Question

1) Calcule o valor numérico das expressões algébricas:

a) a²b - ab², para a = -3/2 e b = 2/3

b) (a - 2) (a - 1) (a - 4), para a = -1

c) a² + 2 ab+ b², para a = 5 e b = 3
---------a² - b²----------------------------

d) 3x² - 2x, para x = 3/4
-----5x - 1--------------------------------

e) √x + x², para x = 4 e y = 1/4
------√y----------------------

f) b² - 4 ac, para a = 1,b = 2 e c = - 15.

g) 3x +2 xy, para x = 2 e y = 1/2
-----x+4--------------------------------------

Answer 1

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$\boxed{\begin{array}{l}\tt a)~\sf a^2b-ab^2=\bigg(-\dfrac{3}{2}\bigg)^2\cdot\dfrac{2}{3}-\bigg(-\dfrac{2}{3}\bigg)\cdot\bigg(\dfrac{2}{3}\bigg)^2\\\sf a^2b-ab^2=\dfrac{9}{4}\cdot\dfrac{2}{3}-\bigg(-\dfrac{2}{3}\bigg)\cdot\dfrac{4}{9}\\\sf a^2b-ab^2=\dfrac{18}{12}+\dfrac{8}{27}=\dfrac{162+32}{108}\\\sf a^2b-ab^2=\dfrac{194\div2}{108\div2}\\\\\sf a^2b-ab^2=\dfrac{97}{54}\end{array}}$

$\boxed{\begin{array}{l}\tt b)~\sf(a-2)(a-1)(a-4)=(-1-2)(-1-1)(-1-4)\\\sf(a-2)(a-1)(a-4)=(-3)(-2)(-5)=-30\end{array}}$

$\large\boxed{\begin{array}{l}\tt c)~\sf\dfrac{a^2+2ab+b^2}{a^2-b^2}=\dfrac{\diagdown\!\!\!\!\!\!(a-\diagdown\!\!\!\!\!\!b)(a-b)}{\diagdown\!\!\!\!\!\!(a-\diagdown\!\!\!\!\!\!b)(a+b)}=\dfrac{a-b}{a+b}\\\sf =\dfrac{5-3}{5+3}=\dfrac{2\div2}{8\div2}=\dfrac{1}{4}\end{array}}$

$\large\boxed{\begin{array}{l}\tt d)~\sf\dfrac{3x^2-2x}{5x-1}=\dfrac{3\cdot\bigg(\frac{3}{4}\bigg)^2-\diagdown\!\!\!\!2\cdot\bigg(\frac{3}{\diagdown\!\!\!\!4}\bigg)}{5\cdot\frac{3}{4}-1}\\\sf=\dfrac{3\cdot\frac{9}{16}-\frac{3}{2}}{\frac{15}{4}-1}=\dfrac{\frac{27}{16}-\frac{3}{2}}{\frac{15-4}{4}}\\\sf\dfrac{\frac{27-24}{16}}{\frac{11}{4}}=\dfrac{\frac{3}{16}}{\frac{11}{4}}=\dfrac{3}{\diagdown\!\!\!\!\!\!16_4}\cdot\dfrac{\diagdown\!\!\!\!\!4^1}{11}=\dfrac{3}{44}\end{array}}$

$\large\boxed{\begin{array}{l}\tt e)~\sf\dfrac{\sqrt{x+x^2}}{\sqrt{y}}=\sqrt{\dfrac{x+x^2}{y}}=\sqrt{\dfrac{4+4^2}{\bigg(\frac{1}{4}\bigg)^2}}\\\sf =\sqrt{\dfrac{4+16}{\frac{1}{16}}}=\sqrt{\dfrac{20}{\frac{1}{16}}}=\sqrt{20\cdot16}\\\sf=\sqrt{2^2\cdot5\cdot2^4}=2\cdot2^2\sqrt{5}=8\sqrt{5}\end{array}}$

$\large\boxed{\begin{array}{l}\tt f)~\sf b^2-4ac=2^2-4\cdot1\cdot(-15)\\\sf b^2-4ac=4+60=64\end{array}}$

$\large\boxed{\begin{array}{l}\tt g)~\sf\dfrac{3x+2xy}{x+4}=\dfrac{3\cdot2+2\cdot\backslash\!\!\!2\cdot\frac{1}{\backslash\!\!\!2}}{2+4}\\\sf=\dfrac{6+2}{6}=\dfrac{8\div2}{6\div2}=\dfrac{4}{3}\end{array}}$

Answer 2

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