Junior High School Math Problem x 1 x 1 0 Urgent Solution!!

No solution! x=+[x] then the original formula =x-[x]+ 1=0, which represents the fractional part of rational x, then it is a number greater than one, [x] represents its integer part, then -[x]+ sum is a decimal, then x+ decimal is equal to 1, then x is a number less than one.

Consider only all solutions of x>=1!

Solve the equation +1 (n+ )=1 (a number of 0 1, n integer).

Simplified: 2+(n-1) -n-1)=0(*).

Because n>=1 and is an integer.

When n=1, there is no solution.

When n > 1, (*, no matter what value n takes, there is always a solution in 0 1!) Because when 0 is taken, the left side (*) is less than 0, and when 1 is taken, the left side is greater than 0So there is a solution in 0 1!! (Zero-point existence theorem).

To sum up: there are infinite solutions Please select 111111111111111111111111111111111111111

Due to the symmetry of the number axis, only the case of x>0 is considered, and the same is true for x<0. It is not possible when x is an integer, so let's look at x as a fraction:

When x>1 can be set to x=q p,(q=kp+r,r the equation becomes: x+1 x=[x]+[1 x]+1,x}=1-1 x,r p=1-p q, and r p+p q=1 has no solution.

Evidence r p+p q=1 no solution:

r p = (q-p) q, rq = p(q-p), both sides divided by q to take the remainder, left remainder 0 = right remainder (-p 2), because p, q mutisoprime, (p, q) = 1, thus (p 2, q) = 1, p 2, q) = 1, (p, q) refers to the greatest common divisor of p and q, that is, -p 2 cannot be divided by q (p 2 cannot be a multiple of q), that is, rq=p(q-p) cannot be true (the left is a multiple of q and the right is not a multiple of q).

0 In summary, it can be seen that there is no solution to the equation.

Due to the symmetry of the number axis, only the case of x>0 is considered, and the same is true for x<0. It is not possible when x is an integer, so let's look at x as a fraction:

When x>1 can be set x=q p, (q=kp+r, r proves r p+p q=1 without solution:

r p = (q-p) q, rq = p(q-p), both sides divided by q to take the remainder, left remainder 0 = right remainder (-p 2), because p, q mutisoprime, (p, q) = 1, thus (p 2, q) = 1, p 2, q) = 1, (p, q) refers to the greatest common divisor of p and q, that is, -p 2 cannot be divided by q (p 2 cannot be a multiple of q), that is, rq=p(q-p) cannot be true (the left is a multiple of q and the right is not a multiple of q).

0 In summary, it can be seen that there is no solution to the equation.

Obviously, x is not an integer, when x > 0, let x=m n(m,n coprime) then 1 x=n m,x}=m1 n,=n1 m, because m,n are coprime, so the two do not add to 1

When x<0 is set to x=-m n, it is similar.

So there is no solution.

No solution! Solution: The problem is equivalent to taking an integer a such that x+1 x=a; solution x=[a+(a 2-4) ; To find the solution of a rational number of x, it is equivalent to finding a rational constant b such that a 2-4 = b 2 is understood by Fermat's theorem.

I don't have a solution, the reciprocal of a number plus a number can't be equal to 1.

Either the title is written incorrectly.

Because it's a junior high school math problem.

So x+1 x=1 has no solution.

Negligence, redo.

Dizzy Is this a junior high school math problem It's really a blow -

I wrote a program that all the decimal places in between, down to 7 decimal places, don't fit, and of course won't.

Is this a clear question?

Upstairs, does yours make sense? Despise.

x}+-1=x-[x]+1 x-[1 x]-1=0. x+1 x-[x]-[1 x]-1=0 ream-[x]-[1 x]-1=a, obviously, a is an integer here.

then the original formula can be turned into.

x+1/x+a=0

Then the landlord's problem is transformed into a question of whether x+1 x+a=0 has a solution to a rational number, right?

It's already so simple to get here, the landlord will deal with it himself.

By the way, a=2 is not the solution.

Good luck with a speedy epiphany!

x belongs to (empty set), i.e. there is no solution.

Because x is not equal to 0, the arbitrary equation is multiplied by x, and it can be judged to be less than 0 by Weida's theorem

So there is no solution.

There are infinite solutions.

Rationale: Only consider all solutions of x>=1!

Solve the equation +1 (n+ )=1 ( is a number of 0 1, n integer) and simplify: 2+(n-1) -n-1)=0(*) because n>=1 and is an integer.

When n=1, there is no solution.

When n > 1, (*, no matter what value n takes, there is always a solution in 0 1!) Because when 0 is taken, the left side (*) is less than 0, and when 1 is taken, the left side is greater than 0So there is a solution in 0 1!! (Zero-point existence theorem).

To sum up: there are infinite solutions.

Excuse me. Neglected, I was looking for a real solution. If a rational number solution is required, it is also necessary to satisfy that (n-1) 2+4(n-1) in (*) is a perfect square number.

Such a positive integer does not exist (I verify it with a program). So there is no understanding. Hehe.

There are infinite solutions to real numbers, just list a few, otherwise your efforts will be in vain.

You can use the multiplicative distributive property.

Definition: The multiplicative distributive law refers to the multiplication of the sum of two numbers by one number, and the two additions can be multiplied by this number, and then the sum of the two products is added, and the result remains the same.

Formula: Multiplicative distributive property formula: (a+b)c=ac+bcx+7)(x-1)=0

Let's start by looking at X+7 as a whole and distributing it with X-1.

x+7)x+(x+7)(-1)=0

Then perform the multiplicative distributive property for x+7.

x²+7x-x-7=0

x +6x-7 = 0 from the buried foundation

In terms of the value of the algebraic rent type, the two forms are the same type. However, the factorization form is simpler in the next step of multiplication and division, while the polynomial form is more suitable for the next step of addition and subtraction. If this problem is to solve an equation, of course, the form of factorization is appropriate, but if it is part of a larger equation, the next step is to add and subtract with other equations, and the polynomial form is more appropriate.

Because (x+7)(x-1)=, this way of writing can be directly seen, the answer is x=-7 or 1

Answer: Because (x 7) (x 1) = 0, we need to use each term of the previous factor to imitate the old and don't multiply each calendar of the latter factor, and the limb is Bide: x x 1 x 7 x 1 7 = 0, that is: x 6 x 7 = 0

Using the cross multiplication method, we can dismantle the frontal equation x +6x-7=0 to obtain (x+7) (x-1)=0

Solve the equation x +6x-7 = 0

Solution: (x+7) (x-1)=0

x+7=0 or x-1=0

x=-7 or x=1

The solution of the equation is -7 or 1

It is the multiplication and addition of two things.

x times x + x times (-1) = x -x

7 times x + 7 times (-1) = 7x-7

Then these two add up.

x²-x+7x-7=x²+6x-7=0

Decomposing Factors Finding Roots Quadratic equations are transformed into primary equations.

Using the cross multiplication method to split the equation x +6x-7=0, we get (x+7) (x-1)=0 to solve the equation x +6x-7=0 solution: (x+7) (x-1)=0x+7=0 or x-1=0x=-7 or x=1 The solution of the equation is -7 or 1

x +6x-7=0 is factored by "cross multiplication" to get (x+7)x(x-1)=0.

Cross multiplication: ax +bx+c=0, a=a1 a2, c=c1 c2, then b=a1 c2+a2 c1, this is this law.

Mathematics I think it has certain ideas and methods, and he follows such laws.

Solution: (x-3)(x-4)=0

Get x1 = 3 and x2 = 4

Hey, the cups don't have any points.

Solution: (1).

x-1)(3x+1)-(x+1)²

3x²+x-3x-1-x²-2x-1

2x²-4x-2

2(x²-2) -2

Substituting x -2x=1 to get it.

Original = 0

2) Because x=1, y= -1 and x=2, y= -6 all satisfy ax+by=2, substitution, get:

a-b = 2 (1)

2a-6b = 2 (2)

Simplifying (2), we get a-3b = 1 (3).

1) -3), resulting in 2b = 1

So b=1 2

Substituting (1), we get a = 3 2

Because B copied C incorrectly, B's answer cannot satisfy the correct cx-5y=2, and can only be substituted by A's correct solution.

c + 5 = 2

So c = -3

The correct answer is a=3 2, b=1 2, c= -3

∵x²-2x=1

x-1)（3x+1)-（x+1)²

3x²+x-3x-1-(x²+2x+1)=2x²-4x-2

2(x²-2x)-2

02.x=1 y=-1 is substituted for ax+by=2 to get a-b=2(1).

x=2, y=-6 into ax+by=2

2a-6b=-2

i.e. a-3b = -1 (2).

1)-(2) gives 2b=4

b = 2 substituting (1) to get a-2 = 2

a=4x=1 y=-1 is substituted into cx-5y=-2 to get c+5=-2

c=-7

Solution: (1)(x-1)(3x+1)-(x+1) = 3x +x-3x-1-x -2x-1

2x -4x-2 = 2 (x -2x-1) The problem is known: x -2x-1 = 0

So the original formula = 2 * 0 = 0

2) A is the correct solution, so bring xy in.

Gets: a-b=2 c+5=-2

So you can get c=-7

B copied C wrongly, but AB did not copy it wrong.

Bringing in the first equation: -6a-3b=2 and a-b=2 are connected to the system of equations, and the solution is: a=4 9

b=-14/9

c=-7

x -7x+1=0 (both sides are divided by x,x≠0) x-7+1 x=0

x+1/x)=7

x+1/x)^2=49

x-1/x)^2=(x+1/x)^2-(4*x*(1/x))=49-4=45

So. x-1/x=±√45=±3√5(x≠0)

This type of question is very simple, mainly to get the range of the final result x, which is as follows:

x+1）²+5＞0

x+1>-√5

x>-√5-1；

Analysis: 5 under the root number is equal to about that; At this time, about x> satisfies the condition (x+1) +5 0;

Test: (Result; If x= results as, the condition is obviously not satisfied; It shows that the result obtained in this way is correct, but if the landlord is not at ease, he can test it more, and the general direction is like this;

5 under the root number is equal to 5.

I hope the answer given can make you satisfied.

（x+1）²+5＞0

x+1）²＞5

Because negative numbers cannot be squared.

So (x+1) 0

x+1≥0x≥-1

You can use the formula method and the matching method, x -x-1 = 0 , x -x+1 4 = 5 4 , so (x-1 2) 2 = 5 4, the latter is very simple.