Question

buna!!ma poate ajuta cineva la aceste exercitii?
[tex]1. \sqrt{ \frac{27}{-3} }+ \sqrt{ \frac{125}{64} }



2.log _{3} 15+log3 \frac{8}{5} -log _{3} 8
}


3. \sqrt{5} -x=4

4.8X3 ^{x-2} +2X3 ^{x-1} +3 ^{x}X 23 (X este inmultire)


5.log _{5} (10-x)=2


6.log _{2}(2x+5)=log _{7} (x^{2} +3x+3)

7.log _{5} (3x+4)=2

8.log _{3}(x ^{2} -2x)=log _{3} (2x-3)


9.log _{7} (2x+1)=2

[/tex]

Answer 1

$\displaystyle 1). \sqrt{ \frac{27}{-3} } + \sqrt{ \frac{125}{64} } = \sqrt{-9} + \frac{ \sqrt{125} }{ \sqrt{64} } =3i+ \frac{5 \sqrt{5} }{8} =\boxed{ \frac{24i+5 \sqrt{5} }{8}} \\ 2).log_315+log_3 \frac{8}{5}-log_38=log_3 \left(15 \cdot \frac{8}{5} \right)-log_38=log_3 \frac{120}{5}-log_38= \\ =log_324-log_38=log_3 \frac{24}{8} =log_33=\boxed{1} \\ 3). \sqrt{5} -x=4 \Rightarrow -x=4- \sqrt{5} \Rightarrow x= \boxed{\sqrt{5} -4}$
$\displaystyle 4).8 \cdot 3^{x-2}+2 \cdot 3^{x-1}+3^x \cdot 23=3^x(8 \cdot 3^{-2}+2 \cdot 3^{-1}+1 \cdot 23)= \\ =3^x \left(8 \cdot \frac{1}{3^2} +2 \cdot \frac{1}{3^1} +23 \right)=3^x \left( 8 \cdot \frac{1}{9} +2 \cdot \frac{1}{3} +23 \right)= \\ =3^x \left( \frac{8}{9} + \frac{2}{3} +23 \right)=3^x \left( \frac{8}{9} + \frac{6}{9} + \frac{207}{9} \right)=3^x \cdot \frac{221}{9}= \\ = \frac{3^x \cdot 221}{9} = \frac{3^x \cdot 221}{3^2} =\boxed{3^{x-2} \cdot 221}$
$\displaystyle 5).log_5(10-x)=2 \\ log_5(10-x)=log_55^2 \\ 10-x=5^2 \\ 10-x=25 \\ -x=25-10 \\ -x=15 \\ \boxed{x=-15}$
$\displaystyle 7).log_5(3x+4)=2 \\ log_5(3x+4)=log_55^2 \\ 3x+4=5^2 \\ 3x+4=25 \\ 3x=25-4 \\ 3x=21 \\ \boxed{x=7}$
$\displaystyle 8).log_3(x^2-2x)=log_3(2x-3) \\ x^2-2x=2x-3 \\ x^2-2x-2x=-3 \\ x^2-4x=-3 \\ x^2-4x+3=0 \\ \Delta=16-12=4 \\ x_1= \frac{4+2}{2}= \frac{6}{2} =3 \\ x_2= \frac{4-2}{2}= \frac{2}{2} =1 \\ \boxed{x=3}$
$\displaystyle 9).log_7(2x+1)=2 \\ log_7(2x+1)=log_77^2 \\ 2x+1=7^2 \\ 2x+1=49 \\ 2x=49-1 \\ 2x=48 \\ \boxed{x=24 }$