Решения задач. (компилируется на MinGW GCC 8.1)z1:#include <stdio.h> // Ввод-вывод#include <math.h> // Математические функцииdouble abs_double(double x) { return x < 0 ? -x : x;}int main() { double K, L, x, T, Q; scanf("%lf %lf %lf", &K, &L, &x); T = pow(cos(x)*(pow(K, 2) - pow(L, 2), 2) / (K * L * x); Q = sqrt(pow(T, 2) * abs_double(K - L) / 0.25); printf("T = %f, Q = %f", T, Q); return 0;}z2:#include <stdio.h> // Ввод-вывод#include <math.h> // Математические функцииdouble max(double a, double b) { return a > b ? a : b;}int main() { double a, n, l, x1, x2, y, z; scanf("%lf %lf %lf", &a, &n, &l); x1 = (a + n) / pow(a - l, 2); x2 = a / (n * l); if (x1 == x2) { y = x1 * (a - n) / x2; } else if (x1 < x2) { y = x1 + x2; } else { y = a * x1 + n * x2; } z = max(x1, x2) / (x1 + x2); printf("x1 = %f, x2 = %f, y = %f, z = %f", x1, x2, y, z); return 0;}z3:#include <stdio.h>#include <math.h>int main() { double x; double y = 0; scanf("%lf", &x); for (int i = 1; i <= 11; i += 2) { y += pow((x - 1) / (x + 1), i) / i; } printf("Y = %f", y);}z4:#include <stdio.h>int main() {int n = 0, p;int sum = 0;do {p = n;scanf("%d", &n);sum += n;} while (n != 0);printf("sum = %d, prev = %d", sum, p);return 0;}z5:#include <math.h>#include <stdio.h>int main() { const double PI = 3.141592; double a; int b; scanf("%lf %d", &a, &b); double x, d, S, s; if (b == 1) { x = a; d = x * sqrt(2); S = pow(d / 2, 2) * PI; s = pow(x / 2, 2) * PI; } else if (b == 2) { d = a; x = d / sqrt(2); S = pow(d / 2, 2) * PI; s = pow(x / 2, 2) * PI; } else if (b == 3) { S = a; d = 2 * sqrt(S / PI); x = d / sqrt(2); s = pow(x / 2, 2) * PI; } else if (b == 4) { s = a; x = 2 * sqrt(s / PI); d = x * sqrt(2); S = pow(d / 2, 2) * PI; } else { printf("error"); return 0; } printf("x = %f, d = %f, s = %f, S = %f", x, d, s, S); return 0;}