Mathematics in the first year of junior high school, the process of finding the solution, please!

A: 21x|+|y|=5

23.When x is 2, the minimum value is 0

24.m 0, n 0 on the number line, so their opposite number is m 0, n 0m, n, 0, -m, -n is n -m 0 m -n26If |a|=7，|b|=2, and a b

So when b = 2, a = -7

When b = -2, a = -7

So a=2, b=-7 or a=-2, b=-7.

27.When a value is entered, the output result of the screen is always 1 less than the absolute value of the number, so after entering -7, the result of the screen is 6

Then enter it again and the result of the screen is 5

So the final result of the screen output is 5.

6|-1 = 5 < - This is the final result is the last question.

In the hexagonal abcdef, a= b= c= d= e= f=120°, we can see that this hexagon is a regular hexagon, so the circumference of this hexagon = 6ab = 6

According to the known conditions, this hexagon is a regular hexagon, so the perimeter = 1*6 = 6

Because in the hexagonal abcdef, a= b= c= d= e= f=120°, it is a regular hexagon, and because the side length is 1, the circumference is 6

It is a regular hexagon, with each inner angle equal and the opposite sides equal. The circumference is 6!

n 1 2 3 4 5y 1 3 7 13 21 。

We get n 1 2 3 4 5y 1 1+2 1+2+4 1+2+4+6 1+2+4+6 .

Y=1+2+4+...... found2(n-1)

Summing using a series of equal differences: the number of first and last multipliers divided by 2y=(2+2n-2)(n-1) 2 +1

Simplify y=n 2-n+1

In the nth diagram, this is a question about order, and the final answer should be: n*(n-1)+1=n2-n+1 point.

(1) The left and right sides of each formula are multiplied by the least common multiple of the denominator, for example, in equation 1, the denominator is 2, 3, 4, then the least common multiple is 12, and the left and right sides of the equation are multiplied by 12 to get a simplified equation.

By the way, the right side of Equation 2 should be -48

2) Equation 1x2 + Equation 2, equation 4: 15x-6y=24 equation 3x2 + equation 2, equation 5: 4x-y=10 (3) equation 4 - equation 5x6, we get -9x=-36, x=4, and then find y=6, z=12

First of all, your second formula is simplified wrong, it should be 3x+2y-6z=-48 (the new 2 formula).

Then from 1-3 formula: 4x-y=10 (4 formula) from 2 + 2*3 to get 7x-4y=4 (5 formula) from 4 and 5 to get x=4, y=6

After substituting 1, the solution is z=12

Your simplification second form is wrong, so it can't be calculated.

Solution: a1 + a2 + a3 + a4 + a5 means that the sum of these numbers is twice as large (each number appears twice).

1/2(a1+a2+a+a4+a5)=-2-1+0+1+2+3+4+5+6+7=3+4+5+6+7=25

After swapping positions, the first imitation will not be changed. The file appears twice because each number is the same.

The answer to the third question is b

The fifth problem is to the 2n power of x.

The sixth question is 4, and the seventh question is 0

Question 8: x to the power of 9n+2.

<> ni can be calculated in one fell swoop according to the question on the answer, which is an open-ended question with a clear answer and no fixed answer.

bybj

1) The third points of line segment AB are 2, and together with the 2 endpoints of line segment AB, there are 4 points, which can form (4*3) 2=6 line segments.

There are 3 line segments of length 2, 2 line segments of length 4, and 1 line segment of length 6, and the sum of the length of these line segments is 2*3+4*2+6=20.

2) For the convenience of narration, P is recorded as the equinox in the line segment AB.

Take A as the starting point, and mark the position of the M point with the length of the line segment MA.

The third equinox is ;

The quartile points are;

The sixth equinox is ;

Excluding the number of repeats, there are 1,,2,3,4,5 points, including a and b, a total of 9, divided into line segments 9 * 8 2 = 36.

There are 4 of length, 6 of length 1, 4 of length, 5 of length 2, 2 of length, 5 of length 3, 2 of length, 3 of length, 3 of length 4, 2 of length, 2 of length 5, and 1 of 6.

The total length of these segments.