Solve the following equation, where (1) (2) requires writing the test process. (1) 2Y = 3Y + 5 (2) 2 / 2 X-7 = x (3) 4-2x = 6-3x Solve the following equations, among which (1) (2) require writing the test process (1) 2Y = 3Y + 5 (2) 2 of X-7 = x (3) 4-2x = 6-3x

(1) 2Y = 3Y + 5 test: left = 2 * (- 5) = - 102y-3y = 5, right 3 * (- 5) + 5 = - 10-y = 5, so, left = right y = - 5, so y = - 5 (2) x / 2-7 = x test: left = - 14 / 2-7 = - 14x / 2-x = 7, right = - 14-x / 2 = 7, so left = right x = - 14, so x = - 14 (3) 4-2x = 6-3x-2x + 3x

Simply calculate 3 / 35 + 3 / 63 + 1 / 99 + 1 / 143 =?

3 of 35 + 3 of 63 + 1 of 99 + 1 of 143
=3/2(1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13)
=3/2(1/5-1/13)
=3/2x8/65
=12/65

There is a kind of natural number, which has only three divisors, such as 4.25, etc. (1) can you write three more such natural numbers? (2) natural numbers within 1000
How many such numbers are there?

（1）
These numbers are the squares of prime numbers
So there are 9, 49, 121
（2）
Within 1000, it is within the square of 32
Prime numbers are 2,3,5,7,11,13,17,19,23,29,31
So it's 11

5 / 3x + 4 / x + 3 / 9x = 77
Who can help make it out? The submission should be written clearly

(3/5+1/4+9/3)x=77
(12/60+15/60+180/60)x=77
207x/60=77
x=77*60/207=1540/69

(-2)^50+(-2)^49+(-2)^48+(-2)^47+...+(-2)+1
（x-1)(x^3+x^2+x+1)=x^4-1
Factorization in this way
and so on

Let s = (- 2) ^ 50 + (- 2) ^ 49 + (- 2) ^ 48 + (- 2) ^ 47 +... + (- 2) + 1
2S=(-2)^51+(-2)^50+(-2)^49+(-2)^48+...+(-2)^2-2
2S-S=S=(-2)^51-1=-2^51-1

The limit of (nsin (1 / N)) ^ n ^ 2 when n approaches infinity

n→∞lim(nsin1/n)^n²
=n→∞lim[(sin1/n)/(1/n)]^n²
=x→0lim[(sinx)/x)]^(1/x)²
=x→0lime^ln[(sinx)/x)]^(1/x)²
=x→0lime^[(1/x)²]ln[(sinx)/x)]^(1/x)²
=x→0lime^{[(1/x)²]*ln[(sinx)/x)]}
=x→0lime^{ln[(sinx)/x)]/x²}
=X → 0lime ^ {[(x / SiNx) * (xcosx SiNx) / X & # 178;] / 2x} (Robita's law)
=X → 0lime ^ {[(x / x) * (xcosx-x) / X & # 178;] / 2x} (equivalent substitution)
=x→0lime^{[(cosx-1)/2x²]}
=X → 0lime ^ {[(- SiNx) / 4x]} (Robita's law)
=x→0lime^{[(-x)/4x]}
=X → 0lime ^ {[- 1 / 4]} (equivalent substitution)
=e^(-1/4)

Divide the product of 183 nines by 5, and the remainder is______ ．

Because one 9-bit is 9, two 9-bit is 1, three 9-bit is 9, then 183 9 multiplied by 9, 9 divided by 5, the remainder is 4; so the answer is: 4

Simple calculation: 12 out of 13 times 3.42 times 39 out of 12 (this "." is the decimal point)

12/13×3.42×39/12
= 12/13×39/12×3.42
= 3×3.42
= 10.26
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1. Let F 1 and F 2 be the left and right focal points of ellipse C: X & # 178 / A & # 178; + Y & # 178 / B & # 178; = 1 (a ﹥ B ﹥ 0), the line L passing through F 2 intersects the ellipse at two points a and B, the inclination angle of line L is 60 ° and the distance from F 1 to line L is 2 √ 3. (1) find the focal length of ellipse C (2) if the vector af2 = 2 times the vector f2b, find the equation of ellipse C
2. Let the focus of the ellipse e: X & # 178 / / A & # 178; + Y & # 178; / (1-A & # 178;) = 1 be on the x-axis. (1) if the focal length of the ellipse e is 1, the equation of E is obtained. (2) let F1 and F2 be the left and right focus of e respectively, p be the point in the first quadrant of E, and the line F2P intersects the y-axis at the point Q, and f1p ⊥ F1q, it is proved that when a changes, the point P is on a certain line
3. Ellipse C: X & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1 (a ＞ B ＞ 0), the left and right focal points are F1 and F2 respectively, the eccentricity is √ 3 / 2, the length of the line segment cut by ellipse C through F1 and ⊥ on x-axis is 1 (1). The equation for solving ellipse C (2) point P is any point on ellipse C except the endpoint of major axis, connecting Pf1 and PF2. Let the angular bisector PM of ∠ f1pf2 intersect the major axis of ellipse C at point m (m, 0), and find the value range of M

First question:
Second question:
Third question:
From the Internet, for reference only,
Because PM is the angular bisectoline of / 4; let the coordinates of point p be (x, y), then Pf1 = √ [(x + C) &# 178; + Y & # 178;] = √ [(x + C) &# 178; + B & # 178; - B & # 178; X & # 178; /This is the first time that we are going to be in the world of 178; (a-178; (a-178; (a) = (3 x & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ifx = 0, then M = 0; if 0 & lt; X & lt; a (M & gt; 0), then (3x & # 178 / / 4 + 2cx + A & # 178;) / (3x & # 178;) / 4 - 2cx + A & # 178;) = 1 + [4C / (3x / 4 - 2C + A & # 178;]; /x) The & lt; lt; 1 + [4C / (3a / 4-2c-4-4-2c + a)] & lt; lt; 1 + [4C / (3a / 4-4-2c-4-4-2c + a)] = 1 + [(8-3) / (7-4-4-4-4-3) / (7-4-4-4-4-4-4-4-2c + a) [(8-4-4-4-4-4-4-4-2c-2c + 3) = (7 + [(8-3 / 4) [(3 / 4 (3 / 4 / 4) [(3 (3 / 4-4-4) [(6 + 4 {(6 + 4) [(6 + 4) [(6 + 4) / (6) / (6 [6 + 4) [(6) / (6 (6) [(6 + 4) [(6) / (6) / (6 \\\\] = √ 3C / 2 = 3A / 4; similarly, if - A & lt; X & lt; 0 (M & lt; 0), then M & gt; -So - √ 3C / 2 & lt; M & lt; √ 3C / 2;

Solve the following equation, where (1) (2) requires writing the test process. (1) 2Y = 3Y + 5 (2) 2 / 2 X-7 = x (3) 4-2x = 6-3x Solve the following equations, among which (1) (2) require writing the test process (1) 2Y = 3Y + 5 (2) 2 of X-7 = x (3) 4-2x = 6-3x