int a = 10;
int *p = &a;
cout << *p;
tại sao *p lại = 10 :))?
có using namespace std;
code C++
#include
using namespace std;
int main()
{
for(int i=1;i<=10;i++){
cout << "Vong lap thu " << i << endl;
for(int j=1;j<=10;j++){
if(jj > 10){
break;
}
cout << " Binh phuong cua " << j << " = " << jj << endl;
}
if(i >= 3) {
break;
}
}
}
Đoạn lệnh này kết quả như thế nào?
Tìm các nguyên hàm sau :
a)\(I_1=\int\left(1+\sqrt{x}\right)^{10}dx\)
b) \(I_2=\int\frac{xdx}{\sqrt[3]{x^2+a}}\)
c) \(I_3=\int\frac{x^2}{\sqrt{x^6+6}}\)
a) Ta thực hiện phép đổi biến :
\(1+\sqrt{x}=t\) ; \(x=\left(t-1\right)^2\) ; \(dx=2\left(t-1\right)dt\)
Khi đó \(\left(1+\sqrt{x}\right)^{10}dx=t^{10}.2\left(t-1\right)dt\)
tức là :
\(I_1=2\int\left(t^{11}-t^{10}\right)dt=2\int t^{11}dt-2\int t^{10}dt=2\left(\frac{t^{12}}{12}-\frac{t^{11}}{11}\right)+C\)
\(=\frac{1}{66}t^{11}\left(11t-12\right)+c\)
\(=\frac{1}{66}\left(1+\sqrt{x}\right)^{11}\left[11\sqrt{x}-1\right]+C\)
b) Đặt \(x^2+a=t\)
Ta có \(2xdx=dt\)
\(I_2=\frac{1}{2}\int\frac{dt}{\sqrt[3]{t}}=\frac{1}{2}\int t^{-\frac{1}{3}}dt=\frac{1}{2}.\frac{3}{2}t^{\frac{2}{3}}+C=\frac{3}{4}\sqrt[3]{\left(x^2+a\right)^2+C}\)
c) Đặt \(x^3=t\Rightarrow3x^2dx=dt\)
và \(I_3=\frac{1}{3}\int\frac{dt}{\sqrt{t^2+6}}=\frac{1}{3}\ln\left[t+\sqrt{t^2+6}\right]+C\)
\(=\frac{1}{3}\ln\left[x^2+\sqrt{x^2+6}\right]+C\)
#include <bits/stdc++.h>
using namespace std;
int main()
{
long long n,a,b,c;
cin>>n;
a=(n/100%10);
b=(n/10%10);
c=(n%10);
cout<<max(a,max(b,c));
}
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++)
cin>>a[i];
sort(a,a+n);
int cnt=0;
for(int i=n-1;i>=0;i--)
if(a[i]==a[n-1])
cnt++;
cout<<cnt<<endl;
return 0;
}
#include <iostream>
using namespace std;
int main()
{
int t;
cin >> t;
for (int i = 0; i < t; i++)
{
int a, b;
cin >> a >> b;
int diff = abs(a - b);
cout << diff << endl;
}
return 0;
}
#include<iostream>
using namespace std;
int main()
{
int a,b;
while(cin>>a>>b)
{
if(a>b)
cout << "YES"<<endl;
else
cout<<"NO"<<endl;
}
}
#include<iostream>
using namespace std;
int main()
{
int a,b;
while(cin>>a>>b)
{
int s=abs(a-b);
cout << s << endl;
}
}
Mọi người ơi , giúp e tính tích phân bất định với ạ ! Cảm ơn m.n ạ !
a.\(\int\frac{x+6}{\sqrt{x^2-2x+10}}dx\)
b.\(\int\frac{x}{\sqrt{3-2x-x^2}}dx\)
c.\(\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx\)
d,\(\int\frac{dx}{1+tanx}\)
e.\(\int tan^3xdx\)
f. \(\int cos^3xdx\)
g. \(\int sin^2x.cos^3xdx\)
h. \(\int sinx.cos2xdx\)
i. \(\int\frac{sin2x}{1+cos^2x}dx\)
a.
\(I=\int\frac{\frac{1}{2}\left(2x-2\right)+7}{\sqrt{x^2-2x+10}}dx=\frac{1}{2}\int\frac{2x-2}{\sqrt{x^2-2x+10}}dx+7\int\frac{1}{\sqrt{x^2-2x+10}}dx=\frac{1}{2}I_1+7I_2\)
Xét \(I_1=\int\frac{2x-2}{\sqrt{x^2-2x+10}}dx=\int\frac{d\left(x^2-2x+10\right)}{\sqrt{x^2-2x+10}}=2\sqrt{x^2-2x+10}+C_1\)
Xét \(I_2=\int\frac{dx}{\sqrt{x^2-2x+10}}=\int\frac{dx}{\sqrt{\left(x-1\right)^2+9}}\)
Đặt
\(u=x-1+\sqrt{\left(x-1\right)^2+10}\Rightarrow du=\left(1+\frac{\left(x-1\right)}{\sqrt{\left(x-1\right)^2+10}}\right)dx=\frac{x-1+\sqrt{\left(x-1\right)^2+10}}{\sqrt{\left(x-1\right)^2+10}}dx\)
\(\Rightarrow du=\frac{u}{\sqrt{\left(x-1\right)^2+10}}dx\Rightarrow\frac{dx}{\sqrt{\left(x-1\right)^2+10}}=\frac{du}{u}\)
\(\Rightarrow I_2=\int\frac{du}{u}=ln\left|u\right|+C_2=ln\left|x-1+\sqrt{x^2-2x+10}\right|+C_2\)
\(\Rightarrow I=\sqrt{x^2-2x+10}+7ln\left|x-1+\sqrt{x^2-2x+10}\right|+C\)
2.
\(I=\int\frac{\frac{1}{2}\left(2x+2\right)-1}{\sqrt{3-2x-x^2}}dx=\frac{1}{2}\int\frac{2x+2}{\sqrt{3-2x-x^2}}dx-\int\frac{1}{\sqrt{3-2x-x^2}}dx=\frac{1}{2}I_1-I_2\)
Xét \(I_1=\int\frac{2x+2}{\sqrt{3-2x-x^2}}dx=-\int\frac{d\left(3-2x-x^2\right)}{\sqrt{3-2x-x^2}}=-2\sqrt{3-2x-x^2}+C_1\)
Xét \(I_2=\int\frac{1}{\sqrt{3-2x-x^2}}dx=\int\frac{1}{\sqrt{4-\left(x+1\right)^2}}dx\)
Đặt \(x+1=2sinu\Rightarrow dx=2cosu.du\)
\(\Rightarrow I_2=\int\frac{2cosu.du}{2.cosu}=\int du=u+C_2=arcsin\left(\frac{x+1}{2}\right)+C_2\)
\(\Rightarrow I=-\sqrt{3-2x-x^2}-arcsin\left(\frac{x+1}{2}\right)+C\)
c/
\(I=\int\frac{1-\sqrt{x}}{\sqrt{1-x}}dx\)
Đặt \(\sqrt{x}=sint\Rightarrow x=sin^2t\Rightarrow dx=2sint.cost.dt\)
\(\Rightarrow I=\int\frac{2sint.cost\left(1-sint\right)}{\sqrt{1-sin^2t}}dt=\int\frac{2sint.cost\left(1-sint\right)}{cost}dt=\int\left(2sint-2sin^2t\right)dt\)
\(=\int\left(2sint+cos2t-1\right)dt=-2cost+\frac{1}{2}sin2t-t+C\)
\(=-2\sqrt{1-sin^2t}+\frac{1}{2}sint\sqrt{1-sin^2t}-t+C\)
\(=-2\sqrt{1-x}+\frac{1}{2}\sqrt{x\left(1-x\right)}-arcsin\left(\sqrt{x}\right)+C\)
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
int A[n],B[n],C[2*n];
for (int i=0; i<n;i++)
{
cin>>A[i];
}
for (int i=0;i<n; i++)
{
cin>>B[i];
}
for (int i=0;i<n;i++)
{
C[2*i]=A[i];
C[2*i+1]=B[i];
}
for (int i=0; i<2*n;i++)
{
cout<<C[i]<< " ";
}
}
#include <iostream>
using namespace std;
int main()
{
int n, x;
cin >> n;
int a[n];
for (int i = 0; i < n; i++)
cin >> a[i];
int q;
cin >> q;
while (q--)
{
int cnt = 0;
cin >> x;
for (int i = 0; i < n; i++)
if (a[i] == x)
cnt++;
cout << cnt << endl;
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
long long n, a[100000], i, b[100000], s, j, k;
int main()
{
cin >> n;
for (i = 1; i <= n; i++)
{
cin >> a[i];
}
k = 2;
b[1] = 0;
for (i = 2; i <= n; i++)
{
for (j = 1; j <= i - 1; j++)
{
if (a[j] > a[i])
{
s++;
}
}
b[k] = s;
k++;
s = 0;
}
for (i = 1; i <= k - 1; i++)
{
cout << b[i] << " ";
}
}
Tính \(\int\left(10+\cot^2x\right)dx\)