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Given the polynomial a = x * x-x + B, B = x * x-ax + 3, and A-B = x + 2, find the value of a and B?
No. 1 AAA:
A-B
=(x²-x+b)-(x²-ax+3)
=x²-x+b-x²+ax-3
=(a-1)x+b-3
=x+2
∴a-1=1
b-3=2
The solution is a = 2, B = 5
The values of a and B are 2 and 5 respectively
The polynomial (| a | - 4) X4 - (A-4) X3 + X + 4 is a cubic trinomial of X, and a2-3a + 4 is obtained
The polynomial (| a | - 4) X4 - (A-4) X3 + X + 4 is a cubic trinomial of X
Then: | a | - 4 = 0, and A-4 ≠ 0
The solution is: a = - 4
a²-3a+4
=16+12+4
=32
If the polynomial X4 - (A-1) X3 + 5 x2 + (B + 3) X-1 does not contain X3 and X terms, find the value of 3A + B
Because X4 - (A-1) X3 + 5 x2 + (B + 3) X-1 does not contain X3 and X terms
So A-1 = 0, B + 3 = 0
So a = 1, B = - 3
So 3A + B = 1 × 3 + (- 3) = 0
When a car goes from place a to place B, it takes three times as long as it takes to go back on the original road. When it comes back, it travels 36 kilometers more per hour than when it goes back. It's shared back and forth
Let X be the velocity of a when going, x + 36 when returning, and y be the distance
y：x=3y：（x+36）
x+36=3x
X = 18 km / h
The return speed is 54 km / h
Not enough, right?
What is the known travel time
The round trip time is Y / 18 + Y / 54
If we know that it takes t to go back and forth, it is t / 4
Distance AB = 18t / 4
3x of 5 + 2 of 5 = 5x solution
It's easier to change fractions into decimals
0.6x+0.4=5x
5x-0.6x=0.4
4.4x=0.4
X = 1 / 11 (because 0.4 ÷ 4.4 can't be divided completely, it is expressed as a fraction)
I don't know if you understand,
For a batch of goods, 3 / 7 of them will be transported in the first time, and the remaining 5 / 6 will be transported in the second time
In the first time, 3 / 7 of the goods were transported away, and the remaining 5 / 6 were transported away in the second time
3/7+(1-3/7)×5/6=3/7+10/21=19/21
Once the function y = KX + m passes through the point (- 1, - 2), please write a value that makes the function image pass through the second, third and fourth image limits M
It should be m < - 2
When y = KX + m passes through quadrant two three four, it means that both K and m are less than 0
Take (- 1, - 2) into, - K + M = - 2, - k = - 2-m
Because K < 0
So - k > 0
-2-m＞0
m＜-2
It should be. You're asking your classmates
There are 72 male workers and 48 female workers in a workshop. The number of workers in the workshop accounts for 2 / 5 of the total number of workers in the factory. How many workers are there in the factory?
（72+48）/（2/5)=300
The factory has 300 employees
Given that the value of the algebraic formula 2 (X & sup2; + ax + y) - 2 (BX & sup2; + 3x) - 5y-1 has nothing to do with the value of the letter X, find 2 (A & SUP)
1. Given that the value of the algebraic formula 2 (X & sup2; + ax + y) - 2 (BX & sup2; + 3x) - 5y-1 has nothing to do with the value of the letter X, find the value of 2 (A & sup3; - 2b & sup2; - 1) - 3 (A & sup3; - 2b & sup2; - 1)
2（x²+ax+y）-2（bx²+3x）-5y-1
=2(1-b)x^2+2(a-3)x-3y-1
How do you calculate this step?
2（x²+ax+y）-2（bx²+3x）-5y-1
=2(1-b)x^2+2(a-3)x-3y-1
Because it has nothing to do with the value of X, so
1-b=0
a-3=0
therefore
b=1
a=3
Substituting
2（a³-2b²-1）-3（a³-2b²-1）
=-(a³-2b²-1）
=-(27-2-1)
-24
------------------------------
Merge congeners
2（x²+ax+y）-2（bx²+3x）-5y-1
=2(1-b)x^2+2(a-3)x-3y-1
Boys account for 5 / 6 of girls in the chorus, and then three girls are added. At this time, boys account for 5 / 12 of the total number of chorus. How many men and women are there?
Do with invariants, there must be a list of conditions. There must be clear steps
Divide the girls into 6 parts. The number of boys is the same as that of girls, which is also regarded as 5 parts. There are 11 parts in total, which is 5 / 11 of the total number of boys. Plus 3 girls, boys account for 5 / 12 of the total, that is, the total is divided into 12 parts, boys are still 5 parts, that is to say, the total is changed from 11 parts to 12 parts. The extra part is the 3 people who joined later, so
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