1 did you get

2 has

3 will see

4 am

5 is climbing

\((x+5)^2-4x^2\\=(x+5)^2-(2x)^2\\=(x+5-2x)(x+5+2x)\\=(5-x)(3x+5)\)

Lời giải:

$2x^3y+2xy^3+4x^2y^2-8xy$

$=2xy(x^2+y^2+2xy-4)$

$=2xy[(x^2+2xy+y^2)-4]$

$=2xy[(x+y)^2-2^2]=2xy(x+y-2)(x+y+2)$

P.s: lần sau bạn lưu ý ghi đầy đủ yêu cầu đề.

1 B

2 D

3 C

4 A

5 C

VII

1 Although she is careful, she still forgets her phone when she goes out

\((2x-5)^2-64x^2\\=(2x-5)^2-(8x)^2\\=(2x-5-8x)(2x-5+8x)\\=(-6x-5)(10x-5)\)

Lời giải:

$(2x-5)^2-64x^2=(2x-5)^2-(8x)^2$

$=(2x-5-8x)(2x-5+8x)$

$=(-6x-5)(10x-5)$

$=-5(6x+5)(2x-1)$

Python

Pascal

Java

C++

Arduino

câu a chưa đủ đề em hấy

c, \(x\)(\(x\) - 2022) + 4.(2022 - \(x\)) = 0

       (\(x\) - 2022).(\(x\) - 4) = 0

         \(\left[{}\begin{matrix}x-2022=0\\x+4=0\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=2022\\x=4\end{matrix}\right.\)

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

vector<int> solve(int d) {
    vector<int> res(2, 0);
    int p = 1;
    while (d > 0) {
        int digit = d % 10;
        d /= 10;
        if (digit == 4) {
            res[0] += p * 3;
            res[1] += p;
        } else {
            res[0] += p * digit;
        }
        p *= 10;
    }
    return res;
}

int main() {
    int d;
    cin >> d;
    vector<int> res = solve(d);
    cout << res[0] << " " << res[1] << endl;
    return 0;
}