An 8th grade algebra problem, 8th grade math algebra question

(x+1) (x-1)]-4 (x -1)]=1 solution: remove the parentheses, get:

x +2x+1 x -1)-[4 (x -1)]=1 to remove the denominator, yield: x +2x-3 x -1=1

Remove the denominator and get: x +2x-3=x -1

Shift, merge, get: 2x-3=-1

Move the term to get: 2x=-1+3

Combine similar terms to get: 2x=2

Divide both sides of the equation by 2 to get: x=1

Test: Substituting x=1 into the original equation yields:

The denominator of both fractions is 0, and the fractions are meaningless.

After testing, x=1 is the root increase, so there is no solution to the original equation.

(x+1) (x-1)]-4 (x -1)]=1 solution: x +2x+1 x -1]-[4 (x -1)]=1x +2x-3 x -1=1

x²+2x-3=x²-1

2x-3=-1

2x=-1+3

2x=2 x=1

Substituting this solution into the original equation yields:

The denominator of both fractions is 0, and the fractions are meaningless.

After testing, x=1 is the root increase, so there is no solution to the original equation.

Multiply by (x 2-1) on both sides

x+1)^2-4=x^2-1

then x 2 + 2 x - 3 = x 2-1

2x=2x=1

But x=1 is at the bottom of the original equation x-1=0.

So there should be no solution.

No solution. The values of the deformed solution are substituted into the original equation, and the denominator of the two fractions is 0, and the fractions are meaningless. So there is no solution to the equation.

According to the principle of inequality: a 2 + a (-2) 2, b 2 + b (-2) 2 Therefore, if and only if a 2 + a (-2) = 2 = b 2 + b (-2), the known condition is satisfied.

From the above equation, it is obtained: a=b=1

Substituting the equation asked, the answer is 1.

Answer: Incomplete.

x｜-4）^（x+1）=1

x -4=1 x+1=0 and x -4≠0 x -4=-1 and x+1 is even.

x｜=5 ② x+1=0 ｜x｜-4≠0x=±5 x=-1 ｜x｜=4

x≠ 4 x -4=-1 x+1 is an even number.

x = 3 x is an odd number.

x= 3, x= 5, x=-1, or x= 3

x^2 -4(y+1)x +(6y^2 -4y+6) =0

This equation needs to have a positive integer root, so its discriminant formula:

16(y+1)^2 -4(6y^2 -4y+6)= 8(y^2 -6y+1) ≥0

That is: y 2 -6y + 1 0

Solution: 3-2*2 y 3 +2*2 (the root number is not easy to play, use 2 to represent "root number 2").

And because y is a positive integer, therefore: 1 y 5, take y=1,2,3,4,5 into the original equation for testing (then the original equation becomes a pure quadratic equation about x), and we get:

When y=1,2,4,5, the corresponding quadratic equation for x has no positive integer solution, so it is not in line with the topic;

When y=3, the solution is: x=4 or 12;

Therefore, in summary, it can be seen that the solution of the original equation forms a positive integer pair (x, y) of (4, 3) or (12, 3).

1 all known a(3,y1),b(4,y2) on the image of y=x k, if k>0, then y1 y2, if k<0, then y1 y2 (fill in).<=)

Since a and b are both on the same inverse proportional function y=k x, and both are on the positive and semi-axial axes of the x-axis, they are in the same quadrant.

On the image of the inverse proportional function y=k x, when k is 0, the y value in the same quadrant decreases as the x value increases.

When k is 0, the y value in the same quadrant increases as the x value increases.

k>0 (two, four quadrants), y1>y2, k<0 (one, three quadrants) y1

The specific process has been listed, and this question is self-contained.

To make use of the root of the discriminant bai

equation, first du you have to list the general formula zhi of this formula, let him have the coefficient dao condition, and then use the discriminant formula of the root =a she is b2-4ac to calculate the range of k, because he has a real root, so 0. Hope it helps!

If the solution has a solid du root, then 3x-2>=0 , x>=2 3

Preliminary finishing: 3x-2 = 36x2 +12x(2k-3)+4k2-12k+9

The one-ary quadratic DAO equation for zhix returns to 36x2 + (24k-39) x + 4k2-12k + 11 = 0

If x has a solid root, then the answer is (24k-39)2-4x36x(4k2-12k+11)>=0

The unary quadratic inequality about k is sorted out - 144 in -63 "=0

Let me tell you about my thoughts.

First, make the copy of the sub bai into 2 (3x-2).

du+√（3x-2）+2k+1=0

Let (3x-2)=t 0, then the sub-variant of the formula is 2t +t+2k+1=0.

Then change the zhi formula k=-t -t 2-1 2

Is it a question of a quadratic function dao, if it is the second year of junior high school, -t -t 2-1 2 is used for formulation, and the value is discussed according to the limit of t 0.

The rest is the basics.

This has that root formula, when b2 4ac is greater than 0, that is, there are two different roots, it seems to be this formula, I can't remember clearly.

s squared =

When h = meter, the square of s = = * = then s =

When h = 35 meters, the square of s = = * 35 = then s =

a-a one-part = 4, obviously ≠ 0

a²-1=4a

Get a=2 root number 3

Substituting a - a part = 14

2 Ibid., get 18

1+x+x(1+x)+x(1+x) +x multiplied by (1+x) to the power of 2011.

1+x+x(1+x)+x(1+x) +x multiplied by (1+x) to the power of 2011.

1+x)(1+x)+x(1+x) +x multiplied by (1+x) to the power of 2011.

1+x) +x(1+x) +x multiplied by (1+x) to the power of 2011.

x+1)(1+x) +x multiplied by (1+x) to the power of 2011.

1+x) +x multiplied by (1+x) to the power of 2011.

…= (1+x) to the power of 2012.

1) a-a parts = a-1 a = 4

Then (a-1 a) 2=a 2+1 (a 2)-2=16 then a -a part =16+2=182)x-x one-part = x-1 x=3

Then (x-1 x) 2=x 2+1 (x 2)-2=9 then x -x one-half =9+2=111+x+x(1+x)+x(1+x) +x multiplied by (1+x) to the power of 2011=(1+x) 2012

3) 1+x+x(1+x)+x(1+x) =( 1+x)+x(1+x)+x(1+x) =(1+x)*(x+1)+x(1+x) =(1+x) +x(1+x) =(1+x) *(1+x) =(1+x) 3z.