If LG (a-b) + LG (a + b) = LG2 + LGA + LGB, find the value of a / b

If LG (a-b) + LG (a + b) = LG2 + LGA + LGB, find the value of a / b

lg(a-b)+lg(a+b)
=lg（a-b）(a+b)
=lg(aa-bb)
lg2+lga+lgb
=lg2+lgab
=lg2ab
It's AA BB = 2Ab
aa-2ab+bb-2bb=0
（a-b）² =2b²
（a/b-1）² =2
（a/b-1） =√2
a/b=1+√2

lg6-lg3+lg5

lg6-lg3+lg5=lg(6/3*5)=lg10=1

[(X-Y) square + (x + y) square] (x's Square - Y's Square) how to simplify this?
[(X-Y) squared + (x + y) squared] (x squared - y squared)

[(X-Y) squared + (x + y) squared] (x squared - y squared)
=（x²-2xy+y²+x²+2xy+y²）（x²-y²)
=2(x²+y²）（x²-y²）
=2x^4-2x^4

5 / 7-5 / 9 × 5 / 7 + 4 / 9 × 2 / 7 (simple calculation)

5/7-5/9×5/7+4/9×2/7
=（1-5/9）×5/7+4/9×2/7
=4/9×5/7+4/9×2/7
=4/9×（5/7+2/7）
=4/9

It is known that B is the smallest positive integer and a, B satisfy the quadratic power of (C-5) + A + B = 0 (1) direct
It is known that B is the smallest positive integer and a, B satisfy the quadratic power of (C-5) + A + B = 0 (1) write the value of ABC directly (2) the corresponding points of ABC on the number axis are a moving point between ABC points m, please simplify the process of 2M (3) under the condition of problem 1 or 2, point ABC moves on the number axis, and point a moves to the left at a speed of one unit length per second, At the same time, point B and point C move to the left at the speed of two units of six and five units of length per second respectively. Suppose that after P seconds, the distance between weak current B and point C is BC, and the distance between a and point B is ab, then whether bc-ab changes with time

(1) ∵ B is the smallest positive integer, ∵ B = 1. According to the meaning of the question, we can get: a = - 1, B = 1, C = 5; (2) when 0 ≤ x ≤ 1, x + 1 > 0, X-1 ≥ 0, x + 5 > 0, then: | x + 1 | - X-1 | + 2 | x + 5 | = x + 1 - (1-x) + 2 (x + 5) = x + 1-1 + X + 2x + 10 = 4x + 10; when 1 < x ≤ 2, x + 1 > 0, X-1 > 0, x + 5

When solving the profit equation, a + a (1 + x) + the square of a (1 + x) and the square of a (1 + x) are used for the average growth rate. What kind of average growth rate are they used for? What are the implications in the title

Generally speaking, there will be two years of growth or decrease in this kind of problem. Let's ask for the average annual growth or decrease rate in these two years
For example, the profit of a store in 2011 is 10W yuan, and the profit in 2013 is 13W yuan
I wish you a happy study

Can I have the math problem off form calculation in Volume 2 of grade 5?
I need it badly

30.8÷[14-(9.85+1.07)]
[60-(9.5+28.9)]÷0.18
2.881÷0.43-0.24×3.5
20×[(2.44-1.8)÷0.4+0.15]
28-(3.4+1.25×2.4)
2.55×7.1+2.45×7.1
777×9+1111×3
0.8×[15.5-(3.21+5.79)]
(31.8+3.2×4) ÷ 5
31.5×4÷(6+3)
0.64×25×7.8+2.2
2÷2.5+2.5÷2
36.25÷4.25×9.9
5180－705×6
24÷2.4－2.5×0.8
(4121+2389)÷7
671×15-974
469×12+1492
19.4×6.1×2.3
5.67×0.2-0.62
18.1×0.92+3.93
0.4×0.7×0.25
0.78+5.436+1
4.8× 5÷4+1÷4
9.9x5÷6+5÷6
10.3÷4x8÷9-1÷3
14.31x5÷6-5÷6
16.5÷9x18-14x2÷7
20.3x2÷9+1÷3
35.95÷（64-45）
36.178-145÷5×6+42
7.2÷0.8-1.2×5
6.5×（4.8-1.2×4）
28-(3.4+1.25×2.4)
2.55×7.1+2.45×7.1
777×9+1111×3
（31.8+3.2×4）÷5
31.5×4÷(6+3)
0.64×25×7.8+2.2
2÷2.5+2.5÷2
194－64.8÷1.8×0.9
36.72÷4.25×9.9
5180－705×6
24÷2.4－2.5×0.8
(4121+2389)÷7
671×15-974
469×12+1492
405×(3213-3189)
19.4×6.1×2.3
5.67×0.2-0.62
18.1×0.92+3.93
0.043×24+0.875
0.4×0.7×0.25
0.78+5.436+1
4.07×0.86+9.12
240×78÷（154-115）
1437×27＋27×563
〔75-（12+18）〕÷15
2160÷〔（83-79）×18〕
280＋840÷24×5
325÷13×(266－250)
85×(95－1440÷24)
58870÷(105＋20×2)
1437×27＋27×563
81432÷(13×52＋78)
[37.85-（7.85+6.4）] ×30
156×[(17.7-7.2)÷3]
（947－599）＋76×64
36×(913－276÷23)
[192－（54＋38）]×67
81432÷(13×52＋78)
（947－599）＋76×64
60－(9.5＋28.9)÷0.18
2.881÷0.43－0.24×3.5
0.8×〔15.5-(3.21+5.79)〕
（31.8 3.2×4）÷5
194－64.8÷1.8×0.9
36.72÷4.25×9.9
3.416÷（0.016×35）
0.8×[(10－6.76)÷1.2]
(6.8-6.8×0.55)÷8.5
0.12× 4.8÷0.12×4.8
（58+37）÷（64-9×5）
812-700÷（9+31×11）
（3.2×1.5+2.5）÷1.6
85+14×（14+208÷26）
120-36×4÷18+35
9.72×1.6-18.305÷7
3.2×（1.5+2.5）÷1.6
（58+37）÷（64-9×5）
(6.8-6.8×0.55)÷8.5
0.12× 4.8÷0.12×4.8
（3.2×1.5+2.5）÷1.6
120-36×4÷18+35
.15-10.75×0.4-5.7
347+45×2-4160÷52
（58+37）÷（64-9×5）
[（7.1-5.6）×0.9-1.15] ÷2.5
（3.2×1.5+2.5）÷1.6
5.4÷[2.6×（3.7-2.9）+0.62]
12×6÷（12-7.2）-6
3.2×6+（1.5+2.5）÷1.6
（3.2×1.5+2.5）÷1.6
120-144÷18+35
347+45×2-4160÷52
（58+37）÷（64-9×5）
95÷（64-45）
812-700÷（9+31×11）
120-36×4÷18+35
(6.8-6.8×0.55)÷8.5
0.12× 4.8÷0.12×4.8
0.8×〔15.5-(3.21+5.79)〕
（31.8+3.2×4）÷5
31.5×4÷(6+3)
0.64×25×7.8+2.2
2÷2.5+2.5÷2
194－64.8÷1.8×0.9
36.72÷4.25×9.9
5180－705×6
24÷2.4－2.5×0.8
(4121+2389)÷7
671×15-974
469×12+1492
405×(3213-3189)
0.8×[(10－6.76)÷1.2]
19.4×6.1×24
5.67×0.2-0.62
18.1×0.92+3.93
24+0.875×99
0.4×0.7×0.25
0.75×102
100-56.23
31.50＋160÷40
（58+370）÷（64-45）
32.120-144÷18+35
33.347+45×2-4160÷52
34（58+37）÷（64-9×5）
35.95÷（64-45）
36.178-145÷5×6+42
420+580-64×21÷28
37.812-700÷（9+31×11）
（136+64）×（65-345÷23）
38.85+14×（14+208÷26）
120-36×4÷18+35
（58+37）÷（64-9×5）
(6.8-6.8×0.55)÷8.5
0.12× 4.8÷0.12×4.8
（3.2×1.5+2.5）÷1.6
3.2×（1.5+2.5）÷1.6
31.50＋160÷40
（58+370）÷（64-45）
32.120-144÷18+35
33.347+45×2-4160÷52
34（58+37）÷（64-9×5）
35.95÷（64-45）
36.178-145÷5×6+42
420+580-64×21÷28
812-700÷（9+31×11）
（136+64）×（65-345÷23）
（284+16）×（512-8208÷18）
120-36×4÷18+35
（58+37）÷（64-9×5）
(6.8-6.8×0.55)÷8.5
0.12× 4.8÷0.12×4.8
（3.2×1.5+2.5）÷1.6
（23.2×（1.5+2.5）÷1.6
45.6-1.6÷4=
5.38+7.85-5.37=
46.7÷0.8-1.2×5=
6-1.19×3-0.43=
47.6.5×（4.8-1.2×4）=
0.68×1.9+0.32×1.9
297×999
1983×21+21×3017
(0.123+123) ×1009
34×24+3481
0.8×0.196+0.094
105+895×57
32.47×23+27
39.01×18.3-901
76.5+2-2×184
49+23×14
31×99
52×99+52
25×32×125

If the sequence {an} satisfies a (n + 1) = 1 / 2an-3, A1 = 1, and Liman exists, then Liman=_____ [subscript in brackets]

Let the limit be x, a (n + 1) = 1 / 2an-3, and take the limit on both sides to get x = 1 / 2x-3
x=-6

555555555…… What is the remainder of 5 divided by 13?

4 Analysis: 111111 / 13 = 8547, so 6 5's can be divided by 13, and the integral multiple of 6 5's can be divided by 6 5's, so the integral multiple of 6 5's can be divided by 13. 2005 5 = (334 * 6 5) * 10 + 52008 5 = (334 * 6 5) * 10000 + 5555, so the remainder divided by 13 and 5555 divided by 13 are equal, and 5555 = 427 * 13 + 4

There are four numbers 3, - 5,7, - 13. How to make the result equal to 24
There are four numbers - 5, - 13. How can we make the result equal to 24 If we can't use the square

The second power of (3-5) * [- (7-13)] = 24