Factorization (a + 1) (a + 2) + 1 / 4

Factorization (a + 1) (a + 2) + 1 / 4

(a+1)(a+2)+1/4
=a^2+3a+2+1/4
=a^2+3a+9/4
=(a+3/2)^2
The original formula = A & sup2; + 3A + 9 / 4 = (a + 3 / 2) & sup2;
Factorization 4 (a + b) ^ 2 - (B-A) ^ 2
the sooner the better!
4(a+b)^2-(b-a)^2
=:{2(a+b)}^2-(b-a)^2
=（2（a+b)+(b-a))(2（a+b)-(b-a))
=(a+3b)(3a+b)
Can quadratic trinomial be factorized?
In the range of real numbers, if there is no solution, it can't, if there is a solution, it can
In the complex number range, there is always a solution, so it can always be decomposed
So it depends on what level you learn
Not necessarily: for example: X & sup2; + X + 1 can not be hungry!
For the quadratic trinomial ax ^ 2 + BX + C of X,
If B ^ 2-4ac is a complete square, it can be factorized (in the range of rational numbers)
If it is in the range of imaginary number, it can be decomposed
What is the integer solution of inequality system x-20
The solution of inequality system x-20 is - 1
X-2x0==>x>-1
-1
Mathematical problems. (on quadratic equations of two variables)
If the solution of the system of equations {a1x + b1y = C1, a2x + b2y = C2 is x = 3, y = 4, and the solution of the system of equations {3a1x + 2b1y = 5c1, 3a2x + 2b2y = 5c2, Xiao Ming's idea: first, it seems that the condition is not enough, then it is found that the coefficients of the two systems of equations have certain rules, and he remembered: if the solution of the system of equations {2a-3b = 13, 3A + 5B = 30.9 is {a = 8.3, B = 1.2, and the solution of the system of equations {2 (x + 2) - 3 (Y-1) = 13, 3 (x + 2) - 5 (Y-1) = 30.9 is the substitution idea, Finally, there is a solution, please try to solve this problem
The two sides of the original equations are divided by 5, and then the deformation is as follows:
A1 (3 / 5x) + B1 (2 / 5Y) = C1 (3 / 5x means 3 / 5 times x)
a2(3/5x)+b2(2/5y)=c2
So the solution of the equation system a1x + b1y = C1, a2x + b2y = C2 is x = 3, y = 4
The results show that 3 / 5x = 3, 2 / 5Y = 4, and the solution of the original equations is
x=5 y=10
Known
2a-3b=13,
3a+3b=30.9
Then the system of equations
2(x+2)-3(y-1)=13
3(x+2)+5(y-1)=30.9
We can regard (x + 2), (Y-1) as a and B with known conditions respectively
Then there is
（x+2）=8.3
(y-1) =1.3
The solution is x = 6.3, y = 2.3
Inequality and mathematical problems of inequality group
A chicken farm uses cages to hold chickens. If each cage Holds 36 chickens, the remaining 11 chickens will be filled. If two cages are reduced, all chickens will be filled on average. Given that one cage can hold 45 chickens, how many chickens are there in this batch?
3071
Let x cages hold 36 chickens in each cage, there are 36x chickens in each cage, and the remaining 11 chickens are 36x + 11 chickens in total. If 2 cages are reduced, all chickens will be loaded on average, that is, X-2 cages are just right, so (36x + 11) / (X-2) is an integer (36x + 11) / (X-2) = (36x-72 + 83) / (X-2) = [36 (X-2) + 83] / (X-2) = 36 (X-2) / (X-2) + 83 / (X-2) = 36 + 83 / (X-2), so 83 / (X-2) is an integer, so X-2 is a divisor of 83, and 83 is a prime number, So X-2 = 1 or 83, x = 3, x = 85, if x = 3, there are 36x + 11 = 119, obviously X-2 = 1 cage, more than 45 cages per cage, x = 85, 36x + 11 = 3071, at this time, 85-2 = 83 cages, 3071 / 83 = 37 cages per cage, so there are 3071 cages
Mathematical problems about quadratic equation of two variables
1. If the equation (M + 1) x ^ | m | + (3 + m) y ^ 5m-9 = 4 is a quadratic equation with respect to X and y, find the value of m ^ 2 + n ^ 2
If the equation is quadratic
be
2m+3=1
5n-9=1
therefore
m=1,n=2
therefore
m²+n²=1²+2²=5
Several mathematics problems in the first semester of senior one, including solving inequality, basic inequality, inequality proof and value range,
1. (2-4x) / (X & # 178; - 3x + 2) ≥ x + 1 can not be reduced, x + 1 is too complex to move past, the answer is x ≤ 0 or 2 ≤ x < 3
2. Given that a > 0, find the minimum value of (A & # 178; + 16) / A + A / (A & # 178; + 16), the answer is 65 / 8, not 2
3. If a, B ∈ R +, then the size relation of a times of a, B times of B and (a + b) / 2 times of (a + b)
4 find the range y = (X & # 178; + 5) / root sign (X & # 178; + 4) the answer is x ≥ 5 / 2
I think I must have misunderstood the basic inequality. If all the answers are correct, a reward will be offered
The first answer is x ≤ 0 or 1 ＜ x ＜ 2, the answer is wrong one line ~ the rest is no problem... The reference answer may be wrong, but the probability is very small
1. Solve inequality (2-4x) / (X & # 178; - 3x + 2) ≥ x + 1
Multiply both sides by - 1 to get (4x-2) / (X & # 178; - 3x + 2) ≤ - (x + 1)
The result is [(4x-2) + (x + 1) (X & # 178; - 3x + 2)] / (X & # 178; - 3x + 2) = x (X & # 178; - 2x + 3) / (x-1) (X-2) ≤ 0
Because X & # 178; - 2x + 3 = (x-1) &# 178; + 2 ≥ 2 > 0, we can get the same solution inequality: X / (x-1) (X-2) ≤ 0
So the solution set is obtained quickly by the root axis method: {x ｜ x ≤ 0} ∪ {x ｜ 1
Question 1: it's not easy to make arbitrary reductions in Fractional Inequalities. What if the denominator is 0? You think you have made a reduction, and the result is a big fork... Move the one about X to one side, and then divide it into (...) / x ^ 2-3x + 2 ≥ 0, and then use a pin~
Question 2: you ignore the conditions for the basic inequality to be true. If and only if the former one is equal to the latter one, the basic inequality will be true. I calculated that a ^ 2-A + 16 = 0 a has no solution, so you can't use the basic inequality. You can use the check function to do... Expansion
Question 1: it's not easy to make arbitrary reductions in Fractional Inequalities. What if the denominator is 0? You think you have made a reduction, and the result is a big fork... Move the one about X to one side, and then divide it into (...) / x ^ 2-3x + 2 ≥ 0, and then use a pin~
Question 2: you ignore the conditions for the basic inequality to be true. If and only if the former one is equal to the latter one, the basic inequality will be true. I calculated that a ^ 2-A + 16 = 0 a has no solution, so you can't use the basic inequality. You can use the check function to do what is commonly known as the Nike function. If you are patient enough to find the derivative, you can also
Question 3: what did you write!
Question 4: write the numerator as x ^ 2 + 4 + 1, then divide it, write it as root sign x ^ 2 + 4 + root sign x ^ 2 + 1 / 4, and then... Don't use the basic inequality misunderstanding, just like question 2, if and only if it doesn't hold!! Using check function to solve
Well, you have a lot of questions..... A lot of misunderstandings..... I'll help you figure out questions 2 and 4. You see, question 2 can be written as x + 1 / X. do you understand that the value of a ^ 2 + 16 / A is the definition of X, so x ≥ 8. So if x = 8 is substituted, the original formula will be greater than or equal to 8 + 1 / 8, which is greater than or equal to 65 / 8. OK, haha! Next, let's look at question 4, According to what I said above, split 5 into 1 + 4, and then transform the formula into x + 1 / X. the range of x ^ 2 + 4 under the root sign is the definition field of X, so x ≥ 2, so the original formula becomes ≥ 2 + 1 / 2, that is, greater than or equal to 5 / 2. How about it all done ~ ~ simple, ha ha!! how can the fourth question above be bothered with derivation?... and --! It's still wrong? Put it away
The answer is wrong, students, you see the first question, 2 into the near denominator is 0
Will you copy the wrong answer? It's not that you have misunderstandings. Follow up: I've got the wrong number, the process of seeking, these questions have made me collapse. The key is that the thinking is easy to make mistakes
An elimination problem of quadratic equation with two variables
1, find the value of X satisfying 5x + 3Y = x + 2Y = 7
2. The following equations are solved by substitution method
{3x + 2Y = 14 and x = y + 3
{X-Y = 3 and 3x-8y = 14
1.5x+3y=7
x+2y=7
That is, x = - 1, y = 4
Let f (x) = - 3A ^ 2 + a (6-A) x + B, solve the inequality about a: F (1) > 0
Let f (x) = - 3A ^ 2 + a (6-A) x + B, solve the inequality about a: F (1) > 0
Sorry, wrong number
F (x) = - 3x ^ 2 + a (6-A) x + B, solve the inequality about a: F (1) > 0
f(1)=3a²+6a-a²+b>0
2a²+6a+b>0
If the discriminant is less than 0
36-8b9 / 2, then the parabola is always greater than 0
If discriminant = 0
Then B = 9 / 2
4a²+12a+9>0
(2a+3)²>0
2a+3≠0
If the discriminant is greater than 0
B
--
Is f (x) a new function
f（1）>0
-3a^2+a（6-a）*1+b>0
4a^2-6a-b