A junior high school math problem that does not count and seek detailed solutions, and a junior high

by 3|x|+2y=0, y<0

So assuming x>0, then.

2x+y=-7 ①

3x+2y=0 ②

2* -de, x=-14

So the assumption is not valid.

then x<0

2x+y=-7 ③

3x+2y=0 ④

2* - de, 7x = -14

So x=-2, substitute x=-2 into it.

4+y=-7

So y=-3

So xy=6

There are four scenarios to solve.

1. If x 0 and y 0, then the system of equations is 2x-y=-7 and 3x+2y=0 to obtain x=-2, y=3, which contradicts the hypothetical conditions.

2. If x 0, y<0, then the system of equations is 2x+y=-7 and 3x+2y=0 to solve x=-14, y=21, which contradicts the hypothetical conditions.

3. If x<0 and y 0, then the system of equations is 2x-y=-7 and 3x+2y=0

The solution yields x=-14 and y=-21, which contradict the hypothetical conditions.

4. If x<0, y<0, then the system of equations is 2x+y=-7 and 3x+2y=0 to solve x=-2, y=-3, which is consistent with the assumptions.

So xy=6 in this question

3丨x丨+2y=0,—》y<=0,—》2x-丨y丨=2x+y=-7,x>=0,then:3x+2y=0,—》x=-14,does not meet the requirements,x<0,then:-3x+2y=0,—》x=-2,—》y=-3,—》xy=6.

Because 3|x|It must be a non-negative number, so y is a negative number or 0, but according to the first equation, we can see that x and y will not be 0, so y=, so |y|=, bring in the first become, this time is divided into 2 cases, when x>0, the formula becomes, and the solution is x=-14<0, which is inconsistent with the hypothetical x>0, excluded; When x<0, the equation becomes 2x+, and the solution is x=-2, satisfying; So x=-2, y=-3, so xy=6

by 3|x|+2y=0 gives y<0 so y=-3 2|x|2x+3/2|x|=-7 ①

If x 0, then the equation is 2x-3 2x=1 2x=-7 x=-14 which does not match the assumption.

If x 0, then 2x+3 2x=7 2x=-7 we know that x=-2 is consistent with the assumption.

So y=-3 x=-2

So xy=6

x=y, then x=y=2 9

Brought into the second equation, k = 4519

Substituting x=-3 and y=-2 into :

6+2n=-4

n=1 substitute x=-5, y=2 to get:

5m+4=-6

m=2∴m²+n²+mn=(m+n)²-mn=7

Hello, Purple Wind VV Account:

Solution: 1, x-4) (x-b).

x²-bx-4x+4b

x²-(b+4)x+4b

x²-x+a

(b+4)=-1，4b=a

The solution yields a=-12 and b=-3

-2) +2) -2 (-2) (2) = (-2+2) =0 (-2) +2) =2 (-2) (2) Rule: a +b 2ab

Rationale: (a-b).

a²+b²-2ab≥0

a²+b²≥2ab

1.Left = x 2 - (4 + b) x + 4b

Right = x 2-x + a

So there is 4+b=1

a=4ba=-12 b=-3

2.(2008-1)* (2008+1) and (2008-3)* (2008+3).

That is, 2008 2-1 and 2008 2-9 the former is 3 a^2+b^2-2ab=(a-b)^2>=0a^2+b^2>=2ab

1).(x-4)(x-b)=x -(4+b)x+4b, then 4+b=, a=4b=-12

2) 2007 2009=2008 Whoever is bigger, you understand.

3）＞。Induction(a+b) 2ab (when a=b, the equation holds).

Set t minutes later, B catches up with A for the first time.

Taking A as the starting point, B chases A after 3 sides from point B and point A. So there is:

72t=65t+90*3

Solution: t=270 7

72t = 270 * 72 7, which is about equal to more than 2700;

2700 360=, i.e. 7 and a half turns;

72T is calculated from B according to Jibi Zhizhao B, so when there are many circles, it is on AD.

The diagram attached to the title of Wisdom has been clearly dissolved, please see the diagram I attached.

1.In the first question, the result is -2y

If x= is mistaken for x=- but the result is still correct, then there are three possibilities for the original formula:

1. The simplified formula does not contain x

2. After simplification, the term x containing x is a square term.

3. The simplified formula may be a constant term, so the formula is a fixed value no matter what the unknown number takes.

2.Actual selling price: [A(1+40%)]60%=So it's a loss.

3.According to the title, 1c = 4000 years, so 16c = 64000 years.

So the age of this fossil is 15c = 60,000 years.