Question

1/(x-4) + 3/(x+2) = 4/(x-1)

Answer 1

Ответ:

7

Объяснение:

$\frac{1}{x-4} +\frac{3}{x+2} =\frac{4}{x-1}$

Домножим на (x-4)(x+2)(x-1)

ОДЗ:

(x-4)(x+2)(x-1)≠0

x≠ 4 ;-2 ; 1

$(x+2)(x-1)+3(x-4)(x-1)=4(x-4)(x+2)\\x^2+x-2+3(x^2-5x+4)=4(x^2-2x-8)\\x^2+x-2+3x^2-15x+12=4x^2-8x-32\\x^2+x-2+3x^2-15x+12-4x^2+8x+32=0\\-6x+42=0\\-6x=-42\\x=7$

корень подходит по одз!

Answer 2

To solve the equation $\frac{1}{x-4} + \frac{3}{x+2} = \frac{4}{x-1}$ for $x$, we'll first need to find a common denominator and then solve for $x$. Here's the step-by-step solution:

Step 1: Find a common denominator for the fractions on the left-hand side of the equation. The common denominator is $(x-4)(x+2)(x-1)$.

Step 2: Rewrite each fraction with the common denominator:

$$\frac{(x+2)(x-1)}{(x-4)(x+2)(x-1)} + \frac{3(x-4)(x-1)}{(x-4)(x+2)(x-1)} = \frac{4(x-4)(x+2)}{(x-4)(x+2)(x-1)}$$

Step 3: Combine the fractions on the left-hand side:

$$\frac{(x+2)(x-1) + 3(x-4)(x-1)}{(x-4)(x+2)(x-1)} = \frac{4(x-4)(x+2)}{(x-4)(x+2)(x-1)}$$

Step 4: Expand and simplify the numerators:

$$\frac{x^2 + x - 2 + 3(x^2 - 5x + 4)}{(x-4)(x+2)(x-1)} = \frac{4(x^2 - 2x - 8)}{(x-4)(x+2)(x-1)}$$

Step 5: Combine like terms in the numerators:

$$\frac{4x^2 - 14x + 10}{(x-4)(x+2)(x-1)} = \frac{4x^2 - 8x - 32}{(x-4)(x+2)(x-1)}$$

Step 6: Now, we can subtract the right-hand side expression from the left-hand side expression to isolate the terms:

$$\frac{4x^2 - 14x + 10 - (4x^2 - 8x - 32)}{(x-4)(x+2)(x-1)} = 0$$

Simplify the numerator:

$$\frac{-6x + 42}{(x-4)(x+2)(x-1)} = 0$$

Step 7: Since the numerator is a linear expression ($-6x + 42$), for the fraction to equal zero, the numerator must be zero:

$$-6x + 42 = 0$$

Step 8: Solve for $x$:

$$-6x = -42 \\ x = \frac{-42}{-6} \\ x = 7$$

So, the solution to the equation $\frac{1}{x-4} + \frac{3}{x+2} = \frac{4}{x-1}$ is $x = 7$.