1、25(a-b) ^2-10(a-b)+1 2、(x+2)(x-6)+16 3、(x^2-2x) ^2+2(x^2-2x)+1

1、25(a-b) ^2-10(a-b)+1 2、(x+2)(x-6)+16 3、(x^2-2x) ^2+2(x^2-2x)+1

1 primitive = [5 (a-b)] ^ 2-2 * 5 (a-b) * 1 + 1 ^ 2
=[5(a-b)-1]^2=(5a-5b-1)^2
2 original formula = x ^ 2-4x-12 + 16
=x^2-2*x*2+2^2
=(x-2)^2
3 original formula = [x ^ 2-2x + 1] ^ 2
=(x-1)^4
1=[5(a-b)]^2-2*5(a-b)*1+1^2=[5(a-b)-1]^2=(5a-5b-1)^2
2=x^2-4x-12+16=x^2-2*2x+2^2=(x-2)^2
3=(x^2-2x+1)^2=(x-1)^4
Ask for:
(1)25（a+b)^2-4（2a-b)^2
(2)n^3(m-2)-n(m-2)
(3)3(x-2y)^2-12(x+y)^2
If you can, you'd better explain the steps,
Veda's theorem for quadratic equation of one variable
M is a root of the equation x ^ 2-2007x + 1 = 0. Find the value of m ^ 2-2006m + 2007 / (m ^ 2 + 1)
Given that α and β are two roots of the quadratic equation x ^ 2 + 3x-1 = 0, then the value of α ^ 2 + 2 α - β is_____
If M and N are two unequal real numbers with m ^ 2-2m = 1 and n ^ 2-2n = 1, find the value of 2m ^ 2 + 4N ^ 2-4n + 1994
The first m ^ 2-2006m = m-12007 / (m ^ 2 + 1) = 1 / m, then m ^ 2-2006m + 2007 / (m ^ 2 + 1) = (m ^ 2-m + 1) / M = 2006, the second α ^ 2 = 1-3a α ^ 2 + 2 α - β = 1 - (a + b) = 4, the third m, n are two 2m ^ 2 + 4N ^ 2-4n + 1994 = 2 + 4m + 4 + 8n-4n + 1994 = 2008 of x ^ 2-2x = 1 respectively
m2+1=2007m,2007/(m2+1)=1/m,m2-2006m=m-1,yuanshi=1/m+m-1,weidadingli
So, m times the other root is 1, the other root is 1 / m, and Guangxi is 2007, minus 1 = 2006
A2 + 2A = 1-A, the original formula = 1-a-b = 4
2m2=4m+1,4n2=8n+4,yuanshi=4m+4n+1999=2007
We're going to do it with Veda's theorem, a quadratic equation of one variable
When solving the quadratic equation of one variable, Zhang San misread the constant term and got two solutions of 3 and 4. Xiao Li misread the coefficient of one variable and got two solutions of 1 and 10
2 and 5
2 and 5
The equation of Zhang San's solution is (x-3) (x-4) = x ^ 2-7x + 12
The equation of Xiao Li's solution is (x-1) (X-10) = x ^ 2-11x + 10
Therefore, the correct equation should be x ^ 2-7x + 10 = (X-2) (X-5) = 0
The solution is x = 2 and x = 5. Thank you. Your method is very simple and easy to understand, but what I want is to use the Veda theorem to solve it. Because (x-3) (x-4) = x ^ 2-7x + 12 = 0, Weida theorem has X1 + x2 = 7 and square expansion
The equation of Zhang San's solution is (x-3) (x-4) = x ^ 2-7x + 12
The equation of Xiao Li's solution is (x-1) (X-10) = x ^ 2-11x + 10
Therefore, the correct equation should be x ^ 2-7x + 10 = (X-2) (X-5) = 0
The solution is x = 2 and x = 5. Question: Thank you, your method is very simple and easy to understand, but what I want is to use the Veda theorem to solve it.
Find a problem of solving quadratic equation of one variable by factorization in grade nine
9（X-2)^2=1
Factorization!
9（X-2)^2=1
(3x-6)^2-1=0
(3x-6+1)(3x-6-1)=0
(3x-5)(3x-7)=0
3x-5 = 0 or 3x-7 = 0
x1=5/3,x2=7/3
9（X-2)^2=1
9(x^2-4x+4)=1
9x^2-36x+36-1=0
9x^2-36x+35=0
(3x-5)(3x-7)=0
x1=5/3,x2=7/3
What are the four symbols behind the right bracket?
（x-2)^2=(1/3)^2
X-2 = + 1 / 3 or X-2 = - 1 / 3
therefore
X = 7 / 3 or x = 5 / 3
9(x-2)^2-1=0
[3(x-2)+1][3(x-2)-1]=0
(3x-5)(3x-7)=0
X = 5 / 3 or 7 / 3
The problem of quadratic equation with one variable in Grade 9 Mathematics (factorization method)
（2x-1）²-2（2x-2）=3
（2x-1）²-2（2x-1）=3
（2x-1）²-2（2x-1）-3=0
（2x-1-3）（2x-1+1）=0
2x（2x-4）=0
∴x[1]=0,x[2]=2
It is known that the equation (k-1) x (2k-2) x + K + 1 = 0 has two unequal real roots X1 and x2
(1) Find the value range of K
(2) Is there a real number k so that the two real roots of the equation are opposite to each other? If so, find out the value of K; if not, give the reason
1. The solution shows that: 4 (k ^ 2 + 2x + 1) - 4 (k ^ 2-1) = 8K + 8 > 0, k > - 1 is obtained;
2. When [- (2k-2) + (8K + 8) ^ 0.5] = [- (2k-2) - (8K + 8) ^ 0.5]
We get: K + 1 = - (K + 1), then k = - 1, when k = - 1, there are two equal real roots, so there is no K value, so the two real roots are opposite to each other
Solving the quadratic equation of two variables
The solution of the equations 3x + 5Y = 8 2x-3y = - 1
Equations x + y = 9 3 (x + y) + 2x = 33
Given that the equations 2x-y = 7 ax + y = B and X + by = a 3x + y = 8 have the same solution, find the value of 3a-2b
Given the equations 2x + y = 7, x + 2Y = 6, then X-Y =?
Don't use curly braces
[reference answer]
1、3x+5y=8 ①
2x-3y=-1 ②
① X2 is 6x + 10Y = 16, ③
② X3 is 6x-9y = - 3, ④
③ - 4 gives 19y = 19, i.e. y = 1
Take y = 1 into 2 to get x = 1
The solutions of the original equations are x = 1, y = 1
2、x+y=9 ①
3(x+y)+2x=33 ②
X = 9-y (3) is obtained from (1)
③ Bring it in
3×9+18-2y=33
It is reduced to y = 6
Take y = 6 into 3 to get x = 3
The solutions of the original equations are x = 3, y = 6
3. The simultaneous equations 2x-y = 7 (1) and 3x + y = 8 (2) are obtained
① + 2 gives x = 3
Take x = 3 into ① to get y = - 1
The solution of the equations is x = 3, y = - 1
Take x = 3, y = - 1 into the equations ax + y = B and X + by = A and get the solution
a=1,b=2
∴ 3a-2b=3×1-2×2=-1
4、2x+y=7 ①
x+2y= 6 ②
① - 2
(2x+y)-(x+2y)=7-6
It is reduced to X-Y = 1
∴ x-y=1
Well, is this a topic?
1 x=1 y=1
2 x=3 y=6
3 x=3 y=-1 3a-1=b 3-b=a a= 1 b=2 3a-2b=-1
4 x=8/3 y=5/3 x-y=1
Ax + y = 3x-1 to find the value of a
Ax + y = 3x-1 to find the value of a
ax=3x
y=-1
A=3
Mathematical problems solved by quadratic equation of two variables
Party A and Party B start from 8 am to 10 am, and the distance between them is 36km. By 1 pm, the distance between them is still 36km. If Party A travels 2km more than Party B every hour, how about the distance between a and B?
The solution of quadratic equation of two variables
Suppose that the distance between a and B is YKM, and the speed of B is XKM per hour
y=（2x+2）*2+36
y=（2x+2）*3-36
So y = 180
A. B is 180km away from each other
If the velocity of a is V, then the velocity of B 4 is V-2
Set the distance as S
Then the equations are 2 × (2v-2) + 36 = s
5×（2V-2)-36=S
The solution is s = 84