Solve the equations: 3x − y = 23x = 11 − 2Y

Solve the equations: 3x − y = 23x = 11 − 2Y

3x − y = 2 (1) 3x = 11 − 2Y (2) substitute (2) into (1) to get (11-2y-y = 2), that is, y = 3. Substitute (3) into (2) to get (x = 53), then the solution of the equations is x = 53Y = 3

How many solutions of the quadratic equation 3x + 2Y = 12 in the range of positive integers?

3x+2y=12
2Y = 12-3x Y > 0 2Y > 0 12-3x > 0 x0, so 0

If the function y = loga (x + 2) (a > 0 and a ≠ 1) has Y > 0 on (- 1,0), then the value range of a

Because y = loga (x + 2) is a monotone function, we only need to investigate its maximum and minimum
When x = - 1
loga(x+2)≥0
When x = 0
loga(x+2)≥0
And a ＞ 0, a ≠ 1
The results of solving the above inequalities are as follows
a＞1
If you don't know how to ask, I hope to adopt it

arccos(-x)=∏－arccosx.

pai

-Arctanx = arctan (- x) - arcsinx = arcsin (- x) arccosx = arccos (- x), right
-arctanx=arctan(-x)
-arcsinx=arcsin（-x）
arccosx=arccos（-x）

The first and the second are right
The third is wrong

If arccos x > arccos (1-x) is true, then the value range of X is

arccosx>arccos(1－x)
－1≤x

Find ∫ (1 → - 1) ([x] + x) e ^ - [x] DX, [] as absolute value sign

Piecewise integration:
When x is integral from - 1 to 0, x + ︱ x = x-x = 0, there is no need to calculate;
When x is integrated from 0 to 1, remove all absolute values

|When x | ≤ 1, arcsinx arccos (- x)=______

In conclusion, arcsinx arccos (- x) = - π / 2

How to derive f (x) = ln (x + 1) - x

f'(x)=1/(x+1)-1

F (x) = ln (x + a) + x ^ 2-x derivation process

f'(x)=[ln(x+a)]'+(x²)'-x'
=(x+a)'/(x+a) +2x-1
=1/(x+a) +2x-1