How to find the Thévenin voltage?

The Thévenin equivalent consists of a single voltage source in series with a single resistor, together between points A and B. To find the voltage of the voltage source and the resistor value you consider two different load situations.

1.
No load at all, like it's drawn. The circuit consists of voltage source, R1, R2 and R5. There's no current through R3 or R4. We calculate the current: \$\dfrac{V+}{R1+R2+R5}=\dfrac{10V}{3k + 4k + 3k}=1mA\$. Then the voltage over R2 is 1mA * 4k = 4V, and since there's no voltage drop over R3 or R4 that's also the voltage between A and B.
In the Thévenin equivalent, when A-B is open there won't flow any current, so no voltage drop over the internal resistor. If we want 4V between A and B, the voltage source has to be 4V.

2.
Short-circuit A and B. Now R2 is parallel to the series resistance of R3 and R4. We need to know the equivalent of these (call it R6): \$\dfrac{1}{R6}=\dfrac{1}{R2}+\dfrac{1}{R3+R4}=\dfrac{1}{4k}+\dfrac{1}{6k}=\dfrac{0.417}{1k}\$ so \$R6=\dfrac{1k}{0.417}=2k4\$.
Again we calculate the current: \$\dfrac{V+}{R1+R6+R5}=\dfrac{10V}{3k + 2k4 + 3k}=1.19mA\$. The voltage over R6 is \$10V - 1.19mA \times (R1 + R5) = 2.85V\$, hence the current through R3 and R4 (and the short-circuit A-B) is \$\dfrac{2.85V}{R3 + R4}=\dfrac{2.85V}{6k}=476\mu A\$.
Our Thévenin circuit had a voltage source of 4V. To have 476\$\mu\$A through a short-circuited A-B the internal resistance has to be \$\dfrac{4V}{476\mu A}=8k4\$.

And that's our solution:

Equivalent voltage = 4V,
Equivalent series resistance = 8k4

For Rth，first short the power supply of 10V, then calculate resistance.
R1 is in series with R5, 3k+3k=6k, the result is in parallel with R2 => 6k || 4k = (6k x 4k )/(6k+4K)=2k4, then that is in series with R3 and R4.
2k4+3k+3k=8k4.