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1. Write the conditions for the following equation to hold: （1）√y-3/y-6=√y-3/√y-6 Note: √ Y-3 / y-6 is a whole, Y-3 / y-6 is in the tag √ is the root sign
y≠6
It's the equation （ ）1（ ）2（ ）3（ ）4（ ）5（ ）6（ ）7（ ）8＝0 If you fill in "+" - "the equation holds
1-2+3-4-5+6-7+8=0
Observe the following set of equations: Observe the following set of equations: (a + 1) (a? - A + 1) = a? + 1 （a+2）（a²-2a+4）=a³+8 （a+3）（a²-3a+9）=a²+27 From the above equation, do you have and find? Use your discovery rule to add the appropriate formula in the brackets below 1.（x-3）（x²+3x+9）=（ ） 2.（2x+1）（ ）=8x³+1 3、（ ）（x²+xy+y²）=x³-y³ Calculation: (a 2 - B 2) (a 2 + AB + B 2) (a 2 - AB + B 2) I'm a novice. I don't have any points. I'm sorry
1.（x-3）（x²+3x+9）=（x³-27）
2.（2x+1）（4x²-2x+1）=8x³+1
3、（ x-y ）（x²+xy+y²）=x³-y³
（a²-b²）（a²+ab+b²）（a²-ab+b²）=a^6-b^6
Fill in the following equation with 0 12 3 7 8 9 to make it true () + () = () () - () = () () Fill in the following form with 0 1 2 3 7 8 9 to make it true □+□=□□-□=□□
8+9=20-3=17
Match stick mathematical problem 111 + 1 + 1 = 4 move one to make the equation true
Method 1
Moving a match in 111 to the equal sign constitutes an absolute inequality
11+1+1≠4
Method 2
Move the 1 in the plus sign to make it a minus sign, and then increase it to 111
So it is
1111＋1－1＝4
In this equation, the "1" on the left side of the equation can not be treated as a number, but as a match. In the end, we get 1111 = 4, that is, four matches
Otherwise, there is no solution
Proof of mathematical inequality in senior one (basic inequality) Let a, B, C be unequal positive numbers, and prove that LGA + LGB + LGC < lg9 [(a + b) / 2] + LG [(B + C) / 2] + LG [(c + a) / 2]
LGA + LGB + LGC = LG (ABC) lg9 [(a + b) / 2] + LG [(B + C) / 2] + LG [(c + a) / 2] = lg9 [(a + b) (B + C) (c + a)] / 8] = lg9 / 8 (a + b) (B + b) (B + C) (c + a) inequality inequality inequality, we take index at the same time, that is to compare ABC and 9 / 8 (a + b) (B + C) (c + a) relationship (a + B) (B + C) (c + a) relationship (a + b) (b) (b) (c + C + a) = 2abc + a2bc + A2B + A2 + C2C + B2C + BC2 + BC2 + BC2 + BC2 + BC2 + BC2 + BC2 + BC2 + BC2 + BC2 + BC2 + BC
If and N is a positive integer, show that (a ^ (n + 1) + n * B ^ (n + 1)) / (n + 1) is greater than a * B ^ (n) If n is a positive integer, it is proved that (a ^ (n + 1) + n * B ^ (n + 1)) / (n + 1) is greater than a * B ^ (n). We hope to have a reply within 1 day i forgot the equal sign ,srry
N * B ^ (n + 1) is considered as the addition of N B ^ (n + 1)
(a ^ (n + 1) + n * B ^ (n + 1)) / (n + 1), then this is the arithmetic mean of N + 1
A * B ^ (n) is the geometric mean of N B ^ (n + 1) and a ^ (n + 1)
According to the theorem of the basic inequality in high school
(a^(n+1)+n* b^ (n+1)) / (n+1) ≥ a* b^(n）
(the condition of equal sign is a ^ (n + 1) = B ^ (n + 1))
Prove the equation (n + 1) (n + 2) by mathematical induction (n + n) = 2 to the nth power × 1 × 3 × 5 × In the process of (2n-1), when increasing to K + 1, the factor on the left should be increased
When n = k, the equation is (K + 1) (K + 2).. (K + k) = 2 ^ k * (2k-1)!
When n = K + 1, the equation is (K + 2) (K + 3).. (K + 1 + K + 1) = 2 ^ (K + 1) * (2k + 1)!
The increasing factor on the left is (2k + 1) (2k + 2) / (K + 1) = 2 (2k + 1)
The increasing factor on the right is 2 * (2k + 1)
It is proved by mathematical induction that the algebraic formula of multiplication on the left side of "from K to K + 1" when "(n + 1) (n + 2). (n + n) = 1 * 3 *... * (2n-1) * 2 ^ n" is
Let n = K + 1: (K + 1) (K + 2). (K + k) = 1 * 3 *... * (2k-1) * 2 ^ K. let n = K + 1: left = [(K + 1) + 1] [(K + 1) + 2] [（k+1）+（k+1）]＝[（k+1）（k+2）…… （k+k）]（k+1+k）（k+1+k+1）／（k+1）＝[1*3*...*(2k-1)*2^k]...
It is proved by mathematical induction that 1 + A + A ^ 2... + A ^ (n + 1) = [1-A ^ (n + 2)] / (1-A) when n = 1, the left side of the equation is equal to the left side A.1 B.1+a C.1+a+a^2 D.1+a+a^2+a^3 ------------------------------------ I don't understand why I chose C Please be more specific and easy to understand
On the left side, it is added from 1 to a ^ (n + 1). When n = 1, it can be substituted directly
So 1 + A + A ^ (1 + 1), select C
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