Mọi người giúp mình bài tập này với!
(Trích câu 2, đề thi Tin học trẻ bảng B - THCS TP. Hà Nội năm 2023)
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GN
GV Nguyễn Trần Thành Đạt
Giáo viên
4 tháng 12 2023
\(V=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\pi.2^2.3=4\pi\)
Chọn B
3 tháng 6 2019
Có \(a^2+b^2=3-ab\)
Mà \(a^2+b^2\ge2ab\)
\(\Leftrightarrow3\ge3ab\)
\(\Leftrightarrow1\ge ab\left(1\right)\)
Cũng có:\(a^2+b^2\ge-2ab\)
\(\Leftrightarrow3-ab\ge-2ab\)
\(\Leftrightarrow-3\le ab\left(2\right)\)
Từ (1) và (2) \(1\ge ab\ge-3\)
Lại có :
\(\left(a^2+b^2\right)^2=\left(3-ab\right)^2\)
\(\Leftrightarrow a^4+b^4=9-6ab+a^2b^2-2a^2b^2=9-6ab-a^2b^2\)
\(\Rightarrow P=a^4+b^4-ab=9-7ab-a^2b^2=-\left(a^2b^2+7ab-9\right)\)
\(\Leftrightarrow P=-\left(a^2b^2-7ab+8ab\right)\)
\(\Leftrightarrow P=\left(ab+3\right)\left(-ab-4\right)+21\)
Có \(ab\ge-3\Rightarrow ab+3\ge0\)
\(-ab-4< 0\)
\(\Rightarrow P\le21\)
Max P = 21<=> ab=-3;a=-b<=>\(b=\pm\sqrt{3};a=\pm\sqrt{3}\)tương ứng
30 tháng 11 2023
\(A=log_m\left(8m\right)=log_mm+log_m8\)
\(=1+log_m8\)
\(=1+\dfrac{1}{log_8m}=1+\dfrac{1}{log_{2^3}m}=1+\dfrac{1}{\dfrac{1}{3}\cdot log_2m}\)
\(=1+\dfrac{1}{\dfrac{1}{3}a}=1+1:\dfrac{a}{3}=1+\dfrac{3}{a}=\dfrac{a+3}{a}\)
=>Chọn A
17 tháng 1 2021
Câu 4b:
Ta có \(a-\sqrt{a}=\sqrt{b}-b\Leftrightarrow a+b=\sqrt{a}+\sqrt{b}\). (1)
Áp dụng bất đẳng thức Cauchy - Schwarz ta có:
\(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2};\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\).
Kết hợp với (1) ta có:
\(a+b\le\sqrt{2\left(a+b\right)}\Leftrightarrow0\le a+b\le2\).
Ta có: \(P\ge\dfrac{\left(a+b\right)^2}{2}+\dfrac{2020}{\left(\sqrt{a}+\sqrt{b}\right)^2}\) (Do \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\))
\(=\dfrac{\left(a+b\right)^2}{2}+\dfrac{2020}{\left(a+b\right)^2}\) (Theo (1))
\(\Rightarrow P\ge\dfrac{\left(a+b\right)^2}{2}+\dfrac{2020}{\left(a+b\right)^2}\).
Áp dụng bất đẳng thức AM - GM cho hai số thực dương và kết hợp với \(a+b\le2\) ta có:
\(\dfrac{\left(a+b\right)^2}{2}+\dfrac{2020}{\left(a+b\right)^2}=\left[\dfrac{\left(a+b\right)^2}{2}+\dfrac{8}{\left(a+b\right)^2}\right]+\dfrac{2012}{\left(a+b\right)^2}\ge2\sqrt{\dfrac{\left(a+b\right)^2}{2}.\dfrac{8}{\left(a+b\right)^2}}+\dfrac{2012}{2^2}=4+503=507\)
\(\Rightarrow P\ge507\).
Đẳng thức xảy ra khi a = b = 1.
Vậy Min P = 507 khi a = b = 1.
17 tháng 1 2021
Giải nốt câu 4a:
ĐKXĐ: \(x\geq\frac{-1}{2}\).
Phương trình đã cho tương đương:
\(x^2+2x+1=2x+1+2\sqrt{2x+1}+1\)
\(\Leftrightarrow\left(x+1\right)^2=\left(\sqrt{2x+1}+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-\left(\sqrt{2x+1}+1\right)^2=0\)
\(\Leftrightarrow\left(x+1-\sqrt{2x+1}-1\right)\left(x+1+\sqrt{2x+1}+1\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+1}\right)\left(x+\sqrt{2x+1}+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2x+1}=0\left(1\right)\\x+\sqrt{2x+1}+2=0\left(2\right)\end{matrix}\right.\).
Ta thấy \(x+\sqrt{2x+1}+2>0\forall x\ge-\dfrac{1}{2}\).
Do đó phương trình (2) vô nghiệm.
Xét phương trình (1) \(\Leftrightarrow x=\sqrt{2x+1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-1\right)^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=\sqrt{2}+1>0>-\dfrac{1}{2}\left(TM\right)\\x=-\sqrt{2}+1< 0\left(\text{loại}\right)\end{matrix}\right.\end{matrix}\right.\).
Vậy nghiệm của phương trình là \(x=\sqrt{2}+1\).
NN
19 tháng 1 2021
While it is believed by some people that our modern life has been made better by mobile phones, others may claim that there are some problems which people have to face from this trend. Personally, I agree with group one for communication and convenience.
The most benefit gained from digital phones is that today people can easily communicate with others. In the past, when people live far from their family, they just have to send their letters, and it took their family a few days to receive them. But now, by using a car phone, people can send a message in a few seconds and it is even free. Moreover, if people want to see each other, it allows them to chat video. In business, this is a quick and useful way to contact customers and respond to the needs of them.
Another advantage of using cell phones is that they are small and convenient for users. They fit easily people’s pocket or bag, and do not weigh much. People can take it to everywhere, relax by playing games on cell phones and receive phone calls from customers at the same time. In the event of an accident, injury, criminal incident, or other emergencies, mobile phones ensure that emergency services, as well as family and friends, can be contacted immediately.
Regardless of these advantages, it also has some drawbacks that should be considered. Some studies suggest that the radiation produced by mobile phones causes harm to human health. Brain cancer can be led by using digital phones. Besides, it also harms people’s eyes and has bad effects on physical health.
In conclusion, in spite of the disadvantages of cell phones, the advantages outweigh them. From my perspective, cell phones contribute to our life and make it easier and faster.
MN
19 tháng 1 2021
Cell phones have become very common among everyone, thanks to the technological advancements, the handheld device is available starting from few pennies to even millions of dollars. These tools, depending on features that they come along with, can perform multiple operations. While the technology geeks encourage the uptrend, luddites talk about negative impacts that such devices can bring into the society. Both these arguments will be carefully analysed before arriving at a reasoned conclusion.
These portable devices have become an indispensable part of our mind which requires entertainment-on-demand. For instance, such devices accessing high-speed network can stream a high definition video live. These devices are also being constructively used to control the smart appliances at home remotely from anywhere in the world. In addition to this, the cell phone is used for improving one’s fitness through some useful application that can help through different fitness routines replacing the need for a trainer. In short, the mobile phones are replacing many other types of equipment and emerging as a one-stop solution for many requirements.
On the other hand, due to smartphones, people are glued to their smartphone screens even during meetings in office and also while at home. To illustrate, phubbing has become one of the leading bad behaviours in recent years. Specifically, the situation in some of the house is so worse that they chat with their sibling in next room instead of talking to them. To conclude, the socialisation among the social animal is taking place in an unconventional way. This can lead to loss of linguistic speaking skills in a long term.
Considering the points of view of both camps, the cell phones have many advantages than disadvantages but it is advocated to use the technology judiciously and ensure overall development of human society
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
int n, t;
cin >> n >> t;
vector<int> arr(n);
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
sort(arr.begin(), arr.end());
int bestSum = 0;
double bestDiff = 1e9;
for (int i = 0; i < n - 2; i++) {
for (int j = i + 1; j < n - 1; j++) {
for (int k = j + 1; k < n; k++) {
int sum = arr[i] + arr[j] + arr[k];
double average = static_cast<double>(sum) / 3.0;
double diff = abs(average - t);
if (diff < bestDiff || (diff == bestDiff && sum > bestSum)) {
bestDiff = diff;
bestSum = sum;
}
}
}
}
cout << bestSum << endl;
return 0;
}
Nếu bn dùng Python thì mình gửi nha:
A=[] N=int(input("Nhập số phần tử trong list A: ")) while N >100: N=int(input("Nhập số phần tử trong list A: ")) while N < 3: N=int(input("Nhập số phần tử trong list A: ")) i=0 for i in range(N): A.append(int(input('Nhap so thu %d: ' % (i+1)))) print(A) T=int(input("Nhập T: ")) i=0 j=i+1 k=j+1 kq=abs(((A[i]+A[j]+A[k])/3)) chenhlech=0 chenhlech_1=abs(((A[i]+A[j]+A[k])/3)-T) tongcapmoi= A[i]+A[j]+A[k] i=0 tongcapmoi=0 for i in range (0, len(A)-2): for j in range (i+1, len(A)-1): for k in range (j+1, len(A)): tongcap=abs(A[i]+A[j]+A[k]) trungbinh=abs(((tongcap)/3)) chenhlech=abs(trungbinh-T) if chenhlech <= chenhlech_1: if tongcap > tongcapmoi and chenhlech==chenhlech_1: tongcapmoi=tongcap a=A[i] b=A[j] c=A[k] else : if tongcap <tongcapmoi : tongcapmoi=tongcap a=A[i] b=A[j] c=A[k] chenhlech_1=chenhlech print(a+b+c, " là tổng của ba số trong danh sách thỏa mãn yêu cầu nhất.")