If the real numbers x and y satisfy (X-2) 2 + y2 = 3, then the maximum value of YX is () A. 12B. 33C. 32D. 3

If the real numbers x and y satisfy (X-2) 2 + y2 = 3, then the maximum value of YX is () A. 12B. 33C. 32D. 3

The graph satisfying the equation (X-2) 2 + y2 = 3 is shown in the following figure: YX represents the slope of the line connecting the moving point on the circle and the origin O. from the graph, when the moving point coincides with B, ob is tangent to the circle, YX takes the maximum value and connects BC. In RT △ OBC, BC = 3, OC = 2, it is easy to get ∠ BOC = 60 ° and YX = 3
Polynomial ax ^ 5 + BX ^ 5 + cx-5, when x = 3, the value is equal to 7, find the value of this polynomial when x = - 3
The polynomial ax ^ 5 + BX ^ 5 + cx-5, when x = 3, is equal to 7
3^5a+3^5b+3c-5=7
3^5a+3^5b+3c=12
When x = - 3
-3^5a-3^5b-3c-5
=-（3^5a+3^5b+3c)-5
=-12-5
=-17
So, when x = - 3, the value of this polynomial is - 17
Let f (x) = ax ^ 5 + BX ^ 5 + cx-5,
Then: F (- x) = - ax ^ 5-bx ^ 5-cx-5 = - f (x) - 5-5
So: F (- 3) = - f (3) - 10 = - 7-10 = - 17
When x = - 3, the polynomial is - 17
Question: your answer is wrong. It's impossible
How to solve log3 + logx-1?
RT, 3 and 9 are base numbers
log9(x+5)=log(3)^2(x+5)=1/2log3(x+5)=log3√(x+5)
log3(x-1)=log3√(x+5)
x-1=√(x+5)
x^2-2x+1=x+5
x^2-3x-4=0
(x-4)(x+1)=0
x1=4,x2=-1
It is found that x2 = - 1 does not conform to the test, and it is omitted
X=4
If log is removed directly, the solution 3 (x-1) = 9 (x + 5); X = - 8
Solving quadratic equation 2x (x-1) = x-3
Wrong question. It's negative three
2X (x-1) = x-3 2x square - 2x = x-3 2x square - 3x + 3 = 0
The formula a = 2 b = - 3 C = 3 x = - B ± √ B squared - 4ac / 2A
x=-(-3)±√-3×-3-4×2×3 /2×2
X1=(3+3.8729833 )÷4≈1.718
X2=(3-3.8729833)÷4≈-0.218.
Is the answer equal to minus 3? Or what? According to your question, it's a quadratic equation with one variable, a quadratic equation with one variable
The equation has two answers
2*(x-1)=x-3
2x-2=x-3
x=-1
Substitution formula method
One
2x^2-2x+3=0
The discriminant B ^ 2-4ac = 4-4 * 2 * 3 = - 20
On the inequality x-a / x-a with respect to X (where a is any real number)
I've been bothering you^_^
X is not a & # 178;
or
(x-a)(x-a²) ≤0
When A1 a < A & # 178; [a, a & # 178;]
When 0
The inequality is equivalent to (x-a) (x-a & # 178;) ≤ 0 and X is not equal to a & # 178; when a & lt; 0, the range is [a, a & # 178;) when 0 & lt; a & lt; 1, the range is (A & # 178;, a] when a & gt; 1, the range is [a, a & # 178;) when a = 0, 1, the inequality holds
X is not equal to a ^ 2
Inequality can be equivalent to
(x-a)(x-a²)≤0 （x≠a²）
When a > 1, a & # 178; > A, a ≤ x < A & # 178;
When a = 1, a & # 178; = a = 1, (x-1) &# 178; ≤ 0, x = 1, but x ≠ A & # 178; = 1, so there is no solution
When a < 1, a < A & # 178;, a & # 178; ≤ x < a
Because the result is less than or equal to 0, the numerator and denominator are positive and negative, and the square of x-a is not equal to 0, that is, X is not equal to the square of A
(1) When x-a is greater than or equal to 0, the square of x-a will be less than 0. At this time, X is greater than or equal to a and less than the square of A
(2) When x-a is less than or equal to 0, the square of x-a is greater than 0, and X is less than or equal to a and greater than the square of A
Because the result is less than or equal to 0, the numerator and denominator are positive and negative, and the square of x-a is not equal to 0, that is, X is not equal to the square of A
(1) When x-a is greater than or equal to 0, the square of x-a will be less than 0. At this time, X is greater than or equal to a and less than the square of A
(2) When x-a is less than or equal to 0, the square of x-a must be greater than 0, and X is less than or equal to a and greater than the square of A
(x-a)/(x-a²)≤0
x≠a²
a≥a²
a(a-1)≤0
0≤a≤1
a²
By expanding the square of (1 + 2x-3x), what is the sum of the coefficients of the polynomial?
A complete process is required
(1 + 2x-3x & # 178;) &# 178;: after expansion, we get: a + BX + CX & # 178; + DX & # 179; + ex ^ 4
What you want to calculate is: a + B + C + D + E
We just need to substitute the first equation on both sides with x = 1
a＋b＋c＋d＋e=(1＋2－3)²=0
That is to say, the sum of the coefficients of expansion is 0
（1）：｜3－2x｜＜1
(1) : 3-2x ＜ 1 (2) x square + 7x + 12 ≥ 0
｜3－2x｜＜1
-1＜3－2x＜1
1＜x＜2
2x
（x+2）（x+5）≥0
X ≤ - 5 or X ≥ - 2
-1
Finding the negative integer solution of quadratic equation 2x + 3Y = 20
Need to solve the whole process
x=(100-3y)/2100/3
Take the integer solution by yourself
There is no negative integer solution. Obviously, if there is, the left is negative, and Mao Dun is on the right.
If you want to solve a quadratic equation, there is no way! I really can't solve it! That's a one-way equation you gave me? It's a binary equation, right? Somehow
If the inequality (A-2) x & # 178; + 2 (A-2) x-4 < 0 holds for a real number X
So the value range of a is
A (-∞,-2)
B (-2,2]
C (-∞,2]
D (-2,2)
E the above conclusions are not correct
Between 4 and 0 is the less than sign
1. a-2=0 a=2 （a-2）x²+2（a-2）x-4=-4
I can't see. General separation parameters
Obviously, when a = 2, it holds, so ade is excluded. When a = - 2, - 4x & # 178; - 8x-4
Find the sum of coefficients in the expansion of polynomial (2x ^ 3-5X ^ 2 + 1) ^ 3 * (2 / 3x-1) ^ 2
*It's a multiple sign
Substituting x = 1, the sum of the coefficients is (2-5 + 1) ^ 3 * (2 / 3-1) ^ 2 = - 8 * 1 / 9 = - 8 / 9
Substitute x = 1 into the polynomial and calculate it. Do it yourself