Middle School Math Algebra 30, Middle School Math Algebra

1.When x 0, y = 1That is, the constant crossing point of the function (0,1);

The function is no more than the third quadrant.

1) When k-2 0, i.e. k2, there is y=1 to meet the requirements.

2) When k is not equal to 2, y is a subtraction function, so there is k-2<0, so k<2 In summary, k is less than or equal to 2

3.Move k to the left of the equal sign, squared (3-k) 2>=4, so there is 3-k>=2 or <=-2, i.e., k>=5 or <=1 and 3-k=root sign (x 2+4) >0

So k<3

In summary, 1<=k<3

Less than 0, so k is less than 2

So x(x+1)(x+3)=0 x=0, x+1=0, x+3=0x can be equal to 0, -1, -3

3.What does this question mean?

4.For the application of binary linear equations, the key is the process of transformation from practical problems to mathematical problems. Therefore, in the teaching process, we pay attention to the method of analyzing problems, so that students can learn to use the ideas of mathematical modeling and equations to solve problems.

The selection of example problems is also based on reality, so that students can initially experience the close relationship between mathematics and people's daily life, and experience the role of mathematics in social life, stimulate students' interest in mathematics, and enable students to learn to analyze and solve simple practical problems from the perspective of mathematics.

5.When solving problems, if you don't think about it well or ignore special cases, it often leads to missed solutions.

6.The absolute value of a positive number is itself; The absolute value of a negative number is its opposite; The absolute value of zero is zero

7.Attention should be paid to the positive and negative nature of the results, distinguishing between square roots and cubic roots!

Solution:1b in y=kx+b is the ordinate of the intersection of the line and y-axis.

Straight through (0,1).

Since the straight line is no more than the third quadrant, the image is drawn in the coordinate system: the straight line is trending downward.

k-2<0

k<22.∵x^3+2x^2+2x+1=x^3+2x^2+x+(x+1)=x(x^2+2x+1)+(x+1)=x(x+1)^2+(x+1)=(x+1)(x(x+1)+1)=(x+1)(x^2+x+1)=0

x+1=0→x=-1

x 2+x+1=0 has no solution.

x＾2＝k＾2－6k＋5

From the meaning of the title: (k 5) (k 1) 0

k－5≥0，k－1≥0

or k 5 0, k 1 0

k 5 or k 1

4. Analyze the quantity relationship, set the appropriate 2 unknowns, list the equations according to the equal relationship, and solve them

5 The situation is not considered completely, e.g., only the arithmetic square root is found for the open square, and only the positive value is considered for the absolute value sign

6． \a\＝a（a＞0）

0（a＝0）

a（a＜0）

Such as 2 2 0 0 4 4

7．（√a）＾2＝a（a≥0）

a 2) a (a is the total real number).

1.If the image of the primary function y=(k-2)x+1 does not pass through the third quadrant, then what is the range of values of k? Process.

Answer: Because the image does not pass through the third quadrant.

So the intersection of the image and the y-axis is on the positive semi-axis of the y-axis.

So k-2 is greater than 0

Solution: k is greater than 2

1.Just make the slope (k-2) greater than or equal to 0, so k is greater than or equal to 2.

I don't understand the question, I'm sorry.

7。The seventh problem is that the number in the root number should be greater than or equal to 0, and the result is the number itself.

1.Because the image does not pass through the third quadrant, k-2 is less than 0, so k is less than 2

2: x to the third power = 2x to the second power + 2x + 1

You can draw an image of the function on both sides.

There are several intersections and there are several solutions. so easy...

A 2 + C 2 = 10 (2) C 2 + B 2 = 13 (1) 1 - 2.

b^2-a^2=3

a+b)(a-b)=3

ab positive integer a=1b=2

So c=3

Junior, do you see what this is, a system of ternary quadratic equations, you just need to find another equation about ABC, and the synthesis;

a 2 + c 2 = 10, c 2 + b 2 = 13 a, b, c are all positive integers.

A 2+C 2=10 A = 1 C = 3 or A = 3 C=1C 2+B 2=13 When C = 3, B = 2 When C = 1, B is not a positive integer and is rounded.

a = 1 b = 2 c = 3

Since 4 2 = 16, it is impossible to exceed 4a b c in a b c and can only take a 2+c 2=10 (1) c 2+b 2=13 (2) in 1 2 3

2) -1) b 2 - a 2 = 3 and in 1 2 3 only 2 - 1 2 = 3 satisfies this relation

So b = 2, a = 1, and in substitution (1) we get c = 3

c 2 = 10 - a 2<10, so c can only be 1, 2, 3 when c = 1, a = 3;c=2, it is not possible; When c=3 and a=1c=1, b is not an integer, so it is excluded;

In summary, a=1, b=2, c=3

Solution: Because a, b, c are non-negative, and a+b=7, so: b=7-a 0Sold to a 7

0 a=c-5 7Solution: 5 c 12

And s=a+b+c=c+, and max=12+7=19

So: m-n=7

Because b+c=12, it depends on the maximum and minimum values of a.

A has a maximum of 7 and a minimum of 0

Solution: c a 5

c＝a＋5s＝a＋b＋c＝7＋c7＋5＋a＝12＋aa≥0, b≥0

b＝7－a≥0

De: 0 A Jujube line imitation 7

When a 0, with stupid s minimum 12 0 12 n

When a 7, s max 12 7 19 m

m n stool fiber 19 12 7

According to the meaning of the title, you can bring the three points of the quadratic function into the equation with caution, and find a=1 2, b=1, c=-3 2, then the quadratic equation of the original branch is y1=1 2x 2+x-3 2, and only the inverse proportional function y2=k filial piety brother x and the quadratic equation y1 have an intersection point x0 in the first quadrant, and 20, so the value range of k is: 5

a= b=1 c=

From y1, y2 can be solved:

x(x-1)(x+3)=2k

It is easy to judge that on the 2 x 3 interval, k monotonous rent seepage increases.

5 "Late K" shouted paragraph 18

1）x²-4y²=1，x²-2xy=0

So (x + 2y)(x - 2y) =1, x(x -2y)=0

According to the equation on the right, if x=0, then substitute to the left, and the megablocker -4y 2 =1, there is no solution.

If x = 2y, substituting to the left, 0=1, there is still no solution.

2) Vedic theorem.

a + b = 2-m

ab = 3

and A2 + M-2)A + 3 =0, B 2 + M-2)B + 3 =0, A +Ma+3 = 2A, B Potato +MB+3 = 2B

a²+ma+3)(b²+mb+3)= 4ab =12

3) Simplify.

2（a+1-b) =b+3

a, b are both rational numbers, and a+1-b and b+3 are both rational numbers if both sides are equal.

Then 2(a+1-b) is a rational number, so a + 1-b=0, b+3=0, b= -3, a=-4

Vedic Theorem (WEDA's theorem): a quadratic equation ax 2+bx+c (a is not 0).

Let the two roots be x and y

then x+y=-b a

xy=c/a

The Vedic theorem can also be used in higher order equations. In general, for an nth order equation aix i=0

Its roots are denoted as x1, x2....,xn

We have xi=(-1) 1*a(n-1) a(n).

xixj=(-1)^2*a(n-2)/a(n)

xi=(-1)^n*a(0)/a(n)

where is the sum and is the product.

If a quadratic equation.

The root in the complex set is, then.

The French mathematician Veda was the first to discover this relationship between the roots and coefficients of modern number equations, so people call this relationship Vedt's theorem. History is interesting, Veda arrived at this theorem in the 16th century, and proved that this theorem depends on the fundamental theorem of algebra, which was only made by Gauss in 1799.

From the fundamental theorem of algebra can be deduced: any unary equation of n order.

There must be roots in the plural. Thus, the left end of this equation can be decomposed into the product of a factor in the range of complex numbers:

where are the roots of the equation. The coefficient of comparison between the two ends is known as the Vedic theorem.

Vedic theorem has a wide range of applications in equation theory.

Proof of theorem.

Let mathx 1 math, mathx 2 math be the two solutions of the unary quadratic equation mathax 2+bx+c=0 math, and let mathx 1 ge x 2 math. According to the root-finding formula, there is.

mathx_1=\frac}/math，mathx_2=\frac}/math

So mathx 1+x 2= frac + left (-b ight) -sqrt } frac math,mathx 1x 2= frac ight) left (-b - sqrt ight)} frac math

As for the average, may I ask Ranzhou what the average is.

Solution: a x ax 1 7a=0

From Veda's theorem:

x1＋x2=-1/a

x1·x2=1/a²－7

Since the two real roots of this equation are integers, 1 a must be an integer, where a can only be 1 and a 0

So: a=1

There is a two-digit number, the sum of the single digit and the ten-digit number is 14, let the single digit be a, if the ten-digit position is reversed, the resulting new two-digit state is exactly what the digit is.

Solution: Because the single digit is a, then the ten-digit number is (14-a) After changing the position, the single digit is (14-a) and the ten digit is a

Then 14-a+10a=14+9a

I'm a teacher, thank you.

If the single digit is a, then the ten digit is (14-a).

After changing positions, the single digit of Yeji is (14-a), the ten-digit number is always a, the ten-digit number of this number is 10a, and the new number is 14-a+10a=14+9a