-Square of 5x + [square of X - (square of 5x - 2x)] - 3 (square of X - 2x)], where x = - 1

-Square of 5x + [square of X - (square of 5x - 2x)] - 3 (square of X - 2x)], where x = - 1

=-5x^2+x^2-5x^2+2x-3x^2+6x
=-12x^2+8x
=12-8
=4

The square of (x + 5) - (X-5) - 5 (2x + 1) (2x-1) + the square of X * (2x),

Square of (x + 5) - square of (X-5) - 5 (2x + 1) (2x-1) + square of X * (2x)
=(x+5+x-5)(x+5-x+5)-5(4x^2-1)+4x^3
=20x-20x ^ 2 + 5 + 4x ^ 3 x = - 1
=-20-20+5-4
=-39

Simple calculation: 9998 + 3 + 99 + 3 + 998 + 9

9998+3+99+3+998+9 =9998+2+1+99+998+2+1+9=10000+100+1000+10=11110

Given the first n terms of sequence {an} and Sn = 2 ^ n-1, then A1 & # 178; + A2 & # 178; + a3 & # 178; +... + an & # 178;

Sn=2^n-1
Then A1 = S1 = 1
an=Sn-S（n-1）=2^（n-1）
an²=4^（n-1）
a1²+a2²+a3²+...+an²=1*（4^n-1）/（4-1）=1/3（4^n-1）

The polynomial x ^ 3-2x + 1 is a cubic polynomial, and the coefficient of the quadratic term is ()

It's 0

The weight of a barrel of edible oil is 5 kg. After half of the oil is used, the weight of a barrel is 2.6 kg. How many kg of oil is there in the barrel?
How to set up the equation?

There is a total of oil in the barrel x kg
x-1/2x=5-2.6
1/2x=2.4
x=4.8
A: there is 4.8kg of oil in the barrel

Find the maximum value of function f (x) = sin2x-x in the interval [- π / 2, π / 2]

f'(x)=cos2x*(2x)'-1=2cos2x-1=0
cos2x=1/2
-π/2

X + 8 = 13 2.5 + x = 5.3 x-2.7 = 13 x-40 = 15 x-1.6 = 1.4 x = 5.3 = 10 the equations of these problems should be solved in detail

x+8=13
x=13-8
=5
2.5+x=5.3
x+2.5=5.3
x=5.3-2.5
=2.8
x-2.7=13
x=13+2.7
=15.7
X-40=15
x=15+40
=55
x-1.6=1.4
x=1.4+1.6
=3
x-5.3=10
x=10+5.3
=15.3

The water outlet of a water pump is round with a diameter of 8 cm. If the water outlet speed is 8 meters per second, can the water pump fill a pool with a volume of 20 cubic meters in 4 minutes?

8 cm = 0.08 m, 4 min = 240 s, 3.14 × (0.08 △ 2) 2 × 8 × 240 = 3.14 × 0.0016 × 8 × 240 = 0.040192 × 240 = 9.64608 (M3); 9.64608 m3 < 20 m3, a: this pump can't fill a pool with a volume of 20 m3 in 4 min

F (x) = 5 ^ x + 7 ^ X-2, then when x → 0, A. f (x) and X are infinitesimal of the same order but not equivalent, B, f (x) are infinitesimal of higher order than x, please give a small process, thank you

Because Lim {f (x) / X} = Lim {(5 ^ x + 7 ^ X - 2) / X} = Lim {5 ^ x * LN5 + 7 ^ x * ln7-2} = LN5 + ln7-2 = constant;
Therefore, f (x) and X are infinitesimals of the same order, but they are not equivalent, but differ by a coefficient - 2 + ln35, that is, f (x) ~ (- 2 + ln35) X;