\(\left(A\cap B\right)\cup C=[0;\sqrt{5})\)

\(\left(A\cup B\right)\cap C=[0;1)\cup[2;\sqrt{5})\)

\(A\cap\left(B\cup C\right)=[0;1)\)

(A∩B)∪C=[0;5​)

\left(A\cup B\right)\cap C=[0;1)\cup[2;\sqrt{5})(A∪B)∩C=[0;1)∪[2;5​)

$A\cap \left(B\cup C\right)=[0;1)$

Gọi a là diện tích trồng cafe.

Tổng số tiền hộ nông dân thu được trên 10 ha là:

\(a.10000000+\left(10-a\right)12000000=120000000-2000000a\)

Theo bài ra ta có:

\(20a\le100\Rightarrow a\le5.\left(1\right)\)

\(\left(10-a\right)30\le180\Leftrightarrow300-30a\le180\Rightarrow a\ge4.\left(2\right)\)

Kết hợp (1) và (2) suy ra a= 4 hoặc a = 5.

Để số tiền thu được là lớn nhất thì 

\(a.10000000+\left(10-a\right)12000000=120000000-2000000a\) lớn nhất.

Khi đó 2 000 000a bé nhất, suy ra a = 4.

Vậy trồng 4 ha cafe và 6 ha ca cao thì số tiền thu được là lớn nhất.

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Bài 6:

a: Để \(\frac{14}{3\sqrt{x}+6}\) là số nguyên thì \(14\) ⋮\(3\sqrt{x}+6\)

=>\(3\sqrt{x}+6\in\left\lbrace7;14\right\rbrace\)

=>\(3\sqrt{x}\in\left\lbrace1;8\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace\frac13;\frac83\right\rbrace\)

=>x\(\in\left\lbrace\frac19;\frac{64}{9}\right\rbrace\)

=>A={1/9;64/9}

b: Các tập hợp con của A là:

∅; {1/9}; {64/9}; {1/9;64/9}

Bài 7:

a: \(\left(x^4-16\right)\left(x^2-1\right)=0\)

=>\(\left[\begin{array}{l}x^4-16=0\\ x^2-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^4=16\\ x^2=1\end{array}\right.=>\left[\begin{array}{l}x=2\\ x=-2\\ x=1\\ x=-1\end{array}\right.\)

=>A={2;-2;1;-1}

2x-9<=0

=>2x<=9

=>x<=4,5

mà x là số tự nhiên

nên x∈{0;1;2;3;4}

=>B={0;1;2;3;4}

B\A={0;1;2;3;4}\{2;-2;1;-1}

={0;3;4}

X⊂B\A

=>X⊂{0;3;4}

=>X={0;3}; X={0;4}; X={3;4}

b: A\B={2;-2;1;-1}\{0;1;2;3;4}

={-1;-2}

X\(\cap A\) =A\B

=>X\(\cap\) A={-1;-2}

mà X có đúng 2 phần tử

nên X={-1;-2}

Đầu kiện: \(x+y\ne0\Leftrightarrow x\ne-y\)

Ta có:

\(3x^3-y^3=\dfrac{1}{x+y}\\ \Leftrightarrow\left(3x^3-y^3\right)\left(x+y\right)=1\\ \Leftrightarrow\left(3x^3-y^3\right)\left(x+y\right)=\left(x^2+y^2\right)^2\)(*)

Xét \(y=0\Rightarrow x=\pm1\) thay vào phương trình (*) ta thấy không thõa mãn.

Với \(y\ne0\) chia hai vế phương trình (*) cho \(y^4\) ta có:

\(\dfrac{\left(3x^3-y^3\right)\left(x+y\right)}{y^4}=\dfrac{\left(x^2+y^2\right)^2}{y^4}\\ \Leftrightarrow\left(\dfrac{3x^3}{y^3}-1\right)\left(\dfrac{x}{y}+1\right)=\left(\dfrac{x^2}{y^2}+1\right)^2\)

Đặt \(t=\dfrac{x}{y}\) thay vào phương trình trên ta có:

\(\left(3t^3-1\right)\left(t+1\right)=\left(t^2+1\right)^2\)

\(\Leftrightarrow3t^4-t+3t^3-1=t^4+2t^2+1\\ \Leftrightarrow2t^4+3t^3-2t^2-t-2=0\\ \)

\(\Leftrightarrow2t^3\left(t+2\right)-t^2\left(t+2\right)-\left(t+2\right)=0\\ \Leftrightarrow\left(t+2\right)\left(2t^3-t^2-1\right)=0\\ \Leftrightarrow\left(t+2\right)\left(t^3-1+t^3-t^2=0\right)\\ \Leftrightarrow\left(t+2\right)\left(t-1\right)\left(2t^2+t+1\right)=0\\ \)

\(\Rightarrow\left[{}\begin{matrix}t+2=0\\t-1=0\\2t^2+t+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}t=-2\\t=1\\\Delta< 0,vô.nghiệm\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2y\\x=y\end{matrix}\right.\)

Thay x vào phương trình \(x^2+y^2=1\) tìm y => x.

So với đầu kiện bài toán kết luận nghiệm

Ta có: \(\left\{{}\begin{matrix}p+e+n=58\\p=e\\p+e-n=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=19\\n=20\end{matrix}\right.\)

 Every morning, I get up at a quarter past five. As soon as I get up, I do exercise for 10 minutes. After that, I drink a glass of water, brush my teeth and have breakfast at a quarter to six. At 6 am every morning except Sundays, I prepare to go to school. My school starts at 6:45 am and ends at 11:15 am. Right after my school ends, I go straight to home and have lunch at a quarter to twelve. Then, I take a nap, usually for an hour but sometimes it can be 2-3 hours! The reason is, in some afternoons, I have my extra English class at 3 pm and some are my Math lessons, at half past three. I go back home from my extra classes at half past four of at 10 past five depends on what I study in that lesson. After I get home, I take a shower and relax for some minutes, then I have dinner at 6 pm before I have some time to play a little bit. And right now, from 7 pm to a quarter to nine, I am doing my homework. I always listen to music while studying because I think it will help my brain's right hemisphere, which means I will be more creative. Then I spend the 15 last minutes playing chess online. Finally, I go to bed at 9 pm. That is all about my daily life.

 \(Merci\) \(d'avoir\) \(lu!\)

cm: ∀ n ϵ N , n² ≥ n

phương pháp quy nạp toán học

với n =  1 ⇔ 1² = 1 (đúng)

giả sử n² ≥ n đúng với n = k ( k ϵ N) ⇔ k² ≥ k (1)

ta cần chứng minh n² ≥ n đúng với  n = k +  1 

thât vậy với n = k + 1 , k ϵN ta có:

(k+1)² - (k+1) = (k+1)(k+1 -1) =(k+ 1).k = k² + k (2)

thay (1) vào (2) ta có : (k+1)² - (k+1) = k² + k ≥ 2k ≥ 0 vì (kϵN)

⇔ (k+1)² ≥ k +1  

vậy n² ≥ n ∀ n ϵ N (đpcm)