Solve the equation a 2 1 x ax 1 with respect to x

The equation is a 2-a 2x=ax+1 a(a+1)x=(a+1)(a-1).

Divide the two sides by a+1 x=(a-1) a

When a=0 gives 0=-1 the equation has no solution.

When a=-1 gives 0=0 the equation has an infinite number of solutions.

When a is not equal to 0 and a is not equal to -1, it is x=(a-1) a

a^2(1-x)=ax+1.

Obviously, a ≠0, otherwise the original equation becomes 0 1, which is meaningless.

Both sides divide by a 2 : 1-x x a+1 a 2 shift : x a+x 1- a 2

Decomposition factor: (1+1 a) x (1+1 a)(1-1 a) so : x 1-1 a (1+1 a≠0,a≠-1).

The equation is a 2-a 2x=ax+1 a(a+1)x=(a+1)(a-1).

When a=0 gives 0=-1 the equation has no solution.

When a=-1 gives 0=0 the equation has an infinite number of solutions.

When a is not equal to 0 and a is not equal to -1, it is x=(a-1) a

a-1)(a+1)=a(a+1)x

a=0, no solution.

a=-1 and x is any real number.

In all other cases, x=(a-1)a

Dear, this is a very common question in our Yucai Supernormal Experimental Department, not a Olympiad of mathematics!

x+(2 nanocoarse x-1)=a+2 ( a-1)x-a+2 Dongtuanzhen (x-1)-2 (a-1)=0x-a+2(a-1-x+1) [x-1)(a-1)]=0x-a-2(x-a) [x-1)(a-1)]=0x-a)=0

x-a=0 or 1-2[(x-1)(a-1)]=0x-a=0x=a

1-2[(x-1)(a-1)]=0

2[(x-1)(a-1)]=1

x-1)(a-1)=2

x-1=2/(a-1)

x=1+2 (a-1)=(a+1) (a-1) so x=a,x=(a+1) (a-1).

ax-1=2x

a-2)x=1

When a-2=0 i.e. a=2, there is no solution to the original equation.

When a-2≠0 is a≠2, the solution of the original bridge wheel fibrillation equation is x=1 (a-2) The team of mathematics, physics and chemistry for middle school students answers the question for you.

Solution: Bring x=2 into the algebraic formula ax+2=a+1, a 2+2=a+12a+2=a+1

2a-a=1-2

a=-1 i.e.

The equation for x ax+2=a+1, a=-1

Multiply both sides by (x 2-a 2).

x^2-a^2)/(x-a) +x^2-a^2)/(x^2-a^2) = (x^2-a^2)/(x+a)

> (x-a)(x+a)/(x-a) +x^2-a^2)/(x^2-a^2) = (x-a)(x+a)/(x+a)

> x+a+1= x-a

> a+1= -a

> a = 1/2。

It can be seen that regardless of the value of x (but not a), the equation holds when a = 1 2 and does not hold when a ≠ 1 2. Then for the equation of x, the solution should be .

x ≠ a, if a = 1 2;

There is no solution, if a ≠ 1 2.

Write clearly, whether it's 1 x-a or 1 (x-a).

A solution of equation 2 (ax-1) = x+a is x=1, and substituting it into the equation yields:

2(a-1)=1+a

2a-2=1+aa=3

Let f(x)=x+1 x

The image is a checkmark function.

As a straight line y=m, (m is not equal to 0) and f(x) intersect two points, and the product of the abscissa of the two points is 1 and it is easy to know x=a or 1 a