If (x + 3) (x + a) = x & # 178; - 2x-15, then a is equal to

If (x + 3) (x + a) = x & # 178; - 2x-15, then a is equal to

If (x + 3) (x + a) = x & # 178; - 2x-15, then a is equal to
The formulas on both sides are equal;
So x & # 178; + (a + 3) x + 3A = x & # 178; - 2x-15;
a+3=-2;
a=-5;
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
If you don't understand this question, you can ask,
Left expansion = x & # 178; + (a + 3) x + 3A = x & # 178; - 2x-15. According to the corresponding term, a = - 5 can be obtained from a + 3 = - 2 or 3A = - 15
The reduction of (x + 3) (x + a) is X & # 178; + ax + 3x + 3a
And because (x + 3) (x + a) = x & # 178; - 2x-15
So 3A = - 15
So a = - 5
Or because ax + 3x = - 2x
So a + 3 = - 2
So a = - 5
Hope to adopt
The sum of the first n terms of the sequence {an} is Sn = n2-10n (n belongs to n *), and the sequence {BN} satisfies BN = (an + 1) / an (n belongs to n *), (1)
The conclusion of this paper is {an};
(2) Finding the term with the largest value and the term with the smallest value in the sequence {BN}
(3) CN = absolute value an, find the first n terms and TN of sequence
(1) Sn = n ^ 2-10nan = Sn - S (n-1) = (2n-1) - 10 = 2n-11 = > {an} is the equivariant Lou column (2) BN = (an + 1) / an = (2n-10) / (2n-11) max BN = B1 = 8 / 9min BN = B5 = 0 (3) an > 02n-11 > 0n > 11 / 2n = 6cn = | an | for n = 6tn = - (a1 + A2 +.. + A5) + (A6 + a)
If we know that X and y are positive numbers, and one part of x plus nine parts of Y equals one, then the minimum value of X + y is zero
X+y
=1*(x+y)
=(1/x+9/y)(x+y)
=1+9+y/x+9x/y
≥10+2√y/x*9x/y
=10+2*3
=16
The minimum value of X + y is 16
(x + y) = (x + y) × 1 = (x + y) × (1 / x + 1 / y) = 2 + X / y + Y / X ≥ 4, so the minimum value of X + y is 4
Sixteen
Steps:
(1 / x + 9 / y) (x + y) = 1 + 1 / x y + 9 / y x + 9 is greater than or equal to 2 × 3 + 10 = 16
Because X and y are positive numbers, the minimum value of X + y is 16
9+81/8
How much is X & # 178; - 2x-9999 = 0
x²-2x-9999=0
(x+99)(x-101)=0
x1=-99,x2=101
Method of preparation:
x²-2x=9999
x²-2x+1=10000
(x-1)²=10000
∴x-1=±100
∴x1=-99,x2=101
x^2-2x+1=10000
（x-1)^2=10000
X = 101 or x = - 99
Given the first n terms and Sn = 10n-n2 of sequence {an}, and BN = | an}, find the first n terms and TN of {BN}
∵Sn=na1+n(n-1)d/2=d/2n²+﹙a1-d/2﹚n
∵d/2n²+﹙a1-d/2﹚n=-n²+10n
d=-2 a1=9
That's the question. If SN-S (n-1) is used, the actual answer is A1 = 11, but there seems to be nothing wrong with this method logically,
You can calculate A1 = 9 as follows, because the general formula of an is an = - 2n + 11
Given that x, y and Z are positive numbers, if 1 / x + 9 / y = 1, [1] find the minimum value of X + 2Y [2] if x + y + Z = 2, prove that 1 / x + 1 / y + 1 / Z is greater than or equal to 9 / 2
1) Let w = x + 2Y, then x = w-2y has 1 / (w-2y) + 9 / y = 1y + 9 (w-2y) = wy-2y ^ 22y ^ 2 - (17 + W) y + 9W = 0, because the equation y has a solution, the discriminant is > = 0, that is, (17 + W) ^ 2-4 * 2 * 9W = 289 + 34W + W ^ 2-72w = w ^ 2-38w + 289 > = 0 (w-19) ^ 2 > = 72, that is, w-19 > = 72 under root, or w =
Given a = {x 2-2x-3 > 0}, B = {x x x-a ≤ 1}, u = R
Finding a ∩ B when a = 3
If a &; cub, find the value range of real number a
Given the set a = {x 2-2x-3 > 0}, B = {x x-a ≤ 1}, u = RA: (x-3) (x + 1) > 0; x > 3 or X ＜ - 1; - 1 ≤ x-a ≤ 1; A-1 ≤ x ≤ 1 + A; when a = 3B: 2 ≤ x ≤ 4; find a ∩ B = {x | 3 ＜ x ≤ 4}; if a &; cub, Cub = {x | x ＞ 1 + A; or X ＜ A-1} find the value range of real number a ∩ 1 + a ≤ 3; a ≤
Given the first n terms of sequence {an} and Sn = N2 + 4N (n ∈ n *), sequence {BN} satisfies B1 = 1, BN + 1 = 2bn + 1
(1) Find the general formula of sequence {an}, {BN};
(2) Let CN = (an-3) &; (BN + 1) 4, find the first n terms and TN of the sequence {CN}
If x > 0, Y > 0, and 1x + 9y = 1, then the minimum value of X + y is ()
A. 4B. 12C. 16D. 18
∵ 1x + 9y = 1 ∵ x + y = (1x + 9y) (x + y) = 10 + 9xy + YX ≥ 10 + 29xy · YX = 16 if and only if 9xy = YX, take the equal sign. Then the minimum value of X + y is 16. So choose C
Let the complete set be r and a = {X-2 ≤ 0}, B = {x | 10 ^ x ^ 2-2 = 10 ^ x}
A={2}
B={2,-1}
The intersection of complements of a and B = {x | x ≠ 2 and X ≠ - 1}