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Solve 1 by x minus 1 into x minus 2 1 by x minus 2 into x minus 3 is equal to 2 by 3 where X is not equal to 1 2 3?
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Solve 1 by x minus 1 into x minus 2 1 by x minus 2 into x minus 3 is...
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Solve 1 by x minus 1 into x minus 2 1 by x minus 2 into x minus 3 is...
Solution:
Given expression is:
$$\frac{1}{(x-1)(x-2)} \times \frac{1}{(x-2)(x-3)} = \frac{2}{3}$$
Simplifying the expression, we get:
$$\frac{1}{(x-1)(x-2)(x-2)(x-3)} = \frac{2}{3}$$
Multiplying both sides with $(x-1)(x-2)(x-2)(x-3)$, we get:
$$1 = \frac{2(x-1)}{3(x-1)(x-2)(x-2)(x-3)}$$
Simplifying further, we get:
$$3(x-1)(x-2)(x-2)(x-3) = 2(x-1)$$
Expanding the expression, we get:
$$3x^4 - 27x^2 + 54x - 36 = 2x-2$$
Simplifying the expression, we get:
$$3x^4 - 27x^2 + 52x - 34 = 0$$
Using the quadratic formula, we can solve for $x^2$:
$$x^2 = \frac{27 \pm \sqrt{729 - 4(3)(52)}}{6}$$
Simplifying the expression under the square root, we get:
$$x^2 = \frac{27 \pm \sqrt{441}}{6}$$
Simplifying further, we get:
$$x^2 = \frac{27 \pm 21}{6}$$
Therefore, the solutions are:
$$x^2 = \frac{8}{3} \text{ or } x^2 = 6$$
Taking the square root of both sides, we get:
$$x = \pm \sqrt{\frac{8}{3}} \text{ or } x = \pm \sqrt{6}$$
So, the final solution is:
$$x = \pm \sqrt{\frac{8}{3}}, \pm \sqrt{6}$$
Answer: The solution of the given expression is x = ±√(8/3), ±√6.
Given expression is:
$$\frac{1}{(x-1)(x-2)} \times \frac{1}{(x-2)(x-3)} = \frac{2}{3}$$
Simplifying the expression, we get:
$$\frac{1}{(x-1)(x-2)(x-2)(x-3)} = \frac{2}{3}$$
Multiplying both sides with $(x-1)(x-2)(x-2)(x-3)$, we get:
$$1 = \frac{2(x-1)}{3(x-1)(x-2)(x-2)(x-3)}$$
Simplifying further, we get:
$$3(x-1)(x-2)(x-2)(x-3) = 2(x-1)$$
Expanding the expression, we get:
$$3x^4 - 27x^2 + 54x - 36 = 2x-2$$
Simplifying the expression, we get:
$$3x^4 - 27x^2 + 52x - 34 = 0$$
Using the quadratic formula, we can solve for $x^2$:
$$x^2 = \frac{27 \pm \sqrt{729 - 4(3)(52)}}{6}$$
Simplifying the expression under the square root, we get:
$$x^2 = \frac{27 \pm \sqrt{441}}{6}$$
Simplifying further, we get:
$$x^2 = \frac{27 \pm 21}{6}$$
Therefore, the solutions are:
$$x^2 = \frac{8}{3} \text{ or } x^2 = 6$$
Taking the square root of both sides, we get:
$$x = \pm \sqrt{\frac{8}{3}} \text{ or } x = \pm \sqrt{6}$$
So, the final solution is:
$$x = \pm \sqrt{\frac{8}{3}}, \pm \sqrt{6}$$
Answer: The solution of the given expression is x = ±√(8/3), ±√6.
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Solve 1 by x minus 1 into x minus 2 1 by x minus 2 into x minus 3 is equal to 2 by 3 where X is not equal to 1 2 3?
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Solve 1 by x minus 1 into x minus 2 1 by x minus 2 into x minus 3 is equal to 2 by 3 where X is not equal to 1 2 3? for Class 10 2025 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about Solve 1 by x minus 1 into x minus 2 1 by x minus 2 into x minus 3 is equal to 2 by 3 where X is not equal to 1 2 3? covers all topics & solutions for Class 10 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solve 1 by x minus 1 into x minus 2 1 by x minus 2 into x minus 3 is equal to 2 by 3 where X is not equal to 1 2 3?.
Solve 1 by x minus 1 into x minus 2 1 by x minus 2 into x minus 3 is equal to 2 by 3 where X is not equal to 1 2 3? for Class 10 2025 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about Solve 1 by x minus 1 into x minus 2 1 by x minus 2 into x minus 3 is equal to 2 by 3 where X is not equal to 1 2 3? covers all topics & solutions for Class 10 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Solve 1 by x minus 1 into x minus 2 1 by x minus 2 into x minus 3 is equal to 2 by 3 where X is not equal to 1 2 3?.
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