PROSZĘ O ROZWIĄZANIE DAJE NAJ !!!! Ile kto umie niech tyle rozwiaże prosze bardzo o pomoc :) 1.ROZWI.... Question from @Didka96

January 2019 1 17 Report

PROSZĘ O ROZWIĄZANIE DAJE NAJ !!!! Ile kto umie niech tyle rozwiaże prosze bardzo o pomoc :)
1.ROZWIAZ ROWNANIA
a ) 3 do potegi x+1 - 3 do potegi x - 3 do potegi x-1 =15
b )3 do potegi w nawiasie x+1 poza nawiasem w potedze +1, - 5 * 3 do potegi znowu w nawiasie x+1 i po za nawiasem w potedzie -1 =12
c) 7 do potegi 2x + 4 * 7 do potegi x - 5 = 0
D) logdo potegi 2 i potem liczba 128
e) log do pot 9 i liczba 729
f) log do potegi 1/2 i potem 16 i pierwiastek trzeciego stopnia z 2
g/ log do potegi 3 1/6 + + log do potegi 3 z 2/3
PLUS TO CO W ZAŁACZNIKACH TYLKO ŻE TYLKO PRZYKŁADY B INNYCH NIE TRZEBA ROZWIAZYWAC

Epis $a) 3^{x+1}-3^{x}-3^{x-1} = 15\\3^{x}*3-3^{x}-3^{x}*3^{-1} = 15\\3^{x}(3-1-\frac{1}{3}) = 15\\3^{x}*\frac{5}{3}=15\\3^{x} = 9\\3^{x} = 3^{2}\\x = 2$

$b) 3^{(x+1)+1}-5*3^{(x+1)-1}=12\\3^{x+2}-5*3^{x} = 12\\3^{x}*3^{2}-5*3^{x} = 12\\3^{x} (9-5) = 12\\3^{x}*4 = 12\\3^{x} = 3\\x = 1$

$c) 7^{2x}+4*7^{x}-5 = 0\\7^{x} = t\\t^{2}+4t - 5 = 0\\ \Delta = 16+20 = 36\\ \sqrt{\Delta} = 6\\ t_{1} = \frac{-4-6}{2} = -5\\t_{2} = \frac{-4+6}{2}=1\\ 7^{x} = 1\\x = 0$

$d) log_{2}128 =x\\2^{x} = 128\\2^{x} = 2^{7}\\x = 7$

$e) log_{9}729 = a\\9^{a} = 729\\9^{a} = 9^{3}\\a = 3$

$f)log_{\frac{1}{2}}16\sqrt[3]{2}=b\\(\frac{1}{2})^{b} = 16*2^{\frac{1}{3}}\\(\frac{1}{2})^{b} =(\frac{1}{2})^{-4}*(\frac{1}{2})^{\frac{-1}{3}}\\(\frac{1}{2})^{b} =(\frac{1}{2})^{-4\frac{1}{3}} = (\frac{1}{2})^{-\frac{13}{3}} \\b = -\frac{13}{3}}$

$log_{3}\frac{1}{6}+log_{3}\frac{2}{3} = log_{3} \frac{1}{6}*\frac{2}{3}=log_{3} \frac{1}{9} = -2$

Zad. z załączników:
1.57
$2^{3x}-2^{2x+3}-2^{x+4}+128 = 0\\2^{3x}-2^{2x}*2^{3}-2^{x}*2^{4}+128 = 0\\2^{x} = t\\t^{3}-t^{2}*8-t*16+128=0\\t^{2}(t-8)-16(t-8)=0\\(t-8)(t^{2}-16) = 0\\(t-8)(t-8)(t+8) = 0\\t = 8\\ lub\\t = -8\\2^{x} = 8 = 2^{3}\\x = 3$

1.58
$\frac{1}{2^{x}-2}+2^{1-x} =1\\\frac{1}{2^{x}-2}+2*2^{-x} =1\\ 2*(2^{x}-2)*\frac{1}{2^{x}}=2^{x}-2\\2*2^{x}-4*\frac{1}{2^{x}} = 2^{x} -3\\2*2^{x}*2^{x}-4=2^{x}*2^{x}-3*2^{x}\\2*2^{2x}-4=2^{2x}-3*2^{x}\\2^{2x}+3*2^{x}-4 = 0\\a = 2^{x}\\a^{2}+3a-4 = 0\\ \Delta = 9+16 = 25\\ \sqrt{\Delta}=5\\a_{1} =\frac{-3-5}{2} = -4\\a_{2} = \frac{-3+5}{2}=1\\1 = 2^{x}\\x = 0$

1.64
$(\frac{3}{5})^{x} \geq \frac{25}{9}\\(\frac{3}{5})^{x} \geq (\frac{3}{5})^{-2}\\x \geq -2\\\ \textless \ -2, \infty)$

1.65
$(\sqrt{2})^{|x-1|} \ \textless \ 4\\2^{\frac{|x-1|}{2}}\ \textless \ 2^{2}\\ \frac{|x-1|}{2} \ \textless \ 2\\ |x-1| \ \textless \ 4\\ x-1 \ \textless \ 4\\x \ \textless \ 3\\x-1 \ \textgreater \ -4\\x \ \textgreater \ -3\\(-3,3)$

1.115
$\frac{x}{1-x} \ \textgreater \ 0\\x(1-x) \ \textgreater \ 0\\ D = (0,1)$

1.116
3-x > 0
x < 3
x² > 0
x∈(0,3)

1.117
x²+mx+1 > 0
Δ <0
Δ = m²-4 < 0
(m-2)(m+2) <0
m ∈ (-2,2)

1.46
$(\frac{25}{4})^{2x-1} = (\frac{2}{5})^{3x-12}\\(\frac{2}{5})^{-2(2x-1)} = (\frac{2}{5})^{3x-12}\\-4x+2 = 3x - 12\\7x = 14\\x = 2$

1.47
$x \neq 0\\(\frac{3}{4})^{x}=(\frac{3}{4})^{2(\frac{2x-2}{x})}\\x = \frac{4x-4}{x}\\x^{2} = 4x -4\\x^{2}-4x +4 = 0\\(x-2)^{2} = 0\\x = 2$

1.49
$( \frac{5^{2}*5^{\frac{1}{2}}}{5^{-1}} )^{2x}=5^{x^{2}+10}\\(5^{2+1+\frac{1}{2}})^{2x} = 5^{x^{2}+10}\\5^{\frac{7}{2}*2x}=5^{x^{2}+10}\\5^{7x}=5^{x^{2}+10}\\7x = x^{2}+10 \\x^{2}-7x+10 = 0\\ \Delta = 49-40 = 9\\ \sqrt{\Delta} = 6\\x_{1} = \frac{7+6}{2} = \frac{13}{2}\\x_{2} = \frac{7-6}{2 }= \frac{1}{2}$

1.50
$2^{x-x^{2}}*2^{3x} =2^{x-1}*2^{-3}\\2^{4x-x^{2}}=2^{x-4}\\4x-x^{2} = x-4\\x(x-4)+x-4=0\\(x-4)(x+1) = 0\\x_{1} = 4\\ x_{2} = -1$

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