1 5 20 22 x 200 x write process

The original form of the unknown:

Problem solving: Move the unknowns to the left and merge them by moving the terms, move the constant terms to the right and merge them, and finally turn the unknown coefficients to 1

Problem solving: 22x 200-30

x≤85/11

If you have any doubts, please ask and be satisfied.

Finding x Writing Process The original formula of the unknown: Problem solving process: move the unknown to the left and merge it by moving the term, move the constant term to the right and merge it, and finally turn the unknown coefficient into 1 The solution process: If you have any doubts, please ask and be satisfied.

This sentence is just a unary inequality and very simple.

22x <= 200;

22x <= 200 -30;

22x<= 170;

x<= 85/11;

That's how it works.

Pay attention to the order of operations, calculate multiplication first and then solve the equation.

30＋22x≤200

22x≤200-30

22x≤170

x≤85/11

Socks are resistant: Tell the manuscript of the spring respectx+

2x-x=x=

Mergers of the same kind.

15x=x=Note that the decimal family carries a little bit of a mega volt of cherry blossoms.

16x-x＝

15x x Bridge Break.

Find x Minhu 8 75 or x loose matter.

0 0 type, you can use the Robita rule, up and down at the same time to find the derivative, get Vulcos x 1 = -Vulture.

You can also use the trigonometric formula sin--vultures = -sin(v-x-vultures), followed by sinx x important limits, to get -vultures.

When x 1, sin x=sin =0, x-1=0, so it is 0 0 type, and you can use Lopida's rule.

Primitive=limx 1 cos x= cos=-

e xsin xdx = (1 2)e x - 1 10)e xcos2x - 1 5)e xsin2x + is the integral constant.

The process is as follows:

e^xsin²x dx

1/2)∫ e^x(1-cos2x) dx

1/2)e^x - 1/2)∫ e^xcos2x dx (1)

Calculate below: e xcos2x dx

cos2x d(e^x)

Division Points. e^xcos2x + 2∫ e^xsin2x dx

e^xcos2x + 2∫ sin2x d(e^x)

Re-division points.

e^xcos2x + 2e^xsin2x - 4∫ e^xcos2x dx

Divide by the coefficient by merging the -4 e xcos2x dx shift with the left.

Get: e xcos2x dx = 1 5)e xcos2x + 2 5)e xsin2x + c

Substituting the above equation into (1) obtains.

e^xsin²x dx = 1/2)e^x - 1/10)e^xcos2x - 1/5)e^xsin2x + c

Expand Yuanzheng information:

Fractional Credits: UV).'=u'v+uv'

Got:u'v=(uv)'-uv'

Both sides score: u'v dx=∫ uv)' dx - uv' dx

i.e.: u'v dx = uv - uv'd, this is the partial integration formula.

It can also be abbreviated as: V du = UV - U DV

Common Integration Formulas:

1）∫0dx=c

2）∫x^udx=(x^(u+1))/u+1)+c

3) Stare the cavity of the jujube 1 xdx=ln|x|+c

4）∫a^xdx=(a^x)/lna+c

5）∫e^xdx=e^x+c

6）∫sinxdx=-cosx+c

7）∫cosxdx=sinx+c

8）∫1/(cosx)^2dx=tanx+c

9）∫1/(sinx)^2dx=-cotx+c

10）∫1/√（1-x^2) dx=arcsinx+c

11）∫1/(1+x^2)dx=arctanx+c

12）∫1/(a^2-x^2)dx=(1/2a)ln|(a+x)/(a-x)|+c

(1 2) e x(1-cos2x) dx(1 2)e x - 1 2) e xcos2x dx (1)

e^xcos2x dx

cos2x d(e^x)

Division Points. e xcos2x + 2 e xsin2x dxe xcos2x + 2 sin2x d (e x) re-fractional points.

e^xcos2x + 2e^xsin2x - 4∫ e^xcos2x dx

Divide the -4 e xcos2x dx shift term by the coefficient by merging the left defense file edge.

Get: e xcos2x dx = 1 5)e xcos2x + 2 5)e xsin2x + c

Substituting the above formula into (Shichang 1) obtains.

e^xsin²x dx = 1/2)e^x - 1/10)e^xcos2x - 1/5)e^xsin2x + c

e^x(1-cos2x)

dx=(1/2)e^x

e^xcos2xdx

Calculate below: e xcos2x dx

cos2xd(e^x)

Division Points. =e^xcos2x

e^xsin2x

dx=e^xcos2x

sin2xd(e^x)

Re-division points.

e^xcos2x

2e^xsin2x

e xcos2xdx will.

e^xcos2x

The dx shift is combined with the left and divided by the coefficient.

Got: e xcos2xdx

1/5)e^xcos2x

2/5)e^xsin2x

c substitute the above equation into (1).

e^xsin²xdx

1/2)e^x

1/10)e^xcos2x

1/5)e^xsin2x

c [The Beauty of Mathematics] team will answer for you, if you don't understand, please ask, if you solve the problem, please click "Choose as a satisfactory answer" below.

Solution: 2 x = -4

Both sides are in the same seat: 2=-4x

x=-1/2

check, substituting x=-1 2 into :

Left old orange side = 2 (-1 2) = 2 (-2) = -4 = right.

x=-1 2 is the solution of the equation.

x=1lim(x-->1)^1

Both of these limits are a).

1+1 B type, i.e. direct substitution type.

Two important limits, to pay attention to the variables of the manuscript cautious x approach moment key Fu Jing Hall Chun difference.

lim(x-->0)sinx/limx→π(sinx/x)=0/x）^x=(1+1/π=0

limx→1（1+1/