Có: \(\widehat{tOz}=180^o-\widehat{tOy}-\widehat{zOx}=180^o-60^o-30^o=90^o\) (kề bù)

\(\widehat{xOt}=\widehat{xOz}+\widehat{zOt}=30^o+90^o=120^o\)

\(\widehat{zOy}=\widehat{tOz}+\widehat{tOy}=90^o+60^o=150^o\)

Xét 2 tam giác vuông AHD và BKC có:

\(AD=BC\) (gt)

\(\widehat{ADH}=\widehat{BCK}\)

\(\widehat{AHD}=\widehat{BKC}\)

Do đó: ΔAHD = ΔBKC (cạnh huyền - góc nhọn)

=> HD = KC (2 cạnh tương ứng).

#include <iostream>
#include <iomanip>
#include <cmath>
#include <cstdio>
using namespace std;

#include <bits/stdc++.h>

int main() {
    int n;
    cin >> n;
    int total = 0;

    for (int i = 1; i <= n; i++) {
        if (i % 20 == 0 && i % 10 != 0) {
            if (i <= (n / 10)) {
                total += i * i;
            }
        } else {
            total += i;
        }
    }
    cout << "The total is: " << total << endl;

    for (int i = 0; i < n; i++) {
        int du = i % 10;
        int tongcacso = i % 10 * i % 10 * i % 10;
        cout << "du = " << du << endl;
        total += du * du * du;
        cout << "\nsum = " << total << endl;
        cout << "sum = (sum+i*i*i) " << endl;
    }

    return 0;
}

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;

#include <bits/stdc++.h>

int main() {
    int n;
    cin>>n;
    int sum=0;

    for(int i=1;i<=n;i++)
    {
        sum+=i*i;
    }
    cout<<"The total is: "<<sum<<endl;

    for(int j=0;j<=50000;j++)
    {
        int du=j%10;
        int tongcacso=j%10*j%10*j%10;
        cout<<"du="<<du<<endl;
        sum=sum+du*du*du;
        cout<<"\nsum= "<<sum<<endl;
        cout<<"sum= (sum+j*j*j) "<<endl;
    }

    return 0;
}

#include <iostream>
#include <vector>
using namespace std;

int main() {
    int n, k;
    cin >> n >> k;
    int ai[n];
    for (int i = 0; i < n; i++) {
        cin >> ai[i];
    }

    vector<int> pairs(n);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (i == j) continue;
            if (ai[i] + ai[j] == k) {
                pairs.push_back(make_pair(i, j));
            }
        }
    }

    sort(pairs.begin(), pairs.end());
    int count = 0;
    pairs.erase(pairs.begin(), pairs.end());
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (pairs[i].first == i && pairs[i].second == j) {
                count++;
                break;
            }
        }
        if (count == n) {
            break;
        }
        if (pairs[i].second == j) {
            while (pairs[i].first != i) {
                i++;
                count++;
                pairs.erase(pairs.begin() + i);
                for (int j = 0; j < n; j++) {
                    if (ai[j] + ai[i] == k) {
                        pairs.push_back(make_pair(j, i+1));
                        break;
                    }
                }
                for (int j = 1;; j++) {
                    int count_n = 0, count_s = 0;
                    for (int i = j-1; i >= 0; i--) {
                        if (pairs[i].second == j) {
                            j++;
                            count_n++;
                            pairs.erase(pairs.begin() + i + 1);
                            pairs.er

program cho_xuat;

uses
  System.SysUtils, System.Classes, System.Generics.Collections;

type
  Tsum = record
    value: Integer;
  end;

  TsumArray = array of Tsum;

function cho_xuat(N: String; M: Integer): TsumArray;
begin
  SetLength(Result, Length(N));
  for i := 0 to High(Result) do
  begin
    if N[i] >= 58 then
      Result[i].value := Result[i].value + M
    else
      Result[i].value := Result[i].value - M;
  end;
end;

var
  N, M: String;
  a: TsumArray;
begin
    readln(N, M);
    a := cho_xuat(N, M);

    WriteLn('KQ: ' + N, ' -> ' + ConvertIntToString(M) + '');
    for i := 0 to High(a) do
    begin
        WriteLn(' + ', a[i].value, ' +', );
    end;
end.

Để \(\sqrt{\dfrac{2\sqrt{15}-\sqrt{59}}{x-7}}\) được xác định thì \(\dfrac{2\sqrt{15}-\sqrt{59}}{x-7}\ge0\)

\(\Leftrightarrow x-7>0\\ \Leftrightarrow x>7\)

Để \(\sqrt{\dfrac{-2\sqrt{6}+\sqrt{23}}{-x+5}}\) được xác định thì \(\dfrac{-2\sqrt{6}+\sqrt{23}}{-x+5}\ge0\)

\(\Leftrightarrow-x+5< 0\\ \Leftrightarrow x>5\)

Với a, b, c là các số lớn hơn 1, áp dụng BĐT Cô-si có:

\(\dfrac{a^2}{b-1}+4\left(b-1\right)\ge4a\left(1\right)\)

\(\dfrac{b^2}{c-1}+4\left(c-1\right)\ge4b\left(2\right)\)

\(\dfrac{c^2}{a-1}+4\left(a-1\right)\ge4c\left(3\right)\)

Từ (1), (2), (3) suy ra: \(\dfrac{a^2}{b-1}+\dfrac{b^2}{c-1}+\dfrac{c^2}{a-1}\ge12\)

a.

Với \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\) có:

\(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{x-1}\\ =\dfrac{x+\sqrt{x}}{x-1}-\dfrac{\left(\sqrt{x}-1\right)^2}{x-1}\\ =\dfrac{x+\sqrt{x}}{x-1}-\dfrac{x-2\sqrt{x}+1}{x-1}\\ =\dfrac{x+\sqrt{x}-x+2\sqrt{x}-1}{x-1}\\ =\dfrac{3\sqrt{x}-1}{x-1}=VP\)

b.

Với  \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\) có:

\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{1}{x-4}\right)\\ =(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}-\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}).\left(\dfrac{x-4}{1}\right)\\ =(\dfrac{x-2\sqrt{x}}{x-4}-\dfrac{x+2\sqrt{x}}{x-4}).\left(x-4\right)\\ =\left(\dfrac{x-2\sqrt{x}-x-2\sqrt{x}}{x-4}\right)\left(x-4\right)\\ =\dfrac{-4\sqrt{x}\left(x-4\right)}{x-4}\\ =-4\sqrt{x}=VP\)