`@` `\text{Ans}`

`\downarrow`

`2^(2x-3)=32`

`=> 2^(2x-3)=2^5`

`=>2x -3=5`

`=> 2x=8`

`=>x=4` 

Vậy, `x=4.`

2²ˣ⁻³ = 32

2²ˣ⁻³ = 2⁵

2x - 3 = 5

2x = 5 + 3

2x = 8

x = 8 : 2

x = 4

x = 4

`@` `\text {Ans}`

`\downarrow`

\(\left(2\cdot x+2\right)^2=64\)

`\Rightarrow`\(\left(2x+2\right)^2=\left(\pm8\right)^2\)

`\Rightarrow`\(\left[{}\begin{matrix}2x+2=8\\2x+2=-8\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}2x=8+2\\2x=-8+2\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}2x=10\\2x=-6\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=10\div2\\x=-6\div2\end{matrix}\right.\)

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

Vậy, `x \in {5; -3}`

`@` `\text {Kaizuu lv uuu}`

Đáp án là A

Ta có: (x - 4).1000 = 0 nên x - 4 = 0 (vì 1000 ≠ 0)

Suy ra ta có: x - 4 = 0 ⇔ x = 4

Đáp án là A

Ta có: (x - 4).1000 = 0 nên x - 4 = 0 (vì 1000 ≠ 0)

Suy ra ta có: x - 4 = 0 ⇔ x = 4

3 ?

\(\dfrac{13}{4}>x>\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{13}{4}>x>\dfrac{10}{4}\)

\(\Leftrightarrow x\in\left\{\dfrac{11}{4};\dfrac{12}{4}\right\}\)

\(xy+x-y=4\)

\(x\left(y+1\right)-\left(y+1\right)=3\)

\(\left(y+1\right)\left(x-1\right)=3=3\cdot1=1\cdot3=-3\cdot-1=-1\cdot-3\)

lập bảng

xy+x-y=4

x(y+1)-(y+1)=3

(y+1) .(x-1)=3=3.1=1.3--3.(-1)=-1.(-3)

nhớ lập bảng