How to solve the y range, how to get the x y range solved

The problem is with this command, [x y]=solve(' ', '', 'x', 'y'For m, the n value is a specific value, this command is correct, but if you want to use it as a variable, you can only get an analytic expression. To achieve your goal, you should remove the single quotation marks and replace it with [x y]=solve( ,'x', 'y'Then, the program also needs to add a loop to solve each of the statements with m,n for the x,y values. For the specific revisions**, see the annex.

The key is the range of arcsine definitions.

Known: 0<=x<=40, 0<=y<=40

70x + 30y <=3000 30x + 70y <=2000 7 3 - got.

x range, in the same way we get the range of y and then intersect with the beginning range.

y=40-x

Substitute 70x+30y 3000

Get 70x+30(40-x) 3000

70x+1200-30x≤3000

40x≤1800

x 45 to 30x+70y 2000

Get 30x+70(40-x) 2000

30x+2800-70x≤2000

40x≤-800

x 20 is 20 x 45

The same goes for -5 y 20

You can get y=40-x to 70x+30y<=3000 and 30x+70y<=2000.

2^x+2^y=1 ≥2√（2^x×2^y）=2×2^(x+y)/2=2^[(x+y)/2+1];

x+y)/2+1≤0;

x+y≤-2;

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Draw it on the number line.

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Graphs are used in conjunction with classification discussions.

Let's start by constructing a Cartesian coordinate system, discuss:

When x>0 and y>0 draw the image of the function.

...... when x>0 and y<0

...... when x>0 and y<0

...... when x>0 and y<0

It is drawn as a square with a(1,0)b(-1,0)c(0,1)d(0,-1) as vertices.

x|+|y|1 is the area surrounded by the square ABCD is obvious, and the value range is [-1,1].

Because a 2-b 2 = a 3-b 3

So (a-b)(a+b) = (a-b)(a 2+ab+b 2) because a, b are two positive numbers that carry lead unequally.

a+b=a 2+ab+b 2=(a+b) 2-ab because (a+b) 2>4ab

So ab<(a+b) 2 4

So-ab> beam oak-(a+b) 2 4

So (a+b) 2-ab>(a+b) 2-(a+b) 2 4=3(a+b) 2 Aus-Cabein 4

Thus a+b>3(a+b) 2 4

Solution: 0a+b

Solution. a+b>1

Or. a+b<0 (round).

Obtained from (2), (3).

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