If a and B are reciprocal, then a divided by B = ()

If a and B are reciprocal, then a divided by B = ()

If a and B are reciprocal, then a divided by B = (the square of a)

Given the difference between the polynomial 2x equal square + my-12 and the polynomial NX equal square-3y + 6, which does not contain the square and Y terms of X, the value of M + N + Mn is obtained
Thank you. Please write the process clearly

2X ^ 2 + my-12 - (NX ^ 2-3y + 6) = 2x ^ 2 + my-12 - (NX ^ 2-3y + 6) = 2x ^ 2 + my-12-nx ^ 2 + 3y-6 = 2x ^ 2 + my-nx ^ 2 + 3y-18 = x ^ 2 (2-N) + y (M + 3) - 18 remove the square and Y terms of X, get: 2-N = 0, n = 2m + 3 = 0, M = - 3, so m + N + Mn = - 3 + 2 + (- 3) * 2 = - 7

The author of ancient poems about wine
I don't know the basic knowledge

(1) Su Shi's "water melody" when the moon, wine asked the sky. I do not know the sky palace, this night is what year. I want to take the wind back, but also afraid of Qionglouyuyu, high too cold, dance to make clear the shadow, how like in the world. Turn Zhuge, low Qihu, according to sleepless. There should be no hate, why long to other time round

The fourth power of M + the third power of 4m + the square of 4m-9

m^4+4m³+4m²-9
= m²(m²+4m+4) ²-9
=m²(m+2) ²-3²
=(m²+2m) ²-3²
=(m²+2m+3)(m²+2m-3)
=(m²+2m+3)(m-1)(m+3)

What's five out of seven plus three out of eight

Five out of seven plus three out of eight
=(40+21)/56
=61/56

The line passing through the origin intersects the circle x2 + y2-6x + 5 = 0 at two points a and B. the trajectory equation of the midpoint m of the chord AB is obtained

Let the center of the circle x2 + y2-6x + 5 = 0 be C, then the coordinate of C is (3, 0). From the topic meaning, cm ⊥ AB, ① when the slopes of the line cm and ab exist, that is, X ≠ 3, X ≠ 0, then there is kcmkab = - 1, ≁ YX − 3 × YX = - 1 (x ≠ 3, X ≠ 0), which is reduced to x2 + y2-3x = 0 (x ≠ 3, X ≠ 0), ② when x = 3, y = 0, point (3, 0) is suitable for the topic meaning, ③ when x = 0, y = 0, point (0, 0) is not suitable for the topic meaning To solve the equations x2 + Y2 − 3x = 0x2 + Y2 − 6x + 5 = 0, we get x = 53, y = ± 235, and the trajectory equation of point m is x2 + y2-3x = 0 (53 < x ≤ 3)

Solution equation: X-1 / 3 [X-1 / 3 (x + 9)] = 1 / 9 (x + 9)

x－1/3【x－1/3（x+9）】＝1/9（x+9）
x－1/3x+1/9（x+9）】＝1/9（x+9）
x－1/3x = 0
x = 0

Simple operation: (873 × 477-198) / (476 × 874 + 199)

(873 * 477-198) / divided by 476 * 874 + 199)
=[(874-1) (476 + 1) - 198] / divided by 476 * 874 + 199)
=(874 * 476 + 874-476-1-198) / divided by 476 * 874 + 199)
=(874 * 476 + 199) / divided by 476 * 874 + 199)
=1

The simple method of 199.7 * 199.7-19.95 * 1999

199.7*199.7-19.95*1999=1997*0.1*1997*0.1-1995*0.01*1999=0.1*0.1*1997*1997-0.01*1995*1999=0.01*1997*1997-0.01*（1997-2）*（1997+2）=0.01*【1997*1997-（1997*1997-2*2）】=0.01*（1997*1997-1997*1997+4）=0...

When | a + 2 | + (x-3) ^ 2 = 0, find the value of 1 / 3 (- 3ax ^ 2-ax + 3) - (- ax ^ 2-1 / 2ax-1)

When | a + 2 | + (x-3) ^ 2 = 0
a+2=0 x-3=0
a=-2 x=3
1/3(-3ax^2-ax+3)-(-ax^2-1/2ax-1)
=1/3(-3*-2*3^2+2x3+3)-(2x3^2-1/2x-2*3-1)
=1/3(54+6+3)-(18+3-1)
=21-20
=1