The least common multiple of 23 and 69 And the least common multiple of 38 and 76

The least common multiple of 23 and 69 And the least common multiple of 38 and 76

69 76

Solving the fractional equation x / (x-1) = x ^ 2 / (x ^ 2-1)

Solution
Multiply both sides by X & # 178; - 1 = (x-1) (x + 1)
∴x(x+1)=x²
x²+x=x²
x=0
It is proved that x = 0 is the solution of the equation

The fractional equation 3A + 1 x + 1 = a has no solution

If the original fractional equation has no solution, there are two cases: ① when a = 0, the equation AX = 2A + 1 has no solution, so the fractional equation 3A + 1x + 1 = a has no solution; ② when a ≠ 0, the equation AX = 2A + 1, x = 2A + 1A; when the denominator x + 1 = 0, x = - 1, the original fractional equation has no solution. From 2A + 1A = - 1, a = - 13. So when a = 0 or a = - 13, the fractional equation 3A + 1x + 1 = a has no solution

The fractional equation x / x-3 = 2 + A / x-3 has no solution

Multiply x-3 on both sides
x=2(x-3)+a
If there is no solution, the root of the equation is an increasing root
That is, the denominator is 0
x-3=0
x=3
So 3 = 2 × 0 + a
a=3

If the fractional equation x-3 / X-1 = x-3 / MX has no solution, then the value of M is______
On the fractional equation of X, X-1 of x-3 = MX of x-3 has no solution, find the value of M

In order to solve this problem, we must only solve x = 3, let X-1 = MX, and then substitute x = 3 to get m = 2 / 3 (three-thirds)

When a is a, the equation x − ax − 1-3x = 1 has no solution?

To remove the denominator, we get: X (x-a) - 3 (x-1) = x (x-1), x2-ax-3x + 3 = x2-x, (a + 2) x = 3, (1) when a + 2 = 0, a = - 2, the original equation has no solution; (2) when a = 1, x = 1 is the increasing root of the original equation, the original equation has no solution; in conclusion, when a = - 2 or a = 1, the original equation has no solution

It is known that if the fractional equation 1 / X-1 + m / X-2 = 2m + 2 / [X-1] [X-2] has an increasing root, then the value of M is

If there is an increasing root, then x = 1 or 2
Multiply both sides of the equation by [X-1] [X-2]
M [X-1] + [X-2] = 2m + 2
Reduced de m (x-3) + (x-4) = 0
m=（4-x）/（x-3）
Bring in x = 1
The solution is m = - 1.5
X = 2
The solution is m = - 2
Another suggestion is that you should pay attention to one thing when you post questions on the Internet in the future, such as your equation 1 / X-1 + m / X-2 = 2m + 2 / [X-1] [X-2]. When you type it out, you should put brackets in X-1, X-2, 2m + 2, otherwise it will cause ambiguity

If the two square roots of a number are 2m - 6 and 3M + 1 respectively, find the value of M

2m-6 and 3M + 1 are the square roots of the same positive number, then the two numbers are opposite to each other, that is, (2m-6) + (3m + 1) = 0, the solution is: M = 1

How to pass matrix as function parameter in MATLAB

function y=fun(x)
Inside x is a matrix

There are numbers 4 and 9. Try to write another number so that one of the three numbers is the square root of the product of the other two numbers

Write 6.4 × 9 = 36, ± 36 = ± 6, 6 is a square root of the product of 4 and 9