Resposta :
a)
[tex]\dfrac{5\text{x}+2}{4\text{x}+1}=\dfrac{3}{4}[/tex]
[tex]3\cdot(4\text{x}+1)=4\cdot(5\text{x}+2)[/tex]
[tex]12\text{x}+3=20\text{x}+8[/tex]
[tex]8\text{x}=-5[/tex]
[tex]\text{x}=-\dfrac{5}{8}[/tex]
Verificação:
[tex]\dfrac{5\cdot\left(-\frac{5}{8}\right)+2}{4\cdot\left(-\frac{5}{8}\right)+1}=\dfrac{3}{4}[/tex]
[tex]\dfrac{\frac{-25}{8}+2}{\frac{-20}{8}+1}=\dfrac{3}{4}[/tex]
[tex]\dfrac{\frac{-25+16}{8}}{\frac{-20+8}{8}}=\dfrac{3}{4}[/tex]
[tex]\dfrac{-9}{-12}=\dfrac{3}{4}[/tex]
b)
[tex]\dfrac{3}{\text{x}+4}=\dfrac{4}{5\text{x}-2}[/tex]
[tex]3\cdot(5\text{x}-2)=4\cdot(\text{x}+4)[/tex]
[tex]15\text{x}-6=4\text{x}+16[/tex]
[tex]11\text{x}=22[/tex]
[tex]\text{x}=2[/tex]
Verificação:
[tex]\dfrac{3}{2+4}=\dfrac{4}{5\cdot2-2}[/tex]
[tex]\dfrac{3}{6}=\dfrac{4}{8}[/tex]
c)
[tex]\dfrac{2\text{x}-1}{3\text{x}+2}=\dfrac{9}{16}[/tex]
[tex]9\cdot(3\text{x}+2)=16\cdot(2\text{x}-1)[/tex]
[tex]27\text{x}+18=32\text{x}-16[/tex]
[tex]5\text{x}=34[/tex]
[tex]\text{x}=\dfrac{34}{5}[/tex]
Verificação:
[tex]\dfrac{2\cdot\left(\frac{34}{5}\right)-1}{3\cdot\left(\frac{34}{5}\right)+2}=\dfac{9}{16}[/tex]
[tex]\dfrac{\frac{68}{5}-1}{\frac{102}{5}+2}=\dfrac{9}{16}[/tex]
[tex]\dfrac{\frac{68-5}{5}}{\frac{102+10}{5}}=\dfrac{9}{16}[/tex]
[tex]\dfrac{63}{112}=\dfrac{9}{16}[/tex]
d)
[tex]\dfrac{\text{x}+5}{3}=\dfrac{\text{x}-1}{5}[/tex]
[tex]5\cdot(\text{x}+5)=3\cdot(\text{x}-1)[/tex]
[tex]5\text{x}+25=3\text{x}-3[/tex]
[tex]2\text{x}=-28[/tex]
[tex]\text{x}=\dfrac{-28}{2}=-14[/tex]
Verificação:
[tex]\dfrac{-14+5}{3}=\dfrac{-14-1}{5}[/tex]
[tex]\dfrac{-9}{3}=\dfrac{-15}{5}[/tex]
[tex]\left(\dfrac{3}{4}\right)^3=\dfrac{3^3}{4^3}=\dfrac{27}{64}[/tex]
[tex]\left(\dfrac{3}{4}\right)^{-3}=\dfrac{4^3}{3^3}=\dfrac{64}{27}[/tex]
[tex](5^2)^3=5^{2\cdot3}=5^6=5^3\cdot5^3=125\cdot125=15~625[/tex]
[tex](3,2)^5\cdot(3,2)^4=(3,2)^{5-4}=(3,2)^1=3,2[/tex]
[tex]12^0}=\text{a}^0=1[/tex], com [tex]\text{a}\ne0[/tex]
[tex]10^3\cdot10^{-5}=10^{3-5}=10^{-2}=\left(\dfrac{1}{10}\right)^2=\dfrac{1^2}{10^2}=\dfrac{1}{100}[/tex]
[tex]\left(\dfrac{5}{5}\right)^{-3}=1^{-3}=1^3=1[/tex]
[tex]\text{a}^{\frac{\text{b}}{\text{c}}=\sqrt[\text{c}]{\text{a}^{\text{b}}}[/tex]
[tex]4^{\frac{4}{3}}=\sqrt[3]{4^4}=\sqrt[3]{4^3\cdot4}=4\sqrt[3]{4}[/tex]
[tex]9^{\frac{2}{3}}=\sqrt[3]{9^2}=\sqrt[3]{81}=3\sqrt[3]{3}[/tex]
[tex]12^{\frac{1}{2}}=\sqrt[2]{12}=2\sqrt{3}[/tex]
[tex]\sqrt{3}+\sqrt{3}=2\sqrt{3}=\sqrt{12}[/tex]
[tex]\sqrt{4}+\sqrt{32}=2+4\sqrt{2}=2\cdot(1+2\sqrt{2}+1)[/tex]
[tex]\sqrt[3]{27}+\sqrt[3]{81}=3+3\sqrt[3]{3}=3\cdot(1+\sqrt[3]{3}[/tex]
[tex]\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}[/tex]
[tex]\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}[/tex]
[tex]\sqrt[3]{7}\cdot\sqrt{2}=\sqrt[6]{7^2}\cdot\sqrt[6]{2^3}=\sqrt[6]{7^2\cdot2^3}=\sqrt[6]{392}[/tex]
[tex]\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}[/tex]
[tex](\text{a}+\text{b})^2=\text{a}^2+2\text{a}\text{b}+\text{b}^2[/tex]
[tex](\text{a}-\text{b})^2=\text{a}^2-2\text{a}\text{b}+\text{b}^2[/tex]
[tex](\text{a}+\text{b})^2=\text{a}^2+2\text{a}\text{b}+\text{b}^2[/tex]
[tex](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}^2[/tex]
[tex](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[/tex]
[tex](7+\text{m})^2=7^2+2\cdot7\cdot\text{m}+\text{m}^2=49+14\text{m}+\text{m}^2[/tex]
[tex](\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[/tex]