Math Problems for Secondary 2 Quadratic Radicals and Equations

1, observed a = (root number 2) + 1, b = (root number 2) + (root number 3) The original formula is the form of (a + b) ab, which is obtained from the equation.

1 (root number 2+1)+1 (root number 2 + root number 3), and then use the square difference formula to rationalize the denominator (root number 3-1).

2, classification discussion, when a>1992, the equation is (root number a) = 1992 + 1993, calculate the final result first use the perfect square, so simple, when a< = 1992, the equation is 2a - (root number a) + 1 = 0, the root number a as a whole, equivalent to a quadratic equation, discriminant formula = -7<0, no real number solution.

3. Simplify first and then evaluate, and use the factor to decompose the original formula as (x-1) 2 (x-1)-(1-x) (x(x-1))=x-1+1 x=3

4, x+6y=5 times (root number xy) shift factorization to (root number x-2 root number y) * (root number x-3 root number y) = 0

So the root number x = 2 root number y or the root number x = 3 root number y, and the calculated formula subtracts a variable to get 9 8 or 10 7

5, pass to get (m+n) (root number mn) = -5

6, let 1996 (x 3) = 1997 (y 3) = 1998 (z 3) = k, [cubic root number 1996 (x 2) + 1997 (y 2) + 1998 (z 2)] = three root number under (k x + k y + k z) = three root number (1 x + 1 y + 1 z) * three root number under (k) = (third root number 1996) + (third root number 1997) + (third root number 1998), shift the term to get three root number (1 x + 1 y + 1 z) =(1996) (1997) (1997) (1997) (1998) (1998) (k) (1998) (1998) (1998) (1998) (1 x+1 y+1 z) = 1 x+1 y+1 z, respectively, to obtain 1 x+1 y+1 z=0 (rounded), -1 (rounded) or 1

7, this is actually a problem of roots in the range of real numbers, as long as there is a root, the general formula can be decomposed into (x-x1)*(x-x2)=0, called two roots, (1) is equivalent to a solution, discriminant = 16-24k>=0, (2) is equivalent to no solution, discriminant = 16-24k<0, (3) is equivalent to two equal, discriminant = 16-24k=0, is (3x-2) 2 = 0

The fifth, the square root number, and then the second one, is less than or equal to a less than or equal to 1992 and a is greater than 1992

It won't be written on the computer.

I can't think of the second and seventh questions, sorry.

Okay, let's just talk about question 2.

The formula makes sense.

Root number (A-1993) Founded A 1993 1992 So the original formula: a-1992 + root number (a-1993) = a

Root number (a-1993) = 1992

a-1993=1992^2

a-1992^2=1993

1. Simplification 3- 3 (1- 3) = 3- 3 + 3 = 3 so choose A

2. Simplify( x+ y) +x- y) =2x+2y3, calculate 9-4 1 2+ 8

4、a²b-b²a=ab(a-b)

5. Simplify. Original = a+2 a-a

2√a6、x²-2xy+y²

x-y)²

So much, you still don't bother to do it

Mathematics content in the second volume of junior high school: what is a quadratic radical (suitable for students with poor mathematical foundation).

Suppose this radical=x>=0

We found. From left to right, the third radical starts with x.

i.e. x = root number (2 - root number (2+x)).

x 2 = 2 - root number (2 + x).

Root number (2+x) = 2-x 2

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

x^4-4x^2-x+2=0

Obviously x=-1 is a root.

So x 4-4x 2-x+2=x 4+x 3-x 3-x 2-3x 2-3x+2x+2

x+1)(x^3-x^2-3x+2)

Obviously, x=-1 is not a solution (x>=0).

x^3-x^2-3x+2

x^3-2x^2+x^2-2x-x+2

x-2)(x^2+x-1)

x=2 is obviously not a solution (x<2) either

So x 2 + x - 1 = 0

x=(-1+root number 5) 2 (round off negative roots).

1. Double non-negativity: a 0, a 0,2, simplification principle: ab= a* b, (a b)= a b, ( a) 2=a,3, multiplication and division: a b= ab, a b= (a b), 4, the simplest quadratic radical, 5, merge the same quadratic radical.

(9) Original formula = 32 2 + 2 2 = 16 + 1 = 4 + 1 = 5 (10) Original formula = 125 * 5- 20 * 5 = 625 - 100 = 25-10 = 15

11) Original = 8 + 18 = 2 2 + 3 2 = 5 2 (12) Original = ( 2-1) ( 2-1) ( 2-1) ( 2-1) ( 2-1) ( 3 + 2 2).

13) Primitive=2*2*6*2 4+3 6= 6+3 6=4 6(14) Primitive=6-2 3-5 3+5=11-7 3(15)Primitive=12-6=6

16) Original = 1 4+

(1) Root number x 2+2ax+a 0 |x-a|0 and their sum equals 0, which means that they are all equal to 0, so x=a solves a 2+2a 2+a=0

a = 0 or -1 3 (2).

1. The square of 4-x under the root number plus the bisect of x-5 under the root number is not: (4-x) 2+ (x-5) 2

If yes, then: primitive = 4-x|+|x-5|

When x<4 primitive = 4-x + 5-x = 9-2x, when 4 x<5 primitive = x-4 +5-x = 1, when x 5 primitive = x-4 + x-5 =2x-92, when 1 x 5, the square of x-1 under the root number plus the absolute value of x-5 equals =|x-1|+|x-5|=x-1+5-+x =4

3. When simplifying "1-x under the root number plus x-1 under the root number", it is found that 1-x 0 and x-1 0 x =1

That is: the original = 0+0=0

4. If "1-4/x" exists, then x >0

Root number 2-1>0

then (root number 2-1) x >-(root number 2-1).

Divide the root number by 2-1 on both sides

x>-1

(1) Make af bc at the point f, because b=60°, so bf= ab= 10=5, according to the Pythagorean theorem, af=5 3, so the height of dce is 5 3, so s dce=25 3 2, so the required volume of soil and rock is 25 3 2 100=1250 3m

2) DH BC at the point H, in DHC, DHC=90°, DH=AF=5 3, DC=10 3, from the Pythagorean theorem we can get HC=15, so HE=HC CE=15 5=20, in DHE, DH=90°, DH=5 3, HE=20, according to the Pythagorean theorem, DE=5 19

1.The volume to be filled = (5 * 5 3) * 100 2 = 1250 3 cubic meters (5 3 is trapezoidal height, can be found).

2.Let the slope be , tan =5 3 20 = 3 4

arctan√3/4