If the solution of the fractional equation m (x-1) 2 / 5 (x + m) = - 8 / 5 is x = - 1 / 5, what is the value of M

If the solution of the fractional equation m (x-1) 2 / 5 (x + m) = - 8 / 5 is x = - 1 / 5, what is the value of M

The solution of M (x-1) 2 (x + m) = - 5 / 8 is x = - 5 / 1
2（-1/5+m)/m(-1/5-1）=-8/5
2（-1+5m)/(-6m）=-8/5
2*5（-1+5m)=(-6m）*(-8)
-5+25m=24m
m=5

If the solution of the fractional equation X-1 / M + x = 2 / m about X is x = 2, then the value of M is ()
A.-2 B.-4 C.2 D.4

Analysis
m/(x-1)+x=m/2
∴
Substituting x = 2 into
m+2=m/2
2m+4=m
m=-4
B yes

Saw a 3-meter-long cuboid wood into three sections on average, the surface area increased by 2.4 square meters, the volume of this wood is______ Cubic meter

To saw the cuboid wood into three sections averagely, the number of times needed is: 3-1 = 2 (Times) [2.4 ÷ (2 × 2)] × 3 = [2.4 ÷ 4] × 3 = 0.6 × 3 = 1.8 (cubic meter) a: the volume of this wood is 1.8 cubic meter, so the answer is: 1.8

If a six digit number 12 is a multiple of 88, then the quotient of this number divided by 88 is______ ．

If a number is a multiple of 88, it must be a multiple of both 8 and 11. According to the multiple of 8, its last three digits must also be a multiple of 8, so we can know that the number on the six digit single bit is 4. According to the multiple of 11, the difference between the sum of the odd and even digits should be a multiple of 0 or 11. From the known four digits, the sum of the six digit odd and even digits is equal In order to make the sum and difference of the odd and even digits 0, the numbers filled in the two boxes are the same, so there are two possibilities for this six digit number: 124344124344 △ 88 = 1413; so the quotient of this six digit number divided by 88 is 1413

In the arithmetic sequence an, if 3N & sup2; + 2n, then the tolerance D=

Sn=3n^2+2n
S(n-1)=3(n-1)^2+2(n-1)
=3n^2-6n+3+2n-2
=3n^2-4n+1
an=Sn-S(n-1)
=3n^2+3n-(3n^2-4n+1)
=3n^2+3n-3n^2+4n-1
=7n-1
d=7

(11-36 / 11 × 11) + (25-36 / 11) is simple and fast

Original formula = 11 - 11 / 36 × 11 + 25 - 11 / 36
= 11+25 - 11/36 × (11+1)
= 36 - 11/36 ×12
= 36 - 11/3
= 36 - (3 +2/3)
= 33 - 2/3
=32 and 1 / 3

Why is a - (2ab-b & # 178;) / a equal to (A & # 178; - 2Ab + B & # 178;) / a thank you

a-(2ab-b²)/a
=a²/a-(2ab-b²)/a
=(a²-2ab+b²)/a

Fill - 4 - 3 - 2 - 10 1 2 3 4 in a 3 × 3 square to make the sum of 3 numbers in each row, 3 numbers in each column and 3 numbers in diagonal angle zero

The first kind
-3 4 -1
2 0 -2
1 -4 3
The second kind
1 2 -3
-4 0 4
3 -2 -1
The third kind
3 -4 1
-2 0 2
-1 4 -3
The fourth kind
-1 -2 3
4 0 -4
-3 2 1

Junior one binary linear equation addition and subtraction elimination method equations (solution) sister brother help!
{3X4y=16 ①
{5X6y=33 ②
① 1x3, get
9X+12y=48
② X2, get it
10X-12y=66
19X=114
X=6
Substituting x into (1) yields (1)
3x6+4y=16
4y=-2
y=-1/2
So the solution of this equation is {x = 6
{y=-1/2
I don't understand why we have to multiply by three and two, how to calculate it,
And 3x6 + 4Y = 16
4y=-2
How to calculate y = - 1 / 2, how can it be equal to 2 / 1?
3x6 = 18, 18 + 4 = 22y, how can it be 4Y
I know how to find the least common multiple, but I still don't know where the multiplication of 3 and 2 come from? The common multiple is 12, but where did 3 and 2 come from? I'm so stupid. Please, I'm the worst at math. If I don't learn, I'll be a fool~
I'll thank you very much

The result is as follows: {3x + 4Y = 16 ① {5x + 6y = 33 ② ① * 3, 9x + 12Y = 48 12Y = 48-9x ② * 2, 10x-12y = 66 (take the previous 12Y = 48-9x into the 12Y in this step) 10x - (48-9x) = 66 (can you understand it?) 10x-48 + 9x = 66 19x = 66 + 48 19x = 114 x = 6 (the process you gave

Given the polynomial f (x) divided by X + 2 to get the remainder 1, divided by X + 3 to get the remainder - 1, find the remainder of F (x) divided by (x + 2) (x + 3)
Given the polynomial f (x) divided by X + 2 to get the remainder 1, divided by X + 3 to get the remainder - 1, find the remainder formula of F (x) divided by (x + 2) (x + 3)
——I hope there are processes and ideas,

If you take the exam this year, will it be a bit late to start reviewing the comprehensive ability of mathematics now? We are the same. I just started reviewing. But if you take the exam next year, you should grasp the time now. I regret that I didn't grasp it before