Answer: analysis-the factor has four terms and four variables, and each variable appears twice. Obviously, we can see that these four terms are all products of a or b and x or y, so we have two solutions.
& lt 1 & gt； Consider x and y as unknowns, a and b as constants, and merge x and y.
2ax- 10ay+5by-bx = 2ax–bx– 10ay+5by =(2a–b)* x-(2a–b)* 5y =(2a–b)*(x–5y)
& lt2> regards A and B as unknowns, X and Y as constants, and merges A and B..
2ax- 10ay+5by-bx =(2x– 10y)* a+(5y-x)* b = 2(x–5y)* a-(x–5y)* b =(2a–b)*(x–5y)
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(2) Factorization, just like x 2+7x+ 10 = (x+2) (x+5). My teacher calls it cross multiplication, so I want to know the relationship between item C (10) and item B (7), and divide 10 into 2 and 5. Is it regular to subtract which one? (See the top note, thank you)
Answer: The factorization analysis of the unary quadratic equation AX 2+BX+C = 0 is as follows:
& lt 1 & gt； First of all, it depends on whether there is a solution, that is, whether B2–4ac > = 0 holds. If it is true, continue. If not, it can't be decomposed.
& lt2 & gtAx 2+bx+c = 0 = > x 2+(b/a) x+c/a = 0。 Generally speaking, no matter what happens, it can be solved in this way, that is, two solutions of the equation x 1 can be found. 2 =[-b+-(B2–4ac)( 1/2)]/2a =[-7+-(7 2–4 * 1 * 10)]。
x2+7x+ 10 =(x-(-2))*(x-(-5))=(x+2)*(x+5)。
= = = = = = According to your question, I will answer the cross multiplication, assuming that x2+7x+10 = (x+a) (x+b) = x2+(a+b) x+ab.
A+b = 7, ab = 10, obviously a b = 2 or 5. As for how to judge the addition and subtraction, it is very simple, mainly depending on whether the product ab of two factors is greater than 0. There are four situations (it seems complicated, but it is easy to practice more).
ab & gt0，a+b & gt； 0? a & gt0，b & gt0
ab & gt0，a+b & lt； 0? a & lt0，b & lt0
ab & lt0，a+b & gt； 0? A>0, b<0, | a | & gt|b| or a.
ab & lt0，a+b & lt； 0? A>0, b<0, | a | & lt|b| or a.
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(3) Factorization calculation 565 2 * 20-435 2 * 20.
The answer to this question is 2.6 *10 6; But I don't know how the process is worked out. There must be something in his method. It is impossible to directly calculate one by one. That's not scientific. (See the top note, thank you)
Answer: First of all, we all know that the power of 10 is the best. Then we see that both 565 and 435 are multiples of 5, and 20 = 2 2 * 5, so we can get 5 2 * 2 2 =10 2.
565^2*20-435^2*20 = 1 13^2 * 5^2 * 2^2 * 5 - 87^2 * 5^2 * 2^2 * 5
= 5 * 10^2 * ( 1 13^2 - 87^2) = 5 * 10^2 * ( 1 13 + 87) *( 1 13 - 87)
= 5 * 10^2 * 200 * 26 = 26 * 10^5 = 2.6 * 10^6
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(4) Can 47 6-47 5 be divisible by 46?
The answer to this question is yes, but I don't know how. It's impossible to work it out one by one, is it? What I want is a method (see top note, thank you)
Answer: very simple, factorization.
47 6-47 5 = 47 * 47 5–47 5 = (47-1) * 47 5 = 46 * 47 5 is divisible by 46, and the result is 47 5.
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(5) Without changing the fractional value, this big question is calculated by two small questions, and all the coefficients in the numerator and denominator of the following types are turned into integers, and the highest partial coefficient in the numerator and denominator is positive.
The first question (0.3-0.2m)/(0. 15-0.4m)
The answer is (4m-6)/(8m-3) brackets, and my calculation of (6-4m)/(3-8m) is just the opposite. In this case, will the answer be any different? Is my answer correct? Why do you want to do it like him? Thank you. The following questions are similar:
A: The m term is unknown, so the coefficient of m should be greater than 0, and all the coefficients are multiplied by 20 to become an integer. In fact, you can use a stupid method, that is, multiply it by 100 to become an integer, and then drop the score.
(0.3-0.2m)/(0. 15-0.4m)=(0.2m–0.3)/(0.4m-0. 15)=(20m-30)/(40m- 15)
At the same time divide by one and divide by five.
=(4m-6) / (8m -3)
Erti (1/2-0.2m)/(1/3m-1/4)
The answer is (12m-30)/(20m- 15). I added the brackets, and I figured out that (30- 12m)/(20m- 15) is just the opposite. In this case, will the answer be any different? Is my answer correct? Why do you want to do it like him? thank you
Answer: I think you may have written the title wrong. I guess it may be (0.2m–1/2)/(1/3m-1/4).
Multiply up and down by -60, (1/2-0.2m)/(1/4-1/3m) = (12m-30)/(20m-15).
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(7) I think this question is a bit difficult: When y = (x 2/2-3x) is known, what value does x take?
The value of (1)y is positive; (2) the value of y is negative; (3) the value of y is equal to 0; (4) the score is meaningless
Answer: In fact, this is a variant form of solving a quadratic equation with one variable. If x is considered as a variable/unknown, then the equation becomes
X2/2–3x–y = 0, x2–6x–2y = 0. If there is a solution, B2–4ac = (-6) 2–4 *1* (-2y) = 36+8y > = 0,8y & gt =-36, y.
& lt 1 & gt； When y>0, x > 3+ 3 or x
& lt2>y<0, which means -4.5 =
3+0 & lt； 3+(9+2y)^ 1/2 & lt； 3+3,3 < = x < six.
3–0 & gt； 3 -(9+2y)^ 1/2 & gt； 3–3，3 & gt= x & gt0
The range of x is 0.
& lt4> no solution is meaningless, that is, B2–4ac = (-6) 2–4 *1* (-2y) = 36+8y.
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(8) Outward bound training:
Observed value1/(1* 2)+1/(2 * 3)+1/(3 * 4) = (1-1/2)+(/kloc-0)
Calculate1/[x (x+1)]+1[(x+1) (x+2)]+1[(x+2) (x+3)]. ...
And find the value of the algebraic expression when x= 1
I want methods and rules that primary school students can understand and super-detailed explanations! thank you
Answer: Or factorization.
As can be seen from the formula 1,1(1* 2) =1–1/2, 1/(2 * 3) = 1/2. When the denominator is the product of two numbers A and B, and the difference between the two numbers is 1(b-a= 1), we can decompose the factor 1/ (a*b) into1/a–1/b (/kloc-)
It can also be seen that the result is equal to11–1divided by the maximum factor, that is,11/4 = 3/4.
Then our formula 2 is equal to
[ 1/x– 1/(x+ 1)]+[ 1/(x+ 1)- 1/(x+2)]+……。 +[ 1/(x+ 1998)- 1/(x+ 1999)]
= 1/x– 1/(x+ 1999)
When x = 1,
= 1/ 1 – 1/( 1+ 1999) = 1- 1/2000 = 1999/2000
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(9) What is the denominator of equation (0.3x+4)/(0.2)-(0.1x-3)/(0.5) =-9/5 converted into an integer? The answer and process are enough. I don't know if the answer is multiplied by the right on the left or-9,5.
Answer: This topic makes people feel a little puzzling. Is it the denominator of each term or something else?
(0.3x+4)/(0.2)-(0. 1x-3)/(0.5)=-9/5
For each item -> (3x+40)/2-(x-30)/5 =-9/5.
(Solution process:
(0.3x+4)/(0.2)-(0. 1x-3)/(0.5)=-9/5
Multiply both sides by 5 at the same time.
25 *(0.3x-4)- 10(0. 1x-3)=-9
7.5x- 100–x+30 =-9
6.5x = 100–30–9 = 6 1
x = 6 10/65 = 122/ 13)
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(10) I didn't write this knowledge book, but I have to take the exam. I don't know the name of this knowledge point. I'd better give an online tutorial or a name about this knowledge point. Thank you! ) Solve the equation (note that the equation is enclosed in brackets):
(x+y)^2-4(x+y)-45=0(x-y)^2-2(x-y)-3=0
Solution: I don't know how to work out these two formulas later, so I need to explain how to work out what theorem and process to use. Thank you.
Unknown equations (they come from equations, that is, enclosed in braces):
(x+y-9)(x+y+5)= 0(x-y-3)(x-y+ 1)= 0
Then, even more strangely, two unknown equations became four strange equations:
Equations (1)x+y-9=0 x-y-3=0.
Equation set (2)x+y+5=0 x-y+ 1=0.
Equation set (3)x+y+5=0 x-y-3=0.
Equation set (4)x+y-9=0 x-y+ 1=0.
Answer: Actually, it is very simple. You just need to take x+y or x-y as an unknown number. In fact, you still need to solve a quadratic equation.
Assuming that x+y = m and x–y = n, the equation is transformed into
(x+y)^2-4(x+y)-45=0-〉m^2–4m-45 = 0
(x-y)^2-2(x-y)-3=0 -？ n^2–2n–3 = 0
Then factorize
m^2–4m-45 = 0，(m -9)* (m+ 5) = 0，
M–9 = 0 or m+5 = 0.
n^2–2n–3 = 0，(n–3)*(n+ 1)= 0
N–3 = 0 or n+ 1 = 0.
Because x+y = m and x–y = n, so
X+Y–9 = 0 or X+Y+5 = 0-the solution of the first equation.
X-y-3 = 0 or X-Y+ 1 = 0- the solution of the second equation.
So the solutions of the two equations can be solved by any solution of the first equation and any solution of the second equation, and * * * can form four equations.
x+y–9 = 0，x–y–3 = 0-〉x = 6，y =3
x+y–9 = 0，x–y+ 1 = 0-〉x = 4，y =5
x + y + 5 = 0，x–y–3 = 0-〉x =- 1，y =-4
x + y + 5 = 0，x–y+ 1 = 0-〉x =-3，y =-2