Solve the equation x-4 / x2 + X-2 = 1 / (x-1) + (X-6) / (x2-4)

Solve the equation x-4 / x2 + X-2 = 1 / (x-1) + (X-6) / (x2-4)

（x-4）/（x²+x-2）=1/(x-1)+(x-6)/(x²-4)(x-4)/(x-1)(x+2)=1/(x-1)+(x-6)/(x-2)(x+2)(x-4)(x-2)=(x-2)(x+2)+(x-6)(x-1)x²-6x+8=x²-4+x²-7x+6x²-6x+8-2x²+7x-2=0x²+x+6=0△...

Solve the equation 10 / x2 + X-6 + 2 / 2-x = 1
（10/x2+x-6）+（2/2-x）=1

Let 4-x = a
Then 5-x = a + 1
3-x=a-1
The results are as follows:
1/（a+1）+2/a+3/（a-1）=-3
The two sides are multiplied by a (a + 1) (A-1) and simplified
3a^3+6a^2-a-2=0 ...

The solution of equation 1x2 + X − 2 + 1x2 + 7x + 10 = 2 is______ ．

The original equation can be changed into: 1 (x − 1) (x + 2) + 1 (x + 2) (x + 5) = 2, both sides of the equation multiply by (x-1) (x + 2) (x + 5), x + 5 + X-1 = 2 (x-1) (x + 2) (x + 5), the solution is X1 = - 2 + 10, X2 = - 2-10; the root of the equation is x = - 2 ± 10

ABC is a natural number, a × B = 91, B × C = 77, C × a = 143, then a + B + C = ()

ab=91 a=7 b=13 or a=13 b=7
bc=77 b=7 c=11 or b=11 c=7
Combining the above two cases, B can only be 7
Then a = 13
ca=143 c=11
a+b+c=7+13+11=31

In △ ABC, if we know a = 5, B = 6, and S = 15 √ 3 / 2, then how many sides C = a √ 31, B √ 91, C √ 31 or √ 91 D7

C √ 31 or √ 91

Solve the equation 3A + 4x = 7x-6b 10Y + 2 (7y-2) = 5 (4Y + 3) + 3Y about X

3a+6b=3x
x=a+2b
14y-4=20y+3y+15
-9y=19
y=-19/9

On both sides of the equation 7Y = 3Y + 6 at the same time (), we get 4Y = 6 according to ()

Subtracting 3Y at the same time is based on subtracting the same number from both sides of the equation, and the equation still holds

On both sides of the equation 7y-6 = 3Y, we get 4Y = 6

The same solution principle of the equation

If both sides of the equation 7y-6 = 3Y are + (6-3y) at the same time, 4Y = 6 is obtained

7y-6+6-3y=3y+6-3y
(7-3)y+0=(3-3)y+6
4y=6

7y-10y+5=4y-5

10y-7y+4y=5+5
7y=10
y=10/7