High scores are eager to answer some math questions for the connection between junior high school an

1)b (2x+5y)(x-3y)

2) A is divided into 2 cases: 1 is that the denominator is greater than 0, the numerator is less than 0, 2 is the numerator is greater than 0, and the denominator is less than 0, because 0 a 1 you will find that the second case is not true, so the first solution is answer a

3) The denominator is rational and chemical The answer is 2 + root number 3 numerator denominators are multiplied by 2 + root number 3 denominators to become square differences.

4) The enumeration method is to list all possible ones: x is an integer, so there are 5 of them, which are x=-2 -1 0 1 2

So the result is a=

5) Shift the term into (-1 2)x 2+2x-mx>0, then multiply the left and right by -1 2 and change the sign x 2-4x+mx<0 on both sides of the inequality, multiply the negative number to change the sign, and then extract x to get x(x-4+m)<0

0-2 m m+1>-2 cannot be multiplied by m+1 at this time, because the positive and negative of m is not known, so it can only be moved by -2, which is m m+1+2(m+1) (m+1)>0, which is the inequality that is used all the time, so it is (3m+2) (m+1)>0 m<-1 or m>-2 3

10) Use the discriminant formula of the root to have a common point horizontally, that is, there is a solid root existence b 2-4ac -2a 0 ax 2+bx+c corresponds to the abc in the abc in order of quadratic term coefficient primary term coefficient constant term after finishing to obtain 1+4m 2+4ma m 0 and then classify and discuss when m<0 m>0 different cases m=0 (remember that m=0 is not invalid at this time, the previous discriminant formula is constructed by us, in fact, m can be equal to 0).

11) Factorization [x+(a 2+1)][x+2a]<0 Because a 2+1-(-2a)=(a+1) 2 0 so a 2+1 is constant -2a so the inequality is solved as x<-2a or x>a 2+1

There are 2 cases that contain all 2 5 real roots, 1 is 2 5, contained in a 2+1 or in -2a, in fact, just count the endpoints, that is, a 2+1 2 -2a is less than or equal to 5, this should be understandable, less than the smallest endpoint is waiting for the satisfaction of the situation, and the endpoint greater than the largest can also satisfy the situation, so it can be solved to -1 a 1

or a -2 5

I'm counting the same as the last one.

1 b2 a

The value of 3 is 2 + root number 3

4 a=5 x(x-4+m)<0

6 a∈（-0）∪（0，3）∪（3，+∞

7（2x-2y）（3x+y）

8 root number 1 2

9 m<-1 or m>-2 3

10 -1 a 1 or a -2 5

That's the answer.

The process is on the top floor.

Is there a bounty point?

Because ab=ac, the angle abc = angle c

And the angle abc = angle bao, so the angle bao = angle c

And the angle b is a common angle, so the triangle AOB is similar to the triangle CA to get ao ca=ab cb

The carry-on number is worth 4y=x 2

The value of the independent variable x is in the range of (4 roots, 2, 8).

The parentheses are the range of values, you should understand it, there is no mathematical symbol on the keyboard, so it is replaced by text.

。Summer vacation boring panic, solve problems and play...

Since they are on both sides of the straight line, one is above the straight line and the other is below the straight line, so after bringing the two points into the equation of the straight line, we get two formulas about a, and the product after multiplication is less than 0! That is.

3*3-2*1+a)*[3*(-4)-2*6+a]<0 The range of a can be obtained: -7

If (3,1) is above the straight line (-4,6) is below.

9-2+a>0

6-12+a<0 -7 in turn is not solved.

Solution: Original formula = 6x square - (7y+z)x-3y square + 7yz-2z = 6x square - (7y+z)x-(3y-z)(y-2z) = (2x-3y+z)(3x+y-2z) please adopt.

Are you an equation? Make up three flat ways to solve.

Because 1 x1<2, 2 x2<3

So let x1 = 1 and x2 = 2

So 2-2m+n=0

8-4m+n=0

The solution yields m=3 and n=4

Bring x1 and x2 into the equation respectively to get the system of equations: 1) 2-2m+n=0 2)8-4m+n=0 to solve m=3, and bring m into the system of equations to get n=4, so x=[1,2],m=3,n=4 I feel that x is not very accurate, sorry. I'm not good at math.

Are there any other restrictions? Then m and n should be the range!

Solution: y=x +(p+1)x+p 2+1 4, obviously, to make p irrelevant, then the value of x should exactly eliminate p; That is, px=-p 2, i.e. x=-1 2. Substituting the parabola, there is y=(-1 2) +1 2)(p+1)+p 2+1 4=1 4-p 2-1 2+p 2+1 4=0, that is to say, the parabola must pass the point (,0);

Let the parabolic vertices o(x,y), i.e., x=-(p+1) 2,y=-p 4;∴y=-p²/4=-[-p+1)/2]²+p/2+1/4=-x²-[p+1)/2]-1/4=-x²-x-1/4。

i.e. the parabolic analytic formula for those vertices is y=x -x-1 4, I hope it helps you (

Multiplying 0 by any number results in 0, so the essence of the problem is to make the argument (e.g. p represented in the problem) act as any number multiplied by zero, so that it will have a definite value for any number. The next step is how to do it on the topic.

Put the equation and you will get y=x +px+x+p 2+1 4 and then you will get the arguments in a row.

x +(x+1 2)p+x+1 4 If you want to make any value of p, you need to make x=-1 2 and then substitute the value of x into the original to get the value y, so the answer is ( 0) I hope you can understand.