Question

[Hà Mai Thùy]

bài 1

bài 2

Answer 1

[Kiều Vũ Linh]

Bài 1

b) \(3^x.5=405\)

\(3^x=405:5\)

\(3^x=81\)

\(3^x=3^4\)

\(x=4\)

c) \(9.2^x-5.2^x=32\)

\(2^x.\left(9-5\right)=32\)

\(2^x.4=32\)

\(2^x=32:4\)

\(2^x=8\)

\(2^x=2^3\)

\(x=3\)

h) \(5^{x+2}+5^{x+1}=750\)

\(5^{x+1}.\left(5+1\right)=750\)

\(5^{x+1}.6=750\)

\(5^{x+1}=750:6\)

\(5^{x+1}=125\)

\(5^{x+1}=5^3\)

\(x+1=3\)

\(x=3-1\)

\(x=2\)

Answer 2

[『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』]

`2,`

c)

`(x - \frac{1}{2})^3 = -8`

`\Rightarrow (x - \frac{1}{2})^3 = (-2)^3`

`\Rightarrow x - \frac{1}{2} = -2`

`\Rightarrow x = -2 + \frac{1}{2}`

`\Rightarrow x = -\frac{3}{2}`

Vậy, `x = \frac{-3}{2}`

d)

`(5x - 3)^4 = 256`

`\Rightarrow (5x - 3)^4 = (+-4)^4`

`\Rightarrow`\(\left[{}\begin{matrix}5x-3=4\\5x-3=-4\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}5x=7\\5x=-1\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-\dfrac{1}{5};\dfrac{7}{5}\right\}\)

f)

\(\dfrac{\left(x-2\right)^2}{7}=\dfrac{49}{x-2}\)

`\Rightarrow (x-2)^2 * (x - 2) = 7*49`

`\Rightarrow (x - 2)^3 = 7*7^2`

`\Rightarrow (x - 2)^3 = 7^3`

`\Rightarrow x - 2 = 7`

`\Rightarrow x = 7 + 2`

`\Rightarrow x = 9`

Vậy, `x = 9`

g)

`(x - \frac{1}{2})^3 = \frac{1}{27}`

`\Rightarrow (x - \frac{1}{2})^3 = (\frac{1}{3})^3`

`\Rightarrow x - \frac{1}{2} = \frac{1}{3}`

`\Rightarrow x = \frac{1}{3} + \frac{1}{2}`

`\Rightarrow x = \frac{5}{6}`

Vậy, `x = \frac{5}{6}.`