Question

ENCONTRE O VALOR DE X (COM RESOLUÇÃO TUDO CERTO, DEPOIS QUE ACHAR O VALOR DE X POR FAVOR FAÇA A TROCA DO X PELOS LUGARES QUE CONTEM X PARA VERIFICAR O RESULTADO E DEIXA A CONTA A MOSTRA):

 

$\frac{5x+2}{4x+1}=\frac{3}{4} \\ \\ \frac{3}{x+4}=\frac{4}{5x-2} \\ \\ \frac{2x-1}{3x+2}=\frac{9}{16} \\ \\ \frac{x+5}{3}=\frac{x-1}{5}$

 

CALCULE OQUE SE PEDE:

$<var>(\frac{3}{4})^{3} \\ \\ (\frac{3}{4})^{-3} \\ \\ ((5)^{2})^{3} \\ \\ (3,2)^{5}.(3,2)^{-4} \\ \\ 12^{0} \\ \\ 10^{3}.10^{-5} \\ \\ (\frac{5}{5})^{-3}$

 

CALCULE

$\\ \\ (4)^{\frac{4}{3}} \\ \\ (9)^{\frac{2}{3}} \\ \\ (12)^{\frac{1}{2}} \\ \\ \sqrt{3}+\sqrt{3} \\ \\ \sqrt{4}+\sqrt{32} \\ \\ \sqrt[3]{27}+\sqrt[3]{81} \\ \\ \sqrt[3]{2}.\sqrt{8} \\ \\ \\ \sqrt[2]{4}.\sqrt[4]{3} \\ \\ \sqrt[3]{7}.\sqrt[2]{2} \\ \\ \sqrt[3]{27}.\sqrt[3]{81} \\ \\ \sqrt[3]{2}.\sqrt{8}$

CALCULE USANDO PRODUTOS NOTAVEIS

$(5x+2y)^{2} \\ \\ (3m-n)^{2} \\ \\ (7+m)^{2} \\ \\ (\frac{3}{2}+y)^{2} \\ \\ </var>$

Answer 1

a)

 

$\dfrac{5\text{x}+2}{4\text{x}+1}=\dfrac{3}{4}$

 

$3\cdot(4\text{x}+1)=4\cdot(5\text{x}+2)$

 

$12\text{x}+3=20\text{x}+8$

 

$8\text{x}=-5$

 

$\text{x}=-\dfrac{5}{8}$

 

 

Verificação:

 

$\dfrac{5\cdot\left(-\frac{5}{8}\right)+2}{4\cdot\left(-\frac{5}{8}\right)+1}=\dfrac{3}{4}$

 

$\dfrac{\frac{-25}{8}+2}{\frac{-20}{8}+1}=\dfrac{3}{4}$

 

$\dfrac{\frac{-25+16}{8}}{\frac{-20+8}{8}}=\dfrac{3}{4}$

 

$\dfrac{-9}{-12}=\dfrac{3}{4}$

 

 

b)

 

$\dfrac{3}{\text{x}+4}=\dfrac{4}{5\text{x}-2}$

 

$3\cdot(5\text{x}-2)=4\cdot(\text{x}+4)$

 

$15\text{x}-6=4\text{x}+16$

 

$11\text{x}=22$

 

$\text{x}=2$

 

 

Verificação:

 

$\dfrac{3}{2+4}=\dfrac{4}{5\cdot2-2}$

 

$\dfrac{3}{6}=\dfrac{4}{8}$

 

 

c)

 

$\dfrac{2\text{x}-1}{3\text{x}+2}=\dfrac{9}{16}$

 

$9\cdot(3\text{x}+2)=16\cdot(2\text{x}-1)$

 

$27\text{x}+18=32\text{x}-16$

 

$5\text{x}=34$

 

$\text{x}=\dfrac{34}{5}$

 

 

Verificação:

 

$\dfrac{2\cdot\left(\frac{34}{5}\right)-1}{3\cdot\left(\frac{34}{5}\right)+2}=\dfac{9}{16}$

 

$\dfrac{\frac{68}{5}-1}{\frac{102}{5}+2}=\dfrac{9}{16}$

 

$\dfrac{\frac{68-5}{5}}{\frac{102+10}{5}}=\dfrac{9}{16}$

 

$\dfrac{63}{112}=\dfrac{9}{16}$

 

 

d)

 

$\dfrac{\text{x}+5}{3}=\dfrac{\text{x}-1}{5}$

 

$5\cdot(\text{x}+5)=3\cdot(\text{x}-1)$

 

$5\text{x}+25=3\text{x}-3$

 

$2\text{x}=-28$

 

$\text{x}=\dfrac{-28}{2}=-14$

 

 

Verificação:

 

$\dfrac{-14+5}{3}=\dfrac{-14-1}{5}$

 

$\dfrac{-9}{3}=\dfrac{-15}{5}$

 

 

$\left(\dfrac{3}{4}\right)^3=\dfrac{3^3}{4^3}=\dfrac{27}{64}$

 

 

$\left(\dfrac{3}{4}\right)^{-3}=\dfrac{4^3}{3^3}=\dfrac{64}{27}$

 

 

$(5^2)^3=5^{2\cdot3}=5^6=5^3\cdot5^3=125\cdot125=15~625$

 

 

$(3,2)^5\cdot(3,2)^4=(3,2)^{5-4}=(3,2)^1=3,2$

 

 

$12^0}=\text{a}^0=1$, com $\text{a}\ne0$

 

 

$10^3\cdot10^{-5}=10^{3-5}=10^{-2}=\left(\dfrac{1}{10}\right)^2=\dfrac{1^2}{10^2}=\dfrac{1}{100}$

 

 

$\left(\dfrac{5}{5}\right)^{-3}=1^{-3}=1^3=1$

 

 

$\text{a}^{\frac{\text{b}}{\text{c}}=\sqrt[\text{c}]{\text{a}^{\text{b}}}$

 

 

$4^{\frac{4}{3}}=\sqrt[3]{4^4}=\sqrt[3]{4^3\cdot4}=4\sqrt[3]{4}$

 

 

 $9^{\frac{2}{3}}=\sqrt[3]{9^2}=\sqrt[3]{81}=3\sqrt[3]{3}$

 

 

$12^{\frac{1}{2}}=\sqrt[2]{12}=2\sqrt{3}$

 

 

$\sqrt{3}+\sqrt{3}=2\sqrt{3}=\sqrt{12}$

 

 

$\sqrt{4}+\sqrt{32}=2+4\sqrt{2}=2\cdot(1+2\sqrt{2}+1)$

 

 

$\sqrt[3]{27}+\sqrt[3]{81}=3+3\sqrt[3]{3}=3\cdot(1+\sqrt[3]{3}$

 

 

$\sqrt[3]{2}\cdot\sqrt{8}=\sqrt[6]{2^2}\cdot\sqrt[6]{8^3}=\sqrt[6]{2^2\cdot(2^3)^3}=\sqrt[6]{2^2\cdot2^9}=\sqrt[6]{2^{11}}=2\sqrt[6]{2^5}$

 

 

$\sqrt4}\cdot\sqrt[4]{3}=\sqrt[4]{4^2}\cdot\sqrt[4]{3}=\sqrt[4]{4^2\cdot3}=\sqrt[4]{48}=\sqrt[4]{2^4\cdot3}=2\sqrt[4]{3}$

 

 

$\sqrt[3]{7}\cdot\sqrt{2}=\sqrt[6]{7^2}\cdot\sqrt[6]{2^3}=\sqrt[6]{7^2\cdot2^3}=\sqrt[6]{392}$

 

 

$\sqrt[3]{27}\cdot\sqrt[3]{81}=\sqrt[3]{3^3}\cdot\sqrt[3]{3^3\cdot3}=3\cdot3\sqrt[3]{3}=9\sqrt[3]{3}$

 

 

$(\text{a}+\text{b})^2=\text{a}^2+2\text{a}\text{b}+\text{b}^2$

 

 

$(\text{a}-\text{b})^2=\text{a}^2-2\text{a}\text{b}+\text{b}^2$

 

 

$(\text{a}+\text{b})^2=\text{a}^2+2\text{a}\text{b}+\text{b}^2$

 

$(5\text{x}+2\text{y})^2=(5\text{x})^2+2\cdot5\text{x}\cdot2\text{y}+(2\text{y})^2=25\text{x}^2+20\text{xy}+4\text{y}</span><span>^2$

 

 

$(3\text{m}-\text{n})^2=(3\text{m})^2-2\cdot3\text{m}\cdot\text{n}+\text{n}^2=9\text{m}^2-6\text{mn}+\text{n}^2$

 

 

$(7+\text{m})^2=7^2+2\cdot7\cdot\text{m}+\text{m}^2=49+14\text{m}+\text{m}^2$

 

 

$(\frac{3}{2}+\text{y})^2=\left(\frac{3}{2}\right)^2+2\cdot\frac{3}{2}\cdot\text{y}+\text{y}^2=\dfrac{9}{4}+3\text{y}+\text{y}^2$