(2/3)^3-4.(-1 3/4)^2+(-2/3)^3
giúp mình vs ạ :))
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19 tháng 9 2021
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
TQ
1
28 tháng 2 2023
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n,i,kt;
cin>>n;
cout<<pow((n*(n+1)/2),4);
}
TH
2
25 tháng 8 2023
1)
xy + x - 4y = 12
x + y(x - 4) = 12
y(x - 4) = 12 - x
\(y=\dfrac{-x+12}{x-4}\)
Vì \(x,y\inℕ\) nên
\(\left(-x+12\right)⋮\left(x-4\right)\)
\(\left(-x+12\right)-\left(x-4\right)⋮\left(x-4\right)\)
\(16⋮\left(x-4\right)\)
\(\left(x-4\right)\inƯ\left(16\right)\)
\(\left(x-4\right)\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)
\(x\in\left\{5;3;6;2;8;0;12;-4;20;-12\right\}\)
\(y\in\left\{\dfrac{-5+12}{5-4};\dfrac{-3+12}{3-4};\dfrac{-6+12}{6-4};\dfrac{-2+12}{2-4};\dfrac{-8+12}{8-4};\dfrac{-0+12}{0-4};\dfrac{-12+12}{12-4};\dfrac{4+12}{-4-4};\dfrac{-20+12}{20-4};\dfrac{12+12}{-12-4}\right\}\)
\(y\in\left\{7;-9;3;-5;1;-3;0;-2;-\dfrac{1}{2};-\dfrac{7}{5}\right\}\)
\(\left(x;y\right)\in\left\{\left(5;7\right);\left(3;-9\right);\left(6;3\right);\left(2;-5\right);\left(8;1\right);\left(0;-3\right);\left(12;0\right);\left(-4;-2\right);\left(20;-\dfrac{1}{2}\right);\left(-12;-\dfrac{7}{5}\right)\right\}\)
Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)
2)
(2x + 3)(y - 2) = 15
\(\left(2x+3\right)\inƯ\left(15\right)\)
\(\left(2x+3\right)\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
Ta lập bảng
| 2x + 3 | 1 | -1 | 3 | -3 | 5 | -5 | 15 | -15 |
| y - 2 | 15 | -15 | 5 | -5 | 3 | -3 | 1 | -1 |
| (x; y) | (-1; 17) | (-2; -13) | (0; 7) | (-3; -3) | (1; 5) | (-4; -1) | (6; 3) | (-9; 1) |
Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\)
9 tháng 1
\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)
\(=\dfrac{3}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{3}{\dfrac{3\left(3+1\right)}{2}}+...+\dfrac{3}{\dfrac{2022\left(2022+1\right)}{2}}\)
\(=\dfrac{6}{2\left(2+1\right)}+\dfrac{6}{3\left(3+1\right)}+...+\dfrac{6}{2022\cdot2023}\)
\(=\dfrac{6}{2\cdot3}+\dfrac{6}{3\cdot4}+...+\dfrac{6}{2022\cdot2023}\)
\(=6\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\)
\(=6\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)=6\cdot\dfrac{2021}{4046}=\dfrac{12126}{4046}< 3\)
mà \(3< \dfrac{10}{3}\)
nên \(M< \dfrac{10}{3}\)
= \(\left(\dfrac{2}{3}\right)^3-4.\left(-\dfrac{7}{4}\right)^2+\left(-\dfrac{2}{3}\right)^3\)
= \(\dfrac{8}{27}-4.\dfrac{49}{16}+\left(-\dfrac{8}{27}\right)\)
= \(\left[\dfrac{8}{27}+\left(-\dfrac{8}{27}\right)\right]-4.\dfrac{49}{16}\)
= \(-\dfrac{49}{4}\)
\(\left(\dfrac{2}{3}\right)^3-4.\left(-1\dfrac{3}{4}\right)^2+\left(\dfrac{-2}{3}\right)^3\)
= \(\left[\left(\dfrac{2}{3}\right)^3+\left(\dfrac{-2}{3}\right)^3\right]-4.\left(-1\dfrac{3}{4}\right)^2\)
= \(\left(\dfrac{8}{9}+\dfrac{-8}{9}\right)-4.\left(-1\dfrac{3}{4}\right)^2\)
= \(0-4.\left(\dfrac{-7}{4}\right)^2\)
= \(0-4.\dfrac{49}{16}\)
= \(0-\dfrac{49}{4}\)
= \(\dfrac{-49}{4}\)