11 + 1 + 1 - 111 = 4 move a matchstick to make the equation hold!

11 + 1 + 1 - 111 = 4 move a matchstick to make the equation hold!

Take one from the front and put it on the equal sign to make it ≠

Match stick math problem: move a match to make the equation hold 21 + 35 = 68

Change it to 21 + 39 = 60, move the middle of 68 to 35, and change it to 39

Match stick mathematical problem move 909 × 2-6 = 2004 move a match to make the equation true Sorry, it should be + 6

Is there a problem with the title!? it should be 989 × 2-6 = 2004? If it is, then take the one from the bottom left of 8 and turn it into 9, and then change - 6 into + 6, that is, 999 × 2 + 6 = 1998 + 6 = 2004. I don't know if it is correct!? 909 × 2 + 6 = 2004 is easy! Move the vertical root from the bottom left of 0 to the middle of 0 (horizontally), and 0 becomes 9, then 9

Mathematical proof of senior one (basic inequality) Given a, B, C ∈ R +, it is proved that: (a + B + C) [1 / (a + b) + 1 / C] ≥ 4

The original formula = 1 + (a + b) / C + C / (a + b) + 1
>=2 + 2 (under radical sign (a + b) / C * C / (a + b))
=4

Root sign (a ^ 2 + B ^ 2) + root sign (b ^ 2 + C ^ 2) > = root 2 (a + B + C)

On the right side, the transformation is: root 2 (a + b) / 2 + root 2 (B + C) / 2 + radical 2 (a + C) / 2. Then, by using the basic inequality, we get the following conclusion: root 2 (a + b) / 2 > = root a ^ 2 + B ^ 2 (arithmetic mean less than square mean) the same reason: root 2 (a + C) / 2 > = root a ^ 2 + C ^ 2 root 2 (B + C) / 2 > = root B ^ 2 + C ^ 2

Proof of basic inequality It is proved that 1 / 2 (a + b) ^ 2 + 1 / 4 (a + b) is greater than or equal to a radical B + B radical a

a√b+b√a=√ab*(√a+√b)
From the basic inequality, we can get the following conclusions
√ab≤(a+b)/2
So a √ B + B √ a
≤(a+b)*(√a+√b)/2
≤[(a+b)^2+(√a+√b)^2]/4
=[(a+b)^2+2√ab+a+b]/4
≤[(a+b)^2+a+b+a+b]/4
=(a+b)^2/4+(a+b)/2

It is proved that a (a-b) ≥ B (a-b), Let's use the one whose root sign ab ≥ a + B / 2

prove
a(a-b)-b(a-b)
=a^2-ab-ab+b^2
=a^2-2ab+b^2
=a^2+b^2-2ab
=(a+b)^2-4ab
∵(a+b)/2≥√ab
(a+b)≥2√ab
The square of two sides can be obtained
(a+b)^2≥4ab
∴(a+b)^2-4ab≥0
That is, a (a-b) ≥ B (a-b)
Conclusion

1: It is known that a, B, C ∈ R + (a 2 + A + 1) (B 2 + B + 1) (C 2 + C + 1) ≥ 27abc 2: Given that a, b > 0 and a + B = 1, it is proved that (a + 1 / a) 2 + (B + 1 / b) 2 ≥ 25 / 2 3: Let a, B, C ∈ R +, and a + B + C = 1 (1) Verification: (1-a)(1-b)(1-c)≥8abc (2) Verification: a²+b²+c²≥1/3 (3) Verification: √4a+1 +√4b+1 +√4c+1 It is enough to solve (1), (2) and (3) to add 30.3q A ＞ B ＞ 0 C ＞ D ＞ 0 √ AC - √ BD ≥ √ (a-b) (C-D) (all under the radical)

1. (A-1) * (A-1) > = 0A ^ 2-2 * a + 1 > = 0 and 3AA ^ 2 + A + 1 > = 3A, B ^ + B + 1 > = 3b, C ^ 2 + C + 1 > = 3C, so (a ^ 2 + A + 1) (b ^ 2 + B + 1) (C ^ 2 + C + 1) > = 3A * 3B * 3C = 27abc 2, (a + 1 / a) 2 + (B + 1 / b) 2 = a ^ + 2 + 1 / A ^ + 2 + 2 + 1 / b

It is proved by mathematical induction that 1 + 1 / 2 + 1 / 3 + 1 / 4 = + 1 / 2n power - 1 is less than or equal to n

1+1/2+1/3+1/4+...+1/(2^n-1)

It is proved by mathematical induction that the nth power of 1 + 1 / 2 + 1 / 3 +. + 1 / 2 is ≤ n

N = 1, 1 = 1, the inequality holds,
Let n = k = 1 + 1 / 2 + 1 / 3 +. + 1 / 2 to the power of - 1 ≤ K
Then n = K + 1
Left = [1 + 1 / 2 + 1 / 3 +. + 1 / 2 to the K power-1] + [1 / (2 ^ (k-1) + 1) + 1 / (2 ^ (k-1) + 2) + 1 / (2 ^ k)]
Right = K + 1
According to the assumption of 1 + 1 / 2 + 1 / 3 +. + 1 / 2 to the power of - 1 ≤ K
1/（2^(k-1)+1）+1/（2^(k-1)+2）+1/（2^k)≤2^(k-1)*[1/(2^(k-1))]=1
So the left ≤ K + 1
To sum up, the inequality holds for all natural numbers