C language from the keyboard to enter a positive number less than 1000, require the input of its square root (if the square root is not an integer, then enter its integer part) After inputting data, check whether it is a positive number less than 1000, if not, re - input #include #include int main() { int x,y; Printf ("enter an integer x less than 1000", x)); scanf("%d",&x); if(x0) {y=sqrt(x); Printf ("the integral part of the square root of X is Y", x, y) "; } else {printf ("input data does not meet the requirements, re-enter an integer x less than 1000", x) scanf("%d",&x); y=sqrt(x); } return 0 } E: C (6): error c2018: unknown character '0xa3' E: C (6): error c2018: unknown character '0xbb' E: C (7): error c2146: syntax error: missing ';' before identifier 'scanf' E: C (9): warning c4244: '=': conversion from 'double' to 'Int', possible loss of data E: C (14): error c2146: syntax error: missing ';' before identifier 'scanf' E: C (15): warning c4244: '=': conversion from 'double' to 'Int', possible loss of data E: C (18): error c2143: syntax error: missing ';'before'}

C language from the keyboard to enter a positive number less than 1000, require the input of its square root (if the square root is not an integer, then enter its integer part) After inputting data, check whether it is a positive number less than 1000, if not, re - input #include #include int main() { int x,y; Printf ("enter an integer x less than 1000", x)); scanf("%d",&x); if(x0) {y=sqrt(x); Printf ("the integral part of the square root of X is Y", x, y) "; } else {printf ("input data does not meet the requirements, re-enter an integer x less than 1000", x) scanf("%d",&x); y=sqrt(x); } return 0 } E: C (6): error c2018: unknown character '0xa3' E: C (6): error c2018: unknown character '0xbb' E: C (7): error c2146: syntax error: missing ';' before identifier 'scanf' E: C (9): warning c4244: '=': conversion from 'double' to 'Int', possible loss of data E: C (14): error c2146: syntax error: missing ';' before identifier 'scanf' E: C (15): warning c4244: '=': conversion from 'double' to 'Int', possible loss of data E: C (18): error c2143: syntax error: missing ';'before'}

Printf ("enter an integer x less than 1000", x); X does not need to be changed to
Printf ("enter an integer less than 1000)";
Printf ("if the input data does not meet the requirements, re-enter an integer x less than 1000", x)
Add before return 0
Printf ("the square root or integral part of the square root is% D, y)";

Input a positive number of Xiaoyu 1000 from the keyboard, and output its square root (if the square root is not an integer, output its integer part)

Input a positive number less than 1000 from the keyboard and output its square root (if the square root is not an integer, output its integer part)

The first floor already has Pascal and VF versions
I add C and VB version here
C version
#include
int main()
{ int i,n;
scanf("%d",&n);
i=sqrt(n);
printf("%d",i);
system("PAUSE");
}
VB version
Private Sub Form_ Click()
Dim n As Integer
n = InputBox("n=")
Print Int(Sqr(n))
End Sub

1. If a, B and C are trilateral lengths of △ ABC, try to judge the root of the equation CX2 + (a + b) x + C / 4 = 0. 2. It is known that the equation B (x2-1) - 2aX + C (x2 + 1) = 0 has two equal real roots. Can you judge the triangle shape with a, B and C as trilateral lengths? Please explain the reason. 3. Read it first and then solve the problem. (1) the root of the equation x2 – X-12 = 0 is X1 = - 3, X2 = 4 by formula method, (2) using the formula to get the equation 2x2-2-7x + 3, the root of the equation 2x2-2-7x + 3 = 0 is X1 = 1 / 2, X2 = 3, so X1 + x2 = 7 / 2, x1x2 = 3 / 2. (2) using the formula to get the equation 2x2-2x2-3x2-12 = -12. (2) using the formula to get the equation 2x2-2 – 7x-7x + 3, the root of the equation 2x2 – 2 – 7x + 7x + 3 + 3 + 3 + 3 + 3 = 0 is the root of the equation 2x1 = the root of the equation 2x2 = the root of the equation 2x2 = the equation 2x2 = the root of the equation 2x2 = the equation 2x2 = the equation 2x = 1 = the root of the root of the equation 2x2-12 = ((12-12-12-12-12-12-12-12-12-12-12-12-12-12-12-12-12-12-12-12-12& nbsp; & nbsp; &According to (1) (2) (3), please guess: if the two real roots of the univariate quadratic equation MX2 + NX + P = 0 (m ≠ 0, and m, N, P are constants) are x1, X2, then what is the relationship between X1 + X2, x1x2 and the coefficients m, N, P? Please write your conjecture, There are still two more questions to ask. Please write them when you answer them

1. If a, B and C are the three sides of △ ABC, try to judge the root of the equation CX2 + (a + b) x + C / 4 = 0
a. B, C are the three sides of △ ABC, then
a+b＞c
△=（a+b）² -4*c*c/4=（a+b）² -c²＞0
The equation has two unequal real roots
2、b（x2 -1）-2ax+c（x2 +1）=0
（b+c）x²-2ax-b+c=0
△=（2a）²-4（b+c）（-b+c）=4a²-4（c²-b²）=4（a²-c²+b²）=0
Get a & # 178; - C & # 178; + B & # 178; = 0
a²+b²=c²
a. B and C are three sides, and the triangle is right triangle
3. 3) the root of the equation x2 – 3x + 1 = 0 is X1 = (3 - √ 5) / 2, X2 = (3 + √ 5) / 2. Then X1 + x2 = (3), x1x2 = (1)
（4）x1+x2=-n/m,x1x2=p/m
From the root formula
X = - B ± √ B & # 178; - 4ac / 2A

How to divide (36-0.14) by 0.12
Come on, it's urgent

Dismissing
36/0.12-0.14/0.12=300-7/6=1793/6

There are 20 people in a class. If all the people in this class shake hands once, how many times do they shake hands? The sum of shaking hands?

19 times per person, and 19 times 20 divided by two equals 190

153 degrees 19 minutes 41 seconds + 20 degrees 43 minutes 48 seconds =? (expressed in degrees)

=(153+20)+(19+43)/60+(41+48)/3600
=173+1.033+0.025=174.058°

What is the value of integer a when the solution of the equation AX-2 = - X of X is a natural number

ax-2=-x
（a+1）x=2
Only when a = 0 or a = 1, the solution of X is a natural number

Simple calculation of 1.8 × 0.62 + 0.18 × 3.9-0.018

Factorization of 4x ^ 2 + 12xy + 5Y ^ 2-4x + 2y-3

Factorization by principal component method
Let X be the unknowns and y be the coefficients, and let X be arranged in descending order
Original formula = 4x & # 178; + (12y-4) x + 5Y & # 178; + 2Y + 3
=4x²+（12y-4）x+（5y-3）（y+1）
Using cross method: 25y-3
\ /
/ \
2 y+1
Original formula = (2x + 5y-3) (2x + y + 1)