Question

alguém que sabia expressões algébricas pode responder pra mim, não consigo entender as questões (tem que deixar a conta)

Answer 1

Resposta:

$a)\\ \\ \frac{a^{2}-2a}{\sqrt{a}} ~~a=4~~\frac{4^{2}-2.4}{\sqrt{4}} =\frac{16-8}{2} =\frac{8}{2} =4\\ \\ b)\\ \\ m^{2}-2mn+n^{2}~~m=-1~~n=\frac{1}{4}\\ \\ -1^{2} -2.-1.\frac{1}{4} +(\frac{1}{4})^{2}=1-2.(-1).(\frac{1}{4})+\frac{1}{16} =1+\frac{2}{4} +\frac{1}{16}=\\  \\ 1+\frac{1}{2} +\frac{1}{16} ~~mmc=16\\ \\ \frac{16+8+1}{16} =\frac{25}{16}$

$c)\\ \\ \sqrt{\frac{a^{2}+ax}{m}}~~~~a=8,~x=10,~m=9\\  \\ \sqrt{\frac{8^{2}+8.10}{9}} =\sqrt{\frac{64+80}{9}} =\sqrt{\frac{144}{9}}=\sqrt{16} =4$

$d)\\ \\ 3(x^{2}- y^{2})-10(x+y)(x-y)~~~~x=-2,~~y=-2\\  \\ 3x^{2} -3y^{2}-10(x^{2}-y^{2})=3.(-2)^{2}-3.(-2)^{2}-10(-2^{2}-(-2)^{2} )=\\ \\ 3.4-3.4-10(4-4)=12-12-10.0=0$

$e)\\ \\ (a-b)^{2} -c^{2} ~~~~a=\frac{2}{3},~b=1,~c=-1\\ \\ a^{2}-2.a.b+b^{2} -c^{2} =(\frac{2}{3})^{2}-2.(\frac{2}{3}).1+(-1^{2}) =\\ \\ \frac{4}{9} -\frac{4}{3} +1~~~~mmc=9\\ \\ \frac{4-12+9}{9} =\frac{1}{9}$

$f)\\ \\ \frac{1-x^{2} }{xy+1} ~~~~x=0,5;~~y=-8\\ \\ \frac{1-(0,5)^{2}}{-8(0,5)+1}=\frac{1-0,25}{-4+1} =\frac{0,75}{-3} =-0,25$