A factorization problem in grade two of junior high school The square-y-20 factorization of Y

A factorization problem in grade two of junior high school The square-y-20 factorization of Y

y^2-y-20=(y-5)(y+4)
If you multiply by a cross, - 20 = (- 5) * 4, you can't multiply by other numbers, because - 5 + 4 = - 1 (equal to the coefficient of the first term)
y²-y-20
=(y-5)(y+4)
(y-5)(y+4)
y²-y-20
=(y-5)(y+4)
The coefficient of quadratic term multiplied by cross is equal to 1 * 1 = 1 - y = - 5 + 4 - 20 = - 5 * 4
So = (Y-5) (y-4)
(y-5)(y+4)
Factorization problem (grade 2)
（1）（x²+y²）²-4x²y²
There is another difficult question,
Given that P = 3xy-8x + 1, q = x-2xy-2, when x ≠ 0, 3p-2q = 7 holds, find the value of Y
This paper (x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\3xy-8x + 1) - 2 (x-2xy-2) = 7, namely (9xy + 4xy) + (- 24x-2x) + (3
Ask a few math questions, the solution of quadratic equation of two variables
1.｛x+1=5(y+2)
3(2x-5)-4(3y+4)=5
2.｛5(m-1)=2(n+3)
2(m+1)=3(n-3)
3.｛x+y=80
30%x+60%y=20%*90
4.｛2(x-150=5(3y+50)
10%x+3%y=8.5%*800
1. Simplify the original equations
x--5y=9 (1)
x--2y=6 (2)
(1) - (2) get
3y=--3
y=--1
Substituting y = -- 1 into (2) yields
x+2=6
X=4
So the solution of the original equations is: x = 4, y = -- 1
2. Simplify the original equations
5m--2n=11 (1)
2m--3n=--11 (2)
(1) * 3 - (2) * 2
11m=55
M=5
Substituting M = 5 into (1) yields
25--2n=11
--2n=--14
n=7.
So the solution of the original equations is: M = 5, n = 7
3. Simplify the original equations
x+y=80 (1)
x+2y=60 (2)
(2) - (1) get
y=--20
Substituting y = -- 20 into (1) yields
x--20=80
x=100.
So the solution of the original equations is: x = 100, y = -- 20
4. Simplify the original equations
2x--15y=550 (1)
10x+3y=6800 (2）
(2) -- (1) * 5
78y=4050
y=675/13
Substituting y = 675 / 13 into (1) yields
2x--11125/13=550
2x=18275/13
x=18275/26
So the solution of the original equations is: x = 18275 / 26, y = 675 / 13
The known set a = {y | Y & # 178; - (a | 178; + A + 1) y + a (a | 178; + 1) > 0}, B = {y | y = 1 / 2x & # 178; - x + 5 / 2, 0 ≤ x ≤ 3}
If a ∩ B = &;, find the value range of A
(CRA) ∩ B is obtained when a takes the minimum value that makes inequality X & # 178; + 1 ≥ ax hold
The first question: for the set a: (x-a) [x - (a ^ 2 + 1)] > 0 and ∵ a ^ 2 + 1 > a ∪ x ∈ (- ∞, a) ∪ (a ^ 2 + 1, + ∞) for the set B: y = 1 / 2 (x ^ 2-2x + 5), we get: y = 1 / 2 (x-1) ^ 2 + 2 ∵ 0 ≤ x ≤ 3 ∪ y ∈ [2,4] and ∩ B = &; the solution of ∪ A4 is a ∈ (- ∞, - √ 3) ∪ (√ 3,2). The second question: from the question
The first question: for the set a: (x-a) (x - (x-a) (x - (a ? 2 ＾ 2 + 1) for the set a: (x-a) [(x - (x - (a ＾ 2 ＾ 2 ＾ 2 + 1) for the set a: (x-a) [(x - (x - (a ＾ 2 (a ? 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 + 2 ＾ 2 ＾ 2 + 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ? 2 thesolution of a ＾ 2 and a ＾ 2 + 1 ＾ gt; 4 is a The second question: from the question, X ＾ 2-ax + 1 ≥ 0, ＾ Δ = a ＾ 2 -... To expand
The first question: for the set a: (x-a) (x - (x-a) (x - (a ? 2 ＾ 2 + 1) for the set a: (x-a) [(x - (x - (a ＾ 2 ＾ 2 ＾ 2 + 1) for the set a: (x-a) [(x - (x - (a ＾ 2 (a ? 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 + 2 ＾ 2 ＾ 2 + 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ＾ 2 ? 2 thesolution of a ＾ 2 and a ＾ 2 + 1 ＾ gt; 4 is a The second question: the second question: the second question: the second question: the second question: the second question, the second question: the second question, the second question, the second question: the second question, the second question, the second question, the second question, the second question, the second question, the second question, the second question, the second question, the second question, the second question, the second question, the second question, the second question, the second question, the second question, the second question, the second question, X? 2-2-ax ＾ 2-2-ax ＾ 2-ax + 1 ≥ 0, the smallest value of the smallest value of the smallest value of a ? ? ＾ ＾ ＾ ＾＾＾＾＾＾＾∩ B = [2,4] Put it away
If you want to ask three mathematical problems, you must solve them with a quadratic equation of two variables
1. The students in the first group are divided into several pencils. If each student takes 5 pencils, there are still 4 pencils left. If one person only takes 2 pencils, the rest can get 6 pencils. How many pencils are there for each student
2. Now we have to process 400 machine parts. If Party A does it for one day, and then two people do it for two days, there are still 60 unfinished parts. If two people cooperate for three days, there will be 20 overproduced parts. How many can party A and Party B do
3. The number of people in the second workshop of a factory is 30 less than 4 / 5 of that in the first workshop. If 10 people are transferred from the first workshop to the second workshop, the number of people in the second workshop is 3 / 4 of that in the first workshop. How many people are there in each workshop
X students, y pencils
Then: 5x + 4 = y; 6x + 1 = y, solution x = 3, y = 19 pencils
A Does X every day, B does Y every day,
Then: x + 2 (x + y) = 400-60; 3 (x + y) = 400 + 20, solution x = 60, y = 80;
X people in the first workshop, y people in the second workshop,
Then (4 / 5) X-Y = 30; (y + 10) / (X-10) = 3 / 4, x = 250, y = 170
Ha ha, it's so simple
Set M = {y ︱ y = x2 + 2, X belongs to R}, n = {t ︱ t = 5-2x-x2}, then m ∩ n =? M ∪ n =?
MUN=R;M U N=2
A mathematical problem about the quadratic equation of two variables
The school bought three tables and four chairs for 780 yuan. The price of each table is three times that of each chair. How much is each table?
I just want to know how to solve it
Set a table x yuan, a chair y yuan
｛3x+4y=780
｛x=3y
The solution of the equations is {x = 180, y = 60}
A: 180 yuan per table
In plane point set M = {(x, y) | x2-2x + 2 ≤ y ≤ 6x-x2-3, and X, y ∈ Z}, find the number of elements in M
From x2-2x + 2 ≤ 6x-x2-3, we get 2x2-8x + 5 ≤ 0. The solution is: X ∈ [2 − 62, 2 + 62] ? x ∈ Z ? x ∈ {1, 2, 3} when x = 1, 1 ≤ y ≤ 2, then (1, 2), (1, 1) ∈ m, when x = 2, 2 ≤ y ≤ 5, then (2, 2), (2, 3), (2, 4), (2, 5) ∈ m, when x = 3, 5 ≤ y
A mathematical problem of quadratic equation of two variables
It takes 5 hours for a ship to navigate 60 km downstream and 5 hours for a ship to navigate 40 km backward, then the water velocity is?
Using binary linear equation
Let the water velocity be x km / h
The speed is y kilometers per hour
So x + y = 60 △ 5 = 12 (1)
y-x=40÷5=8 (2)
(1)-(2)
x+y-y+x=4
2x=4
X=2
A: the water velocity is 2 km / h
Ship speed x water speed y
5（x+y）=60
5（x-y）=40
x=10
Y=2
X water velocity y ship velocity
5（x+y)=60
5 (y-x)=40
If we want to reduce the two formulas, we get x = 2km / h
Set ship speed V1, water speed v2
Let the current velocity be x and the ship velocity be y
60/5=y+x
40/4=y-x
X=2
Y=8
The flow velocity is 2km / h
x+y=12km/h x-y=8km/h
Ship speed x = (12 + 8) / 2 = 10km / h
Water velocity y = (12-8) / 2 = 2km / 2H
A = {x │ x2 + ax + 1 ≤ 0} B {x │ x2-3x + 2 ≤ 0} set a is a sufficient and unnecessary condition for set B. how to find the value range of a?
Mei
Let a be an empty set, and let 2-4ac be - 2 = 3 or - 2