Finding n examples of factorization / fractional operation

Finding n examples of factorization / fractional operation

1. (2 points) judge right or wrong: decomposition factor: (x2-y2-z2) 2-4y2z2 = (x + Y-Z) (X-Y + Z) (x + y + Z) (X-Y-Z) () 2. (2 points) judge right or wrong: decomposition factor: A2 + b2-2ab-4 = (a-b-2) (a-b + 2) () 3. (2 points) judge right or wrong: decomposition factor: a4-3a2-4 = (A-2) (a + 2) (A2 + 1) () 4. (2 points) judge right or wrong: decomposition factor: a4-3a2-4 = (A-2) (a + 2) (A2 + 1) () 4
Given that the line L: 3x-4y + 2 = 0 and the circle C: (x-4) 2 + (Y-1) 2 = 9, then the positional relationship between the line L and the circle C is ()
A. L is tangent to C. B. l intersects with C and passes through the center of C. C. L is away from C. D. l intersects with C and does not pass through the center of C
From the circle C: (x-4) 2 + (Y-1) 2 = 9, we get the center C (4,1), radius = 3, ∵ the distance d from the center C to the line L: 3x-4y + 2 = 0 = | 12 − 4 + 2 | 5 = 2 < R = 3, ∵ the line L intersects the circle C, and the center (4,1) does not satisfy the equation 3x-4y + 1 = 0, ∵ the line L is not the center C, then the position relationship between the line L and the circle C is the intersection of L and C, but not the center of C
The general steps of solving quadratic equation with one variable by factorization method are: 1__ That is, the right side of the equation is 0, ②__ From one variable quadratic equation to two one variable quadratic equations___
The general steps of solving quadratic equation with one variable by factorization method are as follows
①_ Transference_ That is, the right side of the equation is 0,
②_ Factorization_ From one variable quadratic equation to two one variable linear equations
③___ Solving two linear equations with one variable respectively
The problem of factorization and fraction
1. When (a + b) / (a-b) = 3 / 2, find the value of (a ^ 2-B ^ 2) / ab
2. When x + 2 = 1 / x, find the value of 1 / (x + 1) - (x + 3) / (x ^ 2-1) * (x ^ 2-2x + 1) / (x ^ 2 + 4x = 3)
3. When a, B and C are not 0 and a + B + C = 0, find the value of a (1 / B + 1 / C) + B (1 / A + 1 / C) + C (1 / A + 1 / b)
(the first two questions are fractions, and the last one is factorization.)
(fraction: "/" is division sign, "*" is multiplication sign and "^ 2" is square)
1、∵(a+b)/(a-b)=3/2
2 A / B + 2 = 3A / B-3
A / b = 5
The original formula (a ^ 2-B ^ 2) / AB is divided by B ^ 2
[（a/b）^2-1]/(a/b)
And ∵ A / b = 5
(5 × 5-1) / 5 = 24 / 5
∴(a^2-b^2)/ab ＝24/5
2. I want the building to unfold
1、∵(a+b)/(a-b)=3/2
2 A / B + 2 = 3A / B-3
A / b = 5
The original formula (a ^ 2-B ^ 2) / AB is divided by B ^ 2
[（a/b）^2-1]/(a/b)
And ∵ A / b = 5
(5 × 5-1) / 5 = 24 / 5
∴(a^2-b^2)/ab ＝24/5
2. I think the last "= 3" in the above formula should be + 3, right? Ha ha
It can be concluded from the questions:
1/(x+1)-(x+3)/(x^2-1)*(x^2-2x+1)/(x^2+4x+3)
=1 / (x + 1) - (x + 3) / (x + 1) (x-1) * (x-1) ^ 2 / (x + 1) (x + 3)
=1/(x+1)-（x-1)/(x+1)(x+1)
=(x+1)/(x+1)^2-(x-1)/(x+1)^2
=(x+1-x+1)/(x+1)^2
=2/(x+1)^2
And ∵ x + 2 = 1 / X
Two sides multiply x to get x ^ 2 + 2x = 1
Two sides add one to get: x ^ 2 + 2x + 1 = 2
The simplification is: (x + 1) ^ 2 = 2
In the formula, we can get that:
1/(x+1)-(x+3)/(x^2-1)*(x^2-2x+1)/(x^2+4x+3)＝2/(x+1)^2=2/2=1
The original formula 1 / (x + 1) - (x + 3) / (x ^ 2-1) * (x ^ 2-2x + 1) / (x ^ 2 + 4x + 3) = 1
3、∵a+b+c=0
Two sides divide a, B and C respectively
1+b/a+c/a=0 ∴b/a+c/a=-1
a/b+1+c/b=0 ∴a/b+c/b=-1
a/c+b/c+1=0 ∴a/c+b/c=-1
The original formula is as follows:
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)
=a/b+a/c+b/a+b/c+c/a+c/b
=（a/b+c/b）+（a/c+b/c）+（b/a+c/a）
By substituting B / A + C / a = - 1, a / B + C / b = - 1 and a / C + B / C = - 1 into the formula, we can get that:
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)
=（a/b+c/b）+（a/c+b/c）+（b/a+c/a）
=-1-1-1
=-3
The original formula a (1 / B + 1 / C) + B (1 / A + 1 / C) + C (1 / A + 1 / b) = - 3
1. When (a + b) / (a-b) = 3 / 2, find the value of (a ^ 2-B ^ 2) / ab.
If (a + b) / (a-b) = 3 / 2, a = 5b is obtained, and (a ^ 2-B ^ 2) / AB = 24 / 5 is carried in
2. When x + 2 = 1 / x, find the value of 1 / (x + 1) - (x + 3) / (x ^ 2-1) * (x ^ 2-2x + 1) / (x ^ 2 + 4x = 3).
I didn't understand what I was asking for
3. When a, B and C are not 0 and a + B + C = 0, we can find a (1 / B + 1 / C) + B (1 / a... expansion
1. When (a + b) / (a-b) = 3 / 2, find the value of (a ^ 2-B ^ 2) / ab.
If (a + b) / (a-b) = 3 / 2, a = 5b is obtained, and (a ^ 2-B ^ 2) / AB = 24 / 5 is carried in
2. When x + 2 = 1 / x, find the value of 1 / (x + 1) - (x + 3) / (x ^ 2-1) * (x ^ 2-2x + 1) / (x ^ 2 + 4x = 3).
I didn't understand what I was asking for
3. When a, B and C are not 0 and a + B + C = 0, find the value of a (1 / B + 1 / C) + B (1 / A + 1 / C) + C (1 / A + 1 / b).
a(1/b+1/c)=a(b+c)/bc=-a^2/bc=-a^3/abc
Similarly, B (1 / A + 1 / C) = - B ^ 3 / ABC
c(1/a+1/b)=-c^3/abc
So the original = - (a ^ 3 + B ^ 3 + C ^ 3) / ABC
Because a + B = - C
So a ^ 3 + B ^ 3 + 3AB ^ 2 + 3A ^ 2B = - C ^ 3
So - (a ^ 3 + B ^ 3 + C ^ 3) = 3AB (a + b) = - 3ABC
Therefore, the original formula = - 3
What is the positional relationship between the line 3x-4y-6 = 0 and the circle (x-1) &# 178; + (Y-3) &# 178; = 9
d=|3-12-6|/√(3²+4²)
=15/5
=Circle radius = 3
So the line 3x-4y-6 = 0 is tangent to the circle (x-1) &# 178; + (Y-3) &# 178; = 9
The basic idea of solving binary linear equations is to transform them into unitary linear equations. The elimination methods include elimination method and elimination method
The basic idea of solving the system of linear equations with two variables is to transform it into linear equations with one variable,
The elimination methods include addition and subtraction elimination method and substitution elimination method
Factorization and fraction calculation
1. Factorization 1. X ^ 5 + X-1
2.(a+b)(a+b-ab)+(ab-1)(ab+1)
2. Only: n is a positive integer, and n ^ $- 16N ^ 2 + 100 is a prime number
3. Let a = x / (y + Z), B = Y / (x + Z), C = Z / (x + y), and X + y + Z ≠ 0
Find a / (a + 1) + B / (B + 1) + C / (c + 1)
There should be no way to decompose 2. (a + b) (a + b-ab) + (AB-1) (AB + 1) = (a + b) (1-ab) - (1-ab) (1 + AB) = (1-ab) (a + b-ab-1) = (1-ab) (A-1) (1-B) 3. N ^ 4-16n ^ 2 + 100 = (n ^ 2-6n + 10) (n ^ 2 + 6N + 10) n ^ 4-16n ^ 2 + 100 is prime n ^ 2-6n + 10 = 1 or n ^ 2 + 6N + 10 = 1, n = 3 or
(x^2-x+1)*(x^3+x^2-1)
N
x/(y+z)/(x/(y+z)+1)+y/(x+z)/(y/(x+z)+1)+z/(x+y)/(z/(x+y)+1)
What is the positional relationship between a straight line x + y = 5 and a circle O: X & # 178; + Y & # 178; - 4Y = 0?
The solution is x ^ 2 + * Y-2) ^ 2 = 4 from the circle X & # 178; + Y & # 178; - 4Y = 0, so the center of the circle is (0,2) radius r = 2, the distance from the center of the circle (0,2) to the straight line x + y = 5 d = / 0 + 2-5 / / √ (1 ^ 2 + 1 ^ 2) = 3 / √ 2 = 3 √ 2 / 2 ＜ 2, i.e. D ＜ R, so the position relationship between the straight line x + y = 5 and the circle O: X & # 178; + Y & # 178; - 4Y = 0 is intersection
The standard equation x ^ 2 + (Y-2) ^ 2 = 4
Then we can know that the center of the circle is (0,2)
If d > R, it is outside the circle, d = R, it is tangent, Dr
So we are apart.
The equation 2kx-4y + k = 3, if the equation is a bivariate linear equation, then K? If the equation is a univariate linear equation, then the solution of the equation is?
The equation 2kx-4y + k = 3, if the equation is a bivariate linear equation, then K? If the equation is a univariate linear equation. Then the solution of the equation is (find x)?
1. 2k≠0 k≠0
2. 2k=0 k=0
The solution is y = - 3 / 4
1,k=1
2,k=0 ,y=-3/4
This equation must be a quadratic equation of two variables with K as a parameter
Inequality, factorization and typical problems of fraction
Beijing Normal University version, to some classic types of questions, often test
1. 1 / A + 1 / b = 4, find (a-3ab + b) / (2a + 2b-7ab) 2, ABC ≠ 0, (a + B + C) / C = (a + B + C) / a = (a + B + C) / B, find (a + b) (B + C) (c + a) / Abc3, a / BC + B / AC + C / AB not equal to several 4, X & sup2; + 3x + 1 = 0, find x ^ 4 + 1 / x ^ 45, ABC = 1, find 1 / (AB + A + 1) + 1 / (BC + B + 1) + 1 / (Ca + C + 1)
1.a^4-4a+3
2.(a+x)^m+1*(b+x)^n-1-(a+x)^m*(b+x)^n
3.x^2+(a+1/a)xy+y^2
4.9a^2-4b^2+4bc-c^2
5.(c-a)^2-4(b-c)(a-b)
6.x^2+2x-8
7.x^2+3x-10
8.x^2-x-20
9.x^2+x-6
... unfold
1.a^4-4a+3
2.(a+x)^m+1*(b+x)^n-1-(a+x)^m*(b+x)^n
3.x^2+(a+1/a)xy+y^2
4.9a^2-4b^2+4bc-c^2
5.(c-a)^2-4(b-c)(a-b)
6.x^2+2x-8
7.x^2+3x-10
8.x^2-x-20
9.x^2+x-6
10.2x^2+5x-3
If ͪ (2x3) (ͪ) (2x3) (ͪ (2x9), then
A．2 B． 4 C．6 D．8
12. If 9x2 & # 8722; 12xy + m is the square of the sum of two numbers, then the value of M is ()
A．2y2 B．4y 2 C．±4y2 D．±16y2
13. The factorization result of polynomial A4 &; 2a2b2 + B4 is ()
A．a2(a2−2b2)+b4 B．(a2−b2)2
C．(a−b)4 D．(a+b)2(a−b)2
14. Decompose (a + b) 2 &; 4 (A2 &; B2) + 4 (A &; b) 2 into ()
A．( 3a−b)2 B．(3b+a)2
C．(3b−a)2 D．( 3a+b)2
The results are (∟ 22; ∟) and (∟) respectively
A．(−)2003 B．−(−)2001
C． D．−
16. Given that X and y are arbitrary rational numbers, M = x2 + Y2, n = 2XY, then the relation between M and N is ()
A. M > n B. m ≥ n C. m ≤ n D
17. For any integer m, the polynomial (4m + 5) 2 &; 9 can ()
A. Divisible by 8 B. divisible by M
C. Divisible by (M &; 1) d. divisible by (2n &; 1)
18. Decompose the factor, 3x2n and 6xn, and the result is ()
A．−3xn(xn+2) B．−3(x2n+2xn)
C．−3xn(x2+2) D．3(−x2n−2xn)
19. In the following deformations, the correct factorization is ()
A． 0.09m2− n2 = ( 0.03m+ )( 0.03m−)
B．x2−10 = x2−9−1 = (x+3)(x−3)−1
C．x4−x2 = (x2+x)(x2−x)
D．(x+a)2−(x−a)2 = 4ax
20. The common factor of polynomial (x + Y &; Z) (X &; y + Z) &; (y + Z &; x) (Z &; X &; y) is ()
A. X + Y &; Z b.x &; y + Z C.Y + Z &; X D. does not exist
21. Given that x is any rational number, then the value of the polynomial X &; 1 &; X2 ()
A. It must be negative
B. It can't be a positive number
C. It must be a positive number
D. It can be positive or negative or zero
①：x^2(a-b)+(b-a)
②：x^2y-2xy^2+y^3
③：(b-a)^2+2(a-b)+1
④：(x^2-y^2)^2-(x^2+y^2)^2
⑤：xy^3-6x^2y^2+xy^3
⑥：(3a-b)^2-(b-3a)^2
⑦：x^2(a+b)-36(b+a)
⑧：36b-a^2b
⑨：(a-b)^2+2(b-a)+49
⑩: x^2-4xy+4y^2-9