ĐGTD ĐH Bách khoa - Tư duy Toán học - Phương trình có số mũ

$\frac{-8}{5}$

3

$\frac{8}{5}$

$\frac{12}{5}$

x = -3

x = -2

x = 2

x = 3

{-1;2}

{0;1}

{-1;0}

{-2;1}

S = {1}

S = {3}

S = {2}

S = {5}

0

1

3

4

x = 4

x = 2

x = 5

x = 3

x = 5

x = 4

x = 6

x = 17

3

2

0

1

$x^{\frac{2}{3}}+5=0$

${\left(3x\right)}^{\frac{1}{3}}+{\left(x-4\right)}^{\frac{2}{5}}=0$

$\sqrt{4x-8}+2=0$

$2x^{\frac{1}{2}}-3=0$

$(–\infty ;–3)$

$(–3;+\infty )$

$(3;+\infty )$

$(0;+\infty )$

$\left\{1;{\log }_{3}2\right\}$

{-2;3}

{1}

{3}

2

-1

0

1

-13 < m < -9

3 < m < 9

-9 < m < 3

-13 < m < 3

$\left(-\infty ;1\right)$

$\left[2;+\infty \right)$

$\left(-\infty ;1\right)\cup \left(2;+\infty \right)$

$\left(2;+\infty \right)$

$m\in (\frac{17}{16};\frac{5}{2})$

$m\in [2;4]$

$m\in (\frac{5}{2};6)$

$m\in (1;\frac{5}{2})$

m = 4

m = 1

m = $\frac{5}{2}$

m = 3

m > 1

m < -1

m < 0

-1 < m < 0

$2=3^{{\log }_{5}x}$

$5=x^{{\log }_{2}3}$

$2=x^{{\log }_{3}5}$

$3=5^{\log x}$

$P=\frac{1}{2}$

$P=1-{\log }_{\frac{9}{2}}2$

$P=1$

$P=1-\frac{1}{2}{\log }_{\frac{9}{2}}2$

$-1+2{\log }_{2}3$

$1+2{\log }_{2}3$

-1

$1+1{\log }_{2}3$

$m=\frac{5}{2}$

$m=\frac{3}{2}$

m = 3

m = 2

0,50

0,35

0,40

0,45

1

2

4

3

10

-8

8

-10

m = 3

m = 18

m = $\frac{1}{8}$

m = 1

7

3

5

6