An analysis of the wrong problem (9x & # 178; - 2x + 7) - (X & # 178; + 3x-2) = 9x & # 178; - 2x + 7 + X & # 178; - 3x + 2 = (9x & # 178; + X & # 178;) + (- 2x-3x) + (7 + 2) = 10x & # 178; + (- 5x) + 9

An analysis of the wrong problem (9x & # 178; - 2x + 7) - (X & # 178; + 3x-2) = 9x & # 178; - 2x + 7 + X & # 178; - 3x + 2 = (9x & # 178; + X & # 178;) + (- 2x-3x) + (7 + 2) = 10x & # 178; + (- 5x) + 9

\X100 / x1009x & # 178; - 2x + 7) - (X & # 178; + 3x-2) = 9x & # 178; - 2x + 7 + X & # 178; - 3x + 2 (wrong brackets removed) should be 9x & # 178; - 2x + 7-x & # 178; + 3x-2 = (9x & # 178; - X & # 178;) + (- 2x + 3x) + (7-2) = 8x & # 178; + X + 5
1. Given f (2x-1) = x & # 178; - 3x for f (x) 2. F (√ x + 2) = x - √ x for f (x)
3. It is known that f = (x + 1 / x) = x & # 178; + 1 / X & # 178; - 5 for f (2) 4. F (x) is a quadratic function f (x) + F (x-1) = - 2x & # 178; + X-1 for f (x)
f(2x-1)=x²-3x=(2x-1)²/4-2x+1/4=(2x-1)²/4-(2x-1)-3/4∴f(x)=x²/4-x-3/4f（√x+2）=x-√x=(√x+2)²-5√x-4=(√x+2)²-5(√x+2)+6∴f(x)=x²-5x+6f(x+1/x)=x²+1/x²-5=(x...
Try to determine the value range of real number a so that the system of inequalities (x / 2) + (x + 1 / 3) > 0; X + (5a + 4 / 3) > 4 / 3 (x + 1) + a
Try to determine the value range of real number a so that the system of inequalities (x / 2) + (x + 1 / 3) > 0; X + (5a + 4 / 3) > 4 / 3 (x + 1) + A has exactly two integer solutions
The answer is 1 / 2
The building owner added the title:
The process is as follows:
(x / 2) + (x + 1 / 3) > 0
3x + 2x + 2 > 0;
It can be sorted as (3 + 4) + (3) + (5 a) + (4 a) + (4 a) + (4 a) + (4 a) + (4 a) + (4 a) + (4 a) + (4 a) + (4 a) + (4 a) + (4 a) + (4 a) + (4 a) + (4 a) + (4 a) +
X+(5a+4)/3>4X/3+4/3+a;
The results are as follows: X / 31 / 2;
And because: X
x/2+(x+1)/3>0 5x/6>-1/3
x>-2/5
x+(5a+4)/3>4/3(x+1)+a
x/34/3(x+1)+a
X / 3-1 / 3 is x > - 2 / 9;
After finishing with x + (5a + 4 / 3) > 4 / 3 (x + 1) + A, the following results can be obtained
X+(5a+4)/3>4X/3+4/3+a;
-2/94/3(x+1)+a
X/3
Try to determine the value range of real number a, so that the inequality system x2 + X + 13 ＞ 0x + 5A + 43 ＞ 43 (x + 1) + A has exactly two integer solutions
From x2 + X + 13 ＞ 0, multiply both sides by 6 to get 3x + 2 (x + 1) ＞ 0, the solution is x ＞ - 25, from x + 5A + 43 ＞ 43 (x + 1) + A, multiply both sides by 3 to get 3x + 5A + 4 ＞ 4 (x + 1) + 3a, the solution is x ＜ 2a, the solution set of the original inequality system is - 25 ＜ x ＜ 2A
If x = - 2 is the solution of the equation 3x + 5 = x / 4-m, what is the square of M
Substituting x = - 2 into 3x + 5 = x / 4-m leads to
m=1/2
∴m²=1/4
Substituting x = - 2 into the equation 3x + 5 = x / 4-m
3×（-2）+5=-2/4-m
-1=-1/2-m
m=1/2
m²=1/4
1/4
Taking x = 2 into the equation, we can get m = 1 / 2, so m square equals 1 / 4
Take x = - 2 into the equation 3x + 5 = x / 4-m and get m = 1 / 2, then the square of M is equal to 1 / 4
3X+5=X/4-m X/ 4 - m
X = - 2 is substituted into the original equation
3 * (- 2) + 5 = - 2 / 4-m
-6 +5 = -1/2 -m
m = 1/2
m ² = 1/4
I can't see your denominator is exactly X / 4 - M
Or X /... Expansion
3X+5=X/4-m X/ 4 - m
X = - 2 is substituted into the original equation
3 * (- 2) + 5 = - 2 / 4-m
-6 +5 = -1/2 -m
m = 1/2
m ² = 1/4
I can't see your denominator is exactly X / 4 - M
Or X / (4-m)
X/ （4-m）
3 * （-2）+ 5 = -2 /（ 4-m）
-1 = -2 /（ 4-m）
m = 2
M & sup 2; = 4; put away
Given the complete set u = a ∪ B = {x ∈ n | 0 ≤ x ≤ 10}, a ∩ (B is the complement of U) = {1,3,5,7), try to find the set B
Given the complete set u = a ∪ B = {x ∈ n | 0 ≤ x ≤ 10}, a ∩ (cub) = {1,7), trying to find the set B is actually like this, maybe I don't express it well.
B = {0,2,4,6,8,9,10} complete set u = a ∪ B = {x | x ∈ n, | 0 ≤ x ≤ 10} = {0,1,2,3,4,5,6,7,8,9,10} a ∩ (cub) = {1,3,5,7} indicates that 1,3,5,7 is in the complement of set a, so 1,3,5,7 is not in set B, and a ∪ B = {0,1,2,3,4,5,6,7,8,9,10} so B = {0,2,4,6,8,9
If there is a problem with the title, if (B is the complement of U), then B is not in U, but in front of the title, u = a ∪ B, so the contradiction, let's see the title clearly~
Calculation: (1) (− 32x2y) 2 (2x2 − 4xy + 7y2) (2) (- 4x-3y2) (3y2-4x)
(1) Original formula = 94x4y2 (2x2-4xy + 7y2), = 92x6y2 − 9x5y3 + 634x4y4; (2) original formula = (- 4x-3y2) (- 4x + 3y2), = (- 4x) 2 - (3y2) 2, = 16x2-9y4
What is the circumference formula of a circle?
I want to be more detailed, I just learned circle, I don't know what TTR is
Circumference of circle = 2 × radius × circumference = diameter × circumference
The circumference of the circle is 2 π R
For the equations of X and y, x + 2Y = 5m, X-Y = 9m and 3x + 2Y = 17, the value of M is obtained
If X-Y = 9m, then x = y + 9m, because x + 2Y = 5m, so 3Y + 9m = 5m, then 3Y = - 4m, y = - 4m / 3;
If 2 (X-Y) = 2 * 9m, then (x + 2Y) + 2 (X-Y) = 5m + 2 * 9m, then 3x = 23m, x = 23m / 3;
So 3x + 2Y = 3 * 23m / 3 + 2 * (- 4m / 3) = 23m-8m / 3 = 61m / 3 = 17
Launch: M = 17 * 3 / 61 = 51 / 61
From x + 2Y = 5m
X-Y = 9m, x = (23 / 3) m, y = - (4 / 3) M
Substituting X and Y into 3x + 2Y = 17 gives m = 51 / 61
M = nine fourths
How to solve the equation when the square of one X, 3x plus 7, plus the square of x plus 10 / 3x, equals zero
Let y = x ^ 2 + 3x, then the equation becomes y + 7 + 10 / y = 0, that is, y ^ 2 + 7Y + 10 = 0 (y + 2) (y + 5) = 0, then y = - 2 or y = - 5, then x ^ 2 + 3x = - 2 or x ^ 2 + 3x = - 5, the former gets x = - 1 or x = - 2, the latter has no solution