3-x^2y\right)}{\left(x^2-2xy+y^2\right)\left(x^3+y^3\right)}\\\\=\frac{x^2\left(x-y\right)\left(x^2-y^2\right)}{\left(x^2-2xy+y^2\right)\left(x^3+y^3\right)} =\frac{x^2\left(x-y\right)\left(x^2-y^2\right)}{\left(x-y\right)^2\left(x^3+y^3\right)}\\\\=\frac{x^2\left(x^2-y^2\right)}{\left(x-y\right)\left(x^3+y^3\right)} =\frac{x^2\left(x+y\right)\left(x-y\right)}{\left(x-y\right)\left(x^3+y^3\right)}$

$=\frac{x^2\left(x+y\right)\left(x-y\right)}{\left(x-y\right)\left(x+y\right)\left(x^2-xy+y^2\right)}\\\\=\frac{x^2}{x^2-xy+y^2}$