Question

Ex mate clasa a 10 a : 

    1. $\sqrt[3]{3x+2} = 2$
    2. $\sqrt{x-2} + x = 4$
    3. $5^{2} = \frac{1}{25}$
    4. [tex] 2^{x} + 2^{x+1} = 12
[/tex]
    5. $9^{x} + 3^{x} = 12$
    6. $log_{5} ( 2x-1 ) = 2$
    7. $log_{ 2^{x} } + log_{2} (x+2) = 3$
    8. $log_{3} (x+1 ) - log_{3} (x-1 ) = 1$

Answer 1

1. $(\sqrt[3]{3x+2})^{3} = 2^{3} => 3x+2=8 =>3x=6/:3=> x=2$
2.$\sqrt{x-2} =4-x=> (\sqrt{x-2})^{2} = (4-x)^{2} => x-2=16-8x+ x^{2}$=>$x^{2} -9x+18$
Δ=$b^{2} -4ac=81-72=9$
$x_{1,2} = \frac{-b+- \sqrt{delta} }{2a} => x_{1,2} = \frac{9+-3}{2} => \left \{ {{ x_{1} =6} \atop { x_{2} =2}} \right.$
3.Cred ca ai gresit, aia nu e ecuatie.
4. $2^{x} +2^{x+1} = 3* 2^{2} => x+x+1=2=> 2x=1=> x= \frac{1}{2}$
5.$3^{2x} + 3^{x} =3* 2^{2} => 2x+x=1=> 3x=1=> x= \frac{1}{3}$
6. $2x-1= 5^{2} => 2x=26=> x=13$
7. $\frac{x}{x+2} =8=> x=8x+16=> x=- \frac{16}{7}$
8.$\frac{x+1}{x-1} =3=>x+1=3x-3=> -2x=-4=>x=2$