In the main function, input two positive integers m and N from the keyboard, write two functions to calculate their greatest common divisor and least common multiple respectively

In the main function, input two positive integers m and N from the keyboard, write two functions to calculate their greatest common divisor and least common multiple respectively

input int m,n;
int p=m,q=n,t;
while(t!=0)
{
t=p%q;
p=q;
q=t;
}
int MaxGys=p;
int MinGbs=m*m/p;

Java topic: input two positive integers m and N, find their greatest common divisor and least common multiple
I don't understand deff CD = new deff () from here on
Enter two positive integers m and N to find their greatest common divisor and least common multiple
/**In the loop, as long as the divisor is not equal to 0, divide the larger number by the smaller number, take the smaller number as the larger number of the next cycle, and the remainder as the smaller number of the next cycle. In this way, the loop returns the larger number until the smaller number is 0, which is the greatest common divisor. The least common multiple is the product of two numbers divided by the greatest common divisor*/
import java.util .*;
public class lianxi06 {
public static void main(String[] args) {
int a ,b,m;
Scanner s = new Scanner( System.in );
System.out.print (type an integer:);
a = s.nextInt();
System.out.print (type another integer:);
b = s.nextInt();
deff cd = new deff();
m = cd.deff (a,b);
int n = a * b / m;
System.out.println (greatest common divisor: + m);
System.out.println (LCM: + n);
}
}
class deff{
public int deff(int x,int y) {
int t;
if(x < y) {
t = x;
x = y;
y = t;
}
while(y = 0) {
if(x == y) return x;
else {
int k = x % y;
x = y;
y = k;
}
}
return x;
}
}

import java.util .*;public class lianxi06 { public static void main(String[] args) {int a ,b,m;Scanner s = new Scanner( System.in ); System.out.print ("type an integer):; a = s.nextint(); System.out.print ( ...

Java programming input two positive integers m and N, output their least common multiple and greatest common divisor
Don't change statements related to input and output
import java.util .*;
public class Test40014 {
public static void main(String []args){
\x05 Scanner in =new Scanner( System.in );
\x05 int gcd,lcm,m,n,r;
\x05 int repeat,ri;
\x05 repeat= in.nextInt ();
\x05 for(ri = 1; ri

import java.util .*;
public class Test40014 {
public static void main(String []args){
 Scanner in =new Scanner( System.in );
 int gcd, lcm, m, n,r;
 int repeat, ri;
 repeat= in.nextInt ();
 for(ri = 1; ri

Solutions of differential equation y '= (x + y) ^ 2

Let t = x + y, then y = t-x, dy / DX = DT / DX-1, the original equation is reduced to (DT / DX) - 1 = t? P > so DX / dt = 1 / (t?) integral, x = arctan + C, that is, x = arctan (x + y) + C

Given that the system of inequalities (2x-3) (3x + 2) 0 about X has no real number solution, the value range of real number a is obtained

8.4 ÷ 0.56 the process of vertical calculation

When x tends to 0, find the limit of (cosx-1) / xsinx
When x tends to 0, find the limit of (TaNx SiNx) / (arcsinx to the third power)
When x tends to 0, find the limit of (x-sinx) / (x + SiNx)
Please tell me more about it,

1. When x tends to 0, (cosx-1) / xsinx = (cosx-1) / ((SiNx) squared) = (cosx-1) / (1 - (cosx) squared) = - 1 / (cosx + 1) = - 1 / 22. When x tends to 0, (TaNx SiNx) / (arcsinx cubic) = (TaNx SiNx) / ((SiNx) cubic) = (1 / cosx-1) / ((SiNx) flat

Write out the common solution of the binary linear equation 8x-3y = - 1 and - 7x + 5Y = 8

8x-3y = - 1 times 5 = 40x-15y = - 5
-7x + 5Y = 8 times 3 = - 21x + 15y = 24
40X-15Y+-21X+15y=-5+24
x=1
Take x = 1 into 8x-3y = - 1
Y=3

If the sum of the first four terms of an arithmetic sequence is 40, the sum of the last four terms is 80, and the sum of all terms is 210, then the sequence has ()
A. 12 items B. 14 items C. 16 items D. 18 items

According to the properties of arithmetic sequence, the sum of the first term and the last term is a1 + an = 40 + 804 = 30. According to the sum of all terms is 210 = (a1 + an) · N2, the solution is n = 14, so B

Why does the limit of absolute value of X divided by X not exist when x → 0?
RT

Because the left limit = - X / x = - 1, the right limit = x / x = 1
The left limit is not equal to the right limit, so the limit does not exist