Limx tends to 0 and finds the limit of (e ^ 2 - (1 + 1 / x) x ^ 2) / X How to ask~

Limx tends to 0 and finds the limit of (e ^ 2 - (1 + 1 / x) x ^ 2) / X How to ask~

This is a wrong question
When x tends to 0-0,
1/x->-inf,1+1/x->-inf
X ^ 2 of (1 + 1 / x) is the limit of (- INF) ^ 0 type,
There is no way to ask

4A ^ 2-9y ^ 2 = write process of factorization

4a^2-9y^2
=(2a)^2-(3y)^2
=(2a+3y)(2a-3y)

Qianshan Road Primary School Grade 4 class 1 sports standard number is 24, the number of substandard is 12, the number of standard number of the whole class how many?

24 (24 + 12) = 24 (36) = 23 A: the number of people who reach the standard accounts for 23

Factorization (x2 + 3x) 2 - (2x + 1) 2 = (x2 + x-1) (x2 + 5x + 1)

(x^2+3x)^2-(2x+1)^2=[(x^2+3x)+(2x+1)][(x^2+3x)-(2x+1)]
=(x^2+x-1)(x^2+5x+1)

There are 4 more girls than 23 boys in a class. If there are 3 fewer boys and 4 more girls, then the number of boys and girls is exactly the same. How many boys and girls are there in this class?

Suppose there are x boys, we can get the equation: 23x + 4 + 4 = x-3 & nbsp; 23x + 8 = x-3, & nbsp; & nbsp; & nbsp; 13X = 11, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X = 33.23 × 33 + 4 = 22 + 4, = 26. A: there are 33 boys and 26 girls

(4x + 3) (2x-3) - 6x (3 / 2x + 1) where is 3 / 2,

(4x+3)(2x-3)-6x(2x/3+1)=(4x+3)(2x-3)-2x(2x+3)==8x²+6x-12x-9-9x²-6x=-x²-12x-9

If a is 20% more than B and C is 10% less than a, what percentage of B is C?

Unit a 1,
B: 1 ÷ (1 + 20%) = 5 / 6
C: 1 × (1-10%) = 9 / 10
C is B: 9 / 10 △ 5 / 6 × 100% = 108%

If the denominator of the equation is removed from 2x-1-3 x + 1 = 1, then ()
A.2X-1-X+1=6 B.3(2X-1)-2(X+1)=6 C.2(2X-1)-(X+1)=6 D.3X-3-2X-2=1

The equation 2x-1 of 2-x + 1 of 3 = 1 is de denominator, and 3 (2x-1) - 2 (x + 1) = 6
Do not understand welcome to ask

It is known that 5 / 4 of the number a is equal to 5 / 6 of the number B, and the difference between a and B is 120. Calculate the number a

A × 5 / 4 = B × 5 / 6
A × 3 / 2 = b
B-A = 120
A × 3 / 2-A = 120
A × 1 / 2 = 120
A = 240

It is proved that in a right triangle, four times the sum of the squares of the central lines of two right angled sides is equal to five times the square of the hypotenuse

In CBE and abd, the Pythagorean theorem is used to obtain that the square of CE = the square of be + the square of BC, the square of ad = the square of BD + the square of AB should be BD = 0.5bc, be = 0.5ab