Function f (x) = | X-1 | + | 2x-1 | + | 3x-1 | + | 4x-1 | + | 5x-1 | For detailed A kind of (^ω^) A kind of I can't afford to go up early Clever calculation

Function f (x) = | X-1 | + | 2x-1 | + | 3x-1 | + | 4x-1 | + | 5x-1 | For detailed A kind of (^ω^) A kind of I can't afford to go up early Clever calculation

Inter partition discussion, slow typing can not afford it, negative infinity to 1 / 5], [1 / 5,1 / 4] [1 / 4,1 / 3] [1 / 3,1 / 2] [1 / 2,1] [1, infinity], remove the absolute value, then calculate the minimum is much easier, the results are 2,3 / 5,7 / 4,2,7 / 2,10 respectively, the final minimum is 3 / 5, I didn't calculate carefully, you see
Zero segment method. Five small formulas correspond to five zeros, which are 1,1 / 2,1 / 3,1 / 4,1 / 5 respectively
If the five zeros are listed on the number axis, the whole real number set can be divided into six regions
Next, we just need to discuss the sign of each small formula in each region, so as to get rid of the absolute value sign and turn it into a one-time formula. Then we can find the maximum value in each interval. Finally, we can sum up and take the minimum value
Zero segment method. Five small formulas correspond to five zeros, which are 1,1 / 2,1 / 3,1 / 4,1 / 5 respectively
If the five zeros are listed on the number axis, the whole real number set can be divided into six regions
Next, we just need to discuss the sign of each small formula in each region, so as to remove the absolute value sign and change it into a one-time formula. Then we can find the maximum value in each interval. Finally, we can collect it and take the minimum value
① When x ≥ 1, f (x) = 15x-5 ≥ 10
② When 1 / 2 ≤ x < 1, f (x) = 13x-3 ≥ 7 / 2
③ When 1 / 3 ≤ x < 1 / 2, f (x) = 9x-1 ≥ 2
④ When 1 / 4 ≤ x < 1 / 3, f (x) = 3x + 1 ≥ 7 / 4
⑤ When 1 / 5 ＜ x ＜ 1 / 4, f (x) = - 5x + 3 ＞ 7 / 4
⑥ When x ≤ 1 / 5, f (x) = - 15x + 5 ≥ 2
The minimum value of F (x) is 7 / 4... Expansion
① When x ≥ 1, f (x) = 15x-5 ≥ 10
② When 1 / 2 ≤ x < 1, f (x) = 13x-3 ≥ 7 / 2
③ When 1 / 3 ≤ x < 1 / 2, f (x) = 9x-1 ≥ 2
④ When 1 / 4 ≤ x < 1 / 3, f (x) = 3x + 1 ≥ 7 / 4
⑤ When 1 / 5 ＜ x ＜ 1 / 4, f (x) = - 5x + 3 ＞ 7 / 4
⑥ When x ≤ 1 / 5, f (x) = - 15x + 5 ≥ 2
The minimum value of F (x) is 7 / 4
Given that E1 and E2 are two non collinear vectors, a = 2e1-e2, B = E1 + 3e2, and a + 2B and 2 λ A-B are collinear, then the real number λ = ()
Write the specific process immediately, online and so on, thank you
Please write the specific process, online, etc., thank you.
a+2b = 3e1+2e2
2λa-b = (4λ-1)e1+(-2λ-3)e2
So: 3: (4 λ - 1) = 2: (- 2 λ - 3)
λ = -0.5
I'm a math teacher. You can ask me if you have any questions
If we know that the discriminant of the equation x & sup2; - (a + 2) x + a-2b = 0 is equal to 0 and x = 1 / 2 is a root of the equation, then the value of a + B is
It is obvious that x = 1 / 2 is a quadratic equation with one variable
The simplest formula is x ^ 2-1 / 4x + 1 / 4 = 0
So a + 2 = 1 / 4
a-2b=1/4
a=-7/4
b=-1
a+b=-11/4
It is known that the solution of the system of equations 3x + 4Y = m + 18 4x + 3Y = 2m + 8 is suitable for - 1
3x+4y=m+18 （1）
4x+3y=2m+8（2）
(1) 3 - (2) × 4
9x-16x=3m+54-8m-32;
-7x=22-5m;
x=(5m-22)/7;
∴-1＜(5m-22)/7≤2;
-7＜5m-22≤14;
15＜5m≤36;
∴3＜m≤36/5;
If you don't understand this question, you can ask,
3x+4y=m+18 ①
4x+3y=2m+8②
① The formula is multiplied by 3, subtracted by 2 and multiplied by 4
Now I can handle it myself.
① 3x + 4Y = m + 18, 12Y = - 9x + 3M + 54
② 4X + 3Y = 2m + 8, 12Y = - 16x + 4m + 32
From the above, we can get x = m / 7-22 / 7
Because - 1
8y ^ 3-4y + 1 factorization
8y^3-4y+1
=(8y³－4y²)＋(4y²－4y＋1)
=4y²(2y－1)＋(2y－1)²
=(2y－1)(4y²＋2y－1)
Note: the rational number range decomposition ends here, the real number range decomposition uses the root method to continue decomposition
The method of dividing and adding items
Original formula = 8y ^ 3-4y & # 178; + 4Y & # 178; - 4Y + 1
=(8y³－4y²)＋(4y²－4y＋1)
=4Y & # 178; (2y-1) + (2y-1) &# 178; formula method
=(2y-1) (4Y & # 178; + 2y-1) common factor method
Rational range decomposition completed!
(1) Given x > 1 / 3, find the minimum value of y = 3x + (4 / 3x-1) (2) given X
Is it 4 divided by 3x or 4 / 3 times x
It is known that E1 and E2 are noncollinear vectors, a = E1 + λ E2, B = 2e1-e2. When a ‖ B, the real number λ is equal to
The answer is - 1 / 2. What is the specific process
a‖b
Then a = KB
e1+λe2=2ke1-ke2
therefore
2k=1,k=1/2
λ=-2=-1
Let E1 = (x1, Y1)
e2=(x2,y2)
a=(x1+λx2,y1+λy2)
b=(2x1-x2,2y1-y2)
Because a|b
Then (x1 + λ x2) / (2x1-x2) = (Y1 + λ Y2) / (2y1-y2)
(x1+λx2)(2y1-y2)=(y1+λy2)(2x1-x2)
2x1y1-x1y2 + 2 λ x2y1 - λ x2y2 = 2x1y1-x2y1 +... Expansion
Let E1 = (x1, Y1)
e2=(x2,y2)
a=(x1+λx2,y1+λy2)
b=(2x1-x2,2y1-y2)
Because a|b
Then (x1 + λ x2) / (2x1-x2) = (Y1 + λ Y2) / (2y1-y2)
(x1+λx2)(2y1-y2)=(y1+λy2)(2x1-x2)
2x1y1-x1y2+2λx2y1-λx2y2=2x1y1-x2y1+2λx1y2-λx2y2
2λx2y1+x2y1=2λx1y2+x1y2
(2λ+1)x2y1=(2λ+1)x1y2
(2λ+1)x2y1-(2λ+1)x1y2=0
(2λ+1)(x2y1-x1y2)=0
Because E1, E2 are not collinear vectors
Then X1 / x2y1 / Y2
That is, x1y2y1x2
That is, x2y1-x1y20
SO 2 λ + 1 = 0
λ = - 1 / 2 ﹣ Stow
It is known that the discriminant of the equation X2 - (a + 2) x + a-2b = 0 is equal to 0, and x = 12 is the root of the equation, then the value of a + B is 0___ .
From the meaning of the question, we can get: △ = [- (a + 2)] 2-4 × (a-2b) = 0, that is, A2 + 8b + 4 = 0, then substitute x = 12 into the original equation to get 2a-8b-3 = 0, according to the meaning of the question: A2 + 8b + 4 = 02a-8b-3 = 0, add the two equations to get A2 + 2A + 1 = 0, the solution to get a = - 1, substitute a = - 1 into 2a-8b-3 = 0 to get b = - 58, then a + B = - 138
Given that the solution of the system of equations 3x + 4Y = m + 184x + 3Y = 2m + 8 is suitable for the equation x + y = 8, the value of M is obtained
Substituting y = 8-x into the equations, we can get: 3x + 4 (8 − x) = m + 184x + 3 (8 − x) = 2m + 8, and we can get: − x + 32 = m + 18, ① x + 24 = 2m + 8, ② + 2: M = 10
Factorization: x ^ - 4xy + 4Y ^ - Z^
The original formula = (x-2y) ^ - Z^
=(x-2y-z)(x-2y+z)