True or false: 3.6x-0.3 = 3x, right That's the problem

True or false: 3.6x-0.3 = 3x, right That's the problem

3.6x-0.3 = 3x 3.6x-3x = 0.30.6x = 0.3x = 0.5

Given Liman = 2, find LIM ((n + an) / (n-an))

lim((n+an)/(n-an))
=lim{[(n-an)+2an]/(n-an)}
=lim1+lim(2an/(n-an))
=1+2*lim{1/[(n/an)-1}.
=1+2*1/{lim[(n/an)-1]}.
=1+2*0
=1.

What is the remainder of 2, 000 divided by 13

2^2000=16^500＝(13+3)^500
According to the binomial theorem, it is equivalent to the remainder of 3 ^ 500 divided by 13
3^500=27^166*3＝（26＋1）＾166*3
Therefore, the remainder is 3

Simple calculation of 28 * 15-15 * 8

28×15-15×8
=15×(28-8)
=15×20
=300

Solving ordinary differential equation: y '' + 2Y '/ x + y ^ n = 0, thank you very much!

Let & nbsp; U & nbsp; = & nbsp; X ^ 2, DX & nbsp; = & nbsp; Du & nbsp; / & nbsp; 2x & nbsp; be substituted with:
d( dy / dx ) / dx + 2 dy / ( x dx )  + y^n = 0
d( 2x dy / du ) / dx + 2 dy / ( du / 2 ) + y^n = 0
 2 dx dy / ( du dx ) + 2 x d( dy / du ) / dx + 4 dy / du + y^n = 0
 2 dy / du + (2 x)^2  d( dy / du ) / du + 4 dy / du + y^n = 0
 2 dy + 4 u d( dy / du ) + 4 dy + y^n · du = 0
 2 dy + 4 d( u dy/du ) + y^n · du = 0
Points are as follows:
 2y + 4 u dy/du + ∫ y^n · du = 0    ①
When & nbsp; n & nbsp; = & nbsp; 0 & nbsp;, there are: 2Y & nbsp; + & nbsp; 4U & nbsp; dy / Du & nbsp; + & nbsp; U & nbsp; + & nbsp; C & nbsp; = & nbsp; 0
The general solution is Y & nbsp; = & nbsp; (- 1 / 6) & nbsp; X ^ 2 & nbsp; + & nbsp; C1 & nbsp; / & nbsp; X & nbsp; + & nbsp; C2
Where & nbsp; (- 1 / 6) & nbsp; X ^ 2 & nbsp; + & nbsp; C2 & nbsp; is a special solution and C1 & nbsp; / & nbsp; X & nbsp; is a homogeneous solution;
Observe the form of the special solution, and temporarily guess that the solution is a power function
Y1 (x) & nbsp; = & nbsp; U & nbsp; X ^ (2k) & nbsp; (where u is the undetermined coefficient and 2K is the power of x)
Substituting into the original equation (1), it is obtained that:
If the solution exists, it satisfies the following conditions: n & nbsp; ≠ & nbsp; 1 & nbsp; and & nbsp; n & nbsp; ≠ & nbsp; 3
The general solution is as follows
Y (x) & nbsp; = & nbsp; Y1 (x) & nbsp; = & nbsp; U & nbsp; X ^ (2k) & nbsp; (where the coefficients u, K are defined by the above expression)

If the absolute value of X + 1 and the absolute value of 2Y + 3 are opposite to each other, then what are x, y and X + y equal to?
Given that - 3 is less than or equal to X and less than or equal to 1, we can simplify the absolute value of X-1 and + X + 3

The absolute values of x-3 and 2Y + 3 are opposite to each other
Then the absolute value of x-3 + 2Y + 3 = 0
Because the absolute value of the value is greater than or equal to 0
And the absolute value of x-3 + 2Y + 3 = 0 holds only if
The absolute value of x-3 = 0 and the absolute value of 2Y + 3 = 0 hold at the same time
x-3=0 x=3
2y+3=0 y=-3/2
x+y=3-3/2=3/2

1.035 ÷ 4.5 =? And then check it and list it in vertical form

1.035÷4.5=0.23
Checking calculation: 0.23
× 4.5
———
one hundred and fifteen
ninety-two
———
one point zero three five

Find 1 + 2x under the root of LIM (x tends to 4) and subtract 3 / X-2

The denominator is x-4
Original formula = Lim [√ (1 + 2x) - 3] / (x-4)
= lim [√(1+2x) -3] [√(1+2x) +3] / (x-4)[√(1+2x) +3]
= lim [(1+2x) -9] / (x-4)[√(1+2x) +3]
= lim 2 / [√(1+2x) +3]
= 1/3

How to solve 7x = 5x + 6? Write the process step by step

5x moves to the left of the equal sign and becomes 7x-5x = 6, 2x = 6, x = 3

A barrel of oil weighs 5.8kg, but it weighs 3.3kg after half use. How many kilos of oil and barrel?

Oil (5.8-3.3) x2 = 2.5x2 = 5kg
Barrel 5.8-5 = 0.8kg