Решите уравнения 1)4^(2x) - 7 * 4^x + 16 = 0 2)2^(2x + 1) + 2^(x + 2) - 16 = 0 3)24^2x + 8 = 17 4)2 16^x - 7 * 4^x - 4 = 0 5)4^x + 6 = 4 4^(1/2 - x/2) 6)2^2x - 6 * 2^x + 8 = 0 7)4^x - 5 * 2^x + 4 = 0 8)9^(x+1) + 3^(x+2) - 18 = 0 9)25^x - 24 * 5^(x-1) - 5^(log5 3) + 2 = 0 10)7^(x-2) + 38 * 3^x = 7^(x+1)

Ответы

киса711 01.07.2020 06:56

1) $4^{2x} -7* 4^{x}+16=0$
$4^{x}=t;t0$
t² - 7t + 16 = 0
D= 49-64 <0
корней нет.
2) $2* 2^{2x}+4* 2^{x}-16=0$
$2^{x}=t;t0$
2t² + 4t - 16=0
t² + 2t - 8 = 0
D₁ = 1+8=9
t₁ = 2
t₂ = -4 не удов. усл. t>0
$2^{x}=2$
x=1
3) $2* 4^{2x}+8 = 17$
$2* 4^{2x}=9$
$4^{2x}=4.5$
$4^{2x}= 4^{ log_{4}4.5 }$
2x= log₄4.5
x = log₁₆4.5
4) $2* 16^{x}-7* 4^{x}-4=0$
$2*4^{2x}-7* 4^{x}-4=0$
$4^{x}=t; t0$
2t² - 7t - 4=0
D = 49+32=81
t₁ = 4
t₂ = -1/2 не удов. усл. t>0
$4^{x}=4$
x=1
5) $4^{x}+6* 4^{ \frac{x}{2}- \frac{1}{2} } = 4$
$4^{x}+ \frac{6* 4^{ \frac{x}{2} } }{ 4^{ \frac{1}{2} } }=4$
$4^{x}+3* 4^{ \frac{x}{2} }=4$
$4^{ \frac{x}{2} } =t; t0$
t² + 3t = 4
t² + 3t - 4 = 0
t₁ = 1
t₂ = -4 не удов. усл.t>0
$4^{ \frac{x}{2} } =1$
$\frac{x}{2} =0$
x=0
6) $2^{x}=t; t0$
t² - 6t + 8 = 0
D₁ = 9-8 = 1
t₁ = 4
t₂ = 2
$2^{x}=4$
x=2
$2^{x}=2$
x=1
7) $2^{2x}-5* 2^{x}+4=0$
$2^{x}=t; t0$
t² - 5t + 4= 0
t₁ = 1
t₂ = 4
$2^{x}=1$
x=0
$2^{x}=4$
x=2
8) $9^{x+1}+ 3^{x+2}-18=0$
$3^{2x}*9 + 9* 3^{x} -18=0$
$3^{x}=t; t0$
9t² + 9t - 18 = 0
t₁ = 1
t₁ = -2 не удов.
$3^{x}=1$
x= 0
9) $25^{x}-24* 5^{x-1}-3+2=0$
$5^{2x}- \frac{24* 5^{x} }{5} -1=0$
5* $5^{2x}-24* 5^{x}-5 = 0$
$5^{x}=t; t0$
5t² - 24t - 5 =0
D₁ = 144+25 = 169
t₁ = 5
t₂ = -1/5 не удов.
$5^{x}=5$
x= 1
10) $7^{x-2}+38* 3^{x}= 7^{x+1}$
$\frac{ 7^{x} }{49} +38* 3^{x}= 7^{x}*7$ I: $3^{x}$
$\frac{1}{49}* ( \frac{7}{3}) ^{x}+38 = ( \frac{7}{3}) ^{x}*7$
$7* (\frac{7}{3}) ^{x} - \frac{1}{49}* ( \frac{7}{3}) ^{x}= 38$
$\frac{342}{49}* ( \frac{7}{3}) ^{x} = 38$
$( \frac{7}{3}) ^{x}= \frac{49}{9}$
x = 2

ПОКАЗАТЬ ОТВЕТЫ

Другие вопросы по теме Алгебра