Mathematics for junior 3, quadratic radicals, the following formulas are correct .

Obviously C is correct.

You're in junior high school now, and you've got a non-negative real number under the root number. So Type A doesn't hold.

Formula B is miscalculated.

There is a problem with the d formula, if there is only a root number, the default result is a positive number, so the -x here will appear correct when x is negative, and incorrect when it is positive.

Answer C is parsed as follows:

a, the root is negative in the value, so it is discarded.

b, the value of the radical is greater than 0, and in addition, 4=2

d, the value of the root formula is greater than 0, and the result should be x

Therefore, choose C

The problems discussed in junior high school mathematics are generally in the field of real numbers, and the root number is originally positive and negative, but looking at the options of this problem, it should be required to open the root and take the positive.

a.The minus opening root is meaningless in the field of real numbers.

b.The open root number is equal to 2

c.That's right. d.To discuss the positive or negative of x.

ca cannot be a negative number under the root number.

b。It should be equal to 2

C right. d。It should be x if it is written 0 here. Because the root number cannot be negative. So if that's the case, x can only be equal to 0

ca is wrong, and the number inside the root number cannot be negative.

b is false, -4 is squared to 16 and the value in the root number is 4

d is wrong, because x doesn't know if it's positive or negative, so -x isn't certain either, and any number should be greater than or equal to 0

c。。The root number cannot be a negative number, and the root number cannot be a negative number.

The root number cannot be a negative value A excludes Similarly, the value of the root number must be positive b excludes d d and cannot be judged to be positive or negative, so d excludes The correct answer is c

c。Square first and then open the root number, which is a positive number open root number.

Let the integer parts of 9 + root number 13 and 9 - root number 13 be a and b respectively, and it is easy to know 3 "root number 13<4, therefore:

12<9 + root number 13<13, 5<9 - root number 13<6

Then: a=12, b=5, a=9 + root number 13-12 = root number 13-3, b = 9 - root number 13-5 = 4 - root number 13

So, ab-3a+4b+8=(root number 13-3)(4-root number13)-3(root number 13-3)+4(4-root number13)+8

4 roots: 13-13-12+3, 13-3, 13+9+16-4, 13+8

8 See.

3²<13<4²

So the 9+13 integer part is 12

So the decimal part a=9+ 13-12= 13-33< 13<4

So the decimal part b=9- 13-5=4- 13, so the original formula = a(b-3)+4(b+2).

The decimal part of 9 13.

For: 9 13-12 = 13-3

The decimal part of 9 13.

For: 9- 13-5 = 4- 13

So ab 3a 4b 8

13 is between 3 and 4.

So a = 13-3 b = 4 - 13

Substituting the above equation yields ab 3a 4b 8 = 8

If you don't understand it, ask again.

-3 on the left side of the equal-sign spike and then +3: (x-3) + root number (x-3) -6 + 3 = 0 reams root number (x-3) = a

Original formula: a 2 + a - 3 = 0

The number obtained by solving the root equation a is substituted into the root number (x-3)=a.

That's it.

X-3 is converted to a, then x=a+3 is equal to a+3 + root number a and then -6=0 to solve the shift, put the root number on the left side of the equation.

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So. (1) Original formula = 2 3-3 2- 2 2+ 3 3 = 7 3 3-7 2 2

2) Original = 10 (3 15-5 6) 3 = 30 (3 15-5 6) = 45 2-30 5

3) Original formula = 2 (b ab 5) (3 (2 a 3b) a 3 b = -3 (ab 4 b).

Obviously, a, b, >0 in the question

4) Original = 6 5-10 3

If 2 is squared.

Original = 45 + 75-30 15 = 120-30 15

(1) x takes any number x +2 and is equal to 0, (x-1) is equal to 1

2) x takes any number inside is a perfect flat (x-3) 0

(1) According to the definition of quadratic radicals: the number of squares to be opened must be greater than or equal to 0, so x-square +2 is greater than or equal to 0, and because x-square +2 is everstable at zero.

And since any non-zero to the power of 0 is 0, x-1 is not equal to 0, so it is solved that x is not equal to 1 (all real numbers).

2) Because x-6x+9 is a perfectly squared formula, it is broken down into (x-3) greater than or equal to 0

So x is the whole real number.

x +2 0---x is any real number.

x-1≠0x≠1

Original = x-3

x is any real number.