There are many things we should do to manage our time effectively. Firstly, it's important to prioritize tasks by their urgency and importance. This helps in focusing on what truly matters and prevents wasting time on less significant activities. Secondly, creating a schedule or to-do list can provide a clear roadmap for the day, ensuring that each task is allocated the appropriate amount of time. Thirdly, breaking larger tasks into smaller, manageable chunks can make them feel less daunting and more achievable. Lastly, learning to say no to non-essential commitments helps in avoiding overloading our schedule and preserving time for essential activities and relaxation.

Số nhỏ nhất chia hết cho 5 từ 15 dến 155 là: 15

Số lớn nhất chia hết cho 15 từ 5 đến 155 là: 155

Khoảng cách của 2 số liên tiếp chia hết cho 5 là: 5 

Từ 5 đến 155 có số lượng số chia hết cho 5 là:

(155 - 15) : 5 + 1 = 29 (số) 

@ Phong từ 15 đến 155 mà em. 

Số chẵn lớn nhất có 2 chữ số là: 98

Hay hiệu chúng là 98

Phân số lớn hơn 1 hay tử số lớn hơn mẫu số

Mẫu số là:

  (176 - 98) : 2 = 39

Tử số là:

  39 + 98 = 137

Cách 1:

Tổng số kg giấy vụn và báo cũ lớp 4A thu gom được là:

   108 + 72 = 180 (kg)

Mỗi bạn thu gom được số kg là:

  180 : 36 = 5 (kg)

Cách 2:

Mỗi bạn thu gom được số kg giấy vụn là:

  108 : 36 = 3 (kg giấy vụn)

Mỗi bạn thu gom được số kg báo cũ là:

   72 : 36 = 2 (kg báo cũ)

Vậy mỗi bạn thu gom được số kg vừa báo cũ vừa giấy vụn là:

  3 + 2 = 5 (kg)

a) \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{3}{4}\right)\le x\le\dfrac{1}{24}-\left(\dfrac{1}{8}-\dfrac{1}{3}\right)\)

\(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{3}{4}\le x\le\dfrac{1}{24}-\dfrac{1}{8}+\dfrac{1}{3}\)

\(\dfrac{6}{12}-\dfrac{4}{12}-\dfrac{9}{12}\le x\le\dfrac{1}{24}-\dfrac{3}{24}+\dfrac{8}{24}\)

\(-\dfrac{7}{12}\le x\le\dfrac{6}{24}\)

\(-\dfrac{7}{12}\le x\le\dfrac{1}{4}\) 

Giá trị x nguyên thỏa mãn là: 0  

b) \(-\dfrac{1}{2}-\dfrac{3}{4}+\dfrac{1}{3}\le x\le\dfrac{5}{6}-\left(-\dfrac{3}{4}\right)+\dfrac{7}{12}\)

\(\dfrac{-6}{12}-\dfrac{9}{12}+\dfrac{4}{12}\le x\le\dfrac{5}{6}+\dfrac{3}{4}+\dfrac{7}{12}\)

\(-\dfrac{11}{12}\le x\le\dfrac{10}{12}+\dfrac{9}{12}+\dfrac{7}{12}\)

\(-\dfrac{11}{12}\le x\le\dfrac{26}{12}\)

\(-\dfrac{11}{12}\le x\le\dfrac{13}{6}\)

Các giá trị x nguyên thỏa mãn là: 0; 1; 2

a) Ta có : 

\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{3}{4}\right)\le x\le\dfrac{1}{24}-\left(\dfrac{1}{8}-\dfrac{1}{3}\right)\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{13}{12}\le x\le\dfrac{1}{24}-\dfrac{-5}{24}\)

\(\Leftrightarrow\dfrac{-7}{12}\le x\le\dfrac{1}{4}\)

mà \(x\in Z\)

\(\Rightarrow x=0\)

Vậy \(x=0\)

b) Ta có : 

\(\dfrac{-1}{2}-\dfrac{3}{4}+\dfrac{1}{3}\le x\le\dfrac{5}{6}-\left(\dfrac{-3}{4}\right)+\dfrac{7}{12}\)

\(\Leftrightarrow\dfrac{-6}{12}-\dfrac{9}{12}+\dfrac{4}{12}\le x\le\dfrac{5}{6}+\dfrac{3}{4}+\dfrac{7}{12}\)

\(\Leftrightarrow\dfrac{-11}{12}\le x\le\dfrac{10}{12}+\dfrac{9}{12}+\dfrac{7}{12}\)

\(\Leftrightarrow\dfrac{-11}{12}\le x\le\dfrac{26}{12}\)

\(\Leftrightarrow\dfrac{-11}{12}\le x\le\dfrac{13}{6}\)

mà \(x\in Z\)

\(\Rightarrow x\in\left\{0;1;2\right\}\)

Vậy \(x\in\left\{0;1;2\right\}\)

Số số hạng là: ( 2025 - 5 ) : 5 + 1 = 405

Tổng của dãy số trên là: ( 2025 + 5 ) x 405 : 2 = 411075

Đ/s : 411075

Công thức tính tổng dãy số cháu chưa học rất khó giải thích ah.

a) \(-\dfrac{3}{7}-x=-\dfrac{1}{2}\\ x=-\dfrac{3}{7}-\left(-\dfrac{1}{2}\right)\\ x=\dfrac{-3}{7}+\dfrac{1}{2}=\dfrac{-6}{14}+\dfrac{7}{14}=\dfrac{1}{14}\)

b) \(x-\dfrac{4}{5}=\dfrac{1}{2}-\dfrac{1}{3}\\ x-\dfrac{4}{5}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\\ x=\dfrac{1}{6}+\dfrac{4}{5}=\dfrac{5}{30}+\dfrac{24}{30}\\ x=\dfrac{29}{30}\)

c) \(-x-\dfrac{3}{4}=-\dfrac{8}{11}\\ -x=-\dfrac{8}{11}+\dfrac{3}{4}\\ -x=-\dfrac{32}{44}+\dfrac{33}{44}=\dfrac{1}{44}\\ x=-\dfrac{1}{44}\)

d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\\ \dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\\ x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\\ x=\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}\\ x=-\dfrac{9}{60}\)

a) \(\dfrac{-3}{7}-x=\dfrac{-1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}-\dfrac{3}{7}\)

\(\Rightarrow x=\dfrac{1}{14}\)

Vậy \(x=\dfrac{1}{14}\)

b) \(x-\dfrac{4}{5}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\Rightarrow x-\dfrac{4}{5}=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{1}{6}+\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{29}{30}\)

Vậy \(x=\dfrac{29}{30}\)

c) \(-x-\dfrac{3}{4}=\dfrac{-8}{11}\)

\(\Rightarrow-x=\dfrac{3}{4}-\dfrac{8}{11}\)

\(\Rightarrow-x=\dfrac{1}{44}\)

\(\Rightarrow x-\dfrac{1}{44}\)

Vậy \(x=-\dfrac{1}{44}\)

d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(\Rightarrow x=\dfrac{-3}{20}\)

Vậy \(x=\dfrac{-3}{20}\)

\(a.\left(\dfrac{-4}{5}+\dfrac{3}{7}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{2}{7}\right)\\ =\dfrac{-4}{5}+\dfrac{2}{7}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{2}{7}\\ =\left(-\dfrac{4}{5}-\dfrac{1}{5}\right)+\left(\dfrac{2}{7}+\dfrac{2}{7}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)\\ =-1+\dfrac{4}{7}+0=-\dfrac{3}{7}\)

\(b.\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\\ =7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\\ =\left(7-6-5\right)+\left(\dfrac{7}{4}-\dfrac{3}{4}-\dfrac{5}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\\=1+\dfrac{-1}{4}+0=\dfrac{3}{4}\)

a) 

\(\left(-\dfrac{4}{5}+\dfrac{3}{7}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{2}{7}\right)\\ =\left(-\dfrac{4}{5}-\dfrac{1}{5}\right)+\left(\dfrac{3}{7}+\dfrac{2}{7}\right)+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)\\ =-\dfrac{5}{5}+\dfrac{5}{7}+0\\ =-1+\dfrac{5}{7}\\ =-\dfrac{2}{7}\)

b) 

\(\left(7-\dfrac{3}{4}+\dfrac{1}{2}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\\ =7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\\ =\left(7-6-5\right)+\left(-\dfrac{3}{4}-\dfrac{5}{4}+\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\\ =\left(-4\right)+\left(\dfrac{-1}{4}\right)+0\\ =-\dfrac{17}{4}\)

c) 

\(\left(0,25+\dfrac{7}{9}-\dfrac{1}{7}\right)-\left(0,75-\dfrac{2}{9}-\dfrac{1}{7}\right)\\ =0,25+\dfrac{7}{9}-\dfrac{1}{7}-0,75+\dfrac{2}{9}+\dfrac{1}{7}\\ =\left(0,25-0,75\right)+\left(\dfrac{7}{9}+\dfrac{2}{9}\right)+\left(-\dfrac{1}{7}+\dfrac{1}{7}\right)\\ =-\dfrac{1}{2}+\dfrac{9}{9}+0\\ =-\dfrac{1}{2}+1\\ =\dfrac{1}{2}\)

d) 

\(\dfrac{\dfrac{2}{7}+\dfrac{1}{3}-\dfrac{2}{9}}{\dfrac{3}{7}+\dfrac{1}{2}-\dfrac{1}{3}}\\ =\dfrac{\dfrac{2}{7}+\dfrac{2}{6}-\dfrac{2}{9}}{\dfrac{3}{7}+\dfrac{3}{6}-\dfrac{3}{9}}\\ =\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{9}\right)}{3\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{9}\right)}\\ =\dfrac{2}{3}\)

a) \(\dfrac{-3}{21}+\dfrac{-2}{7}+\dfrac{1}{3}=\dfrac{-3}{21}+\dfrac{-6}{21}+\dfrac{7}{21}=-\dfrac{2}{21}\)

b) \(\dfrac{-13}{15}+\dfrac{5}{-18}+\dfrac{4}{9}=\dfrac{-78}{90}+\dfrac{-25}{90}+\dfrac{40}{90}=\dfrac{63}{90}=\dfrac{7}{10}\)

c) \(\dfrac{-2}{5}-\left(\dfrac{-3}{11}\right)=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)

d) \(\left(-4\right)-\left(\dfrac{-4}{5}\right)-\dfrac{2}{3}=\left(-4\right)+\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{-60}{15}+\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{-58}{15}\)

e) \(\dfrac{3}{5}+\left(\dfrac{-4}{3}\right)-\dfrac{-3}{4}=\dfrac{3}{5}+\left(\dfrac{-4}{3}\right)+\dfrac{3}{4}=\dfrac{36}{60}+\dfrac{-80}{60}+\dfrac{45}{60}=\dfrac{1}{60}\)

g) \(\dfrac{5}{8}-\left(-\dfrac{2}{5}\right)-\dfrac{3}{10}=\dfrac{5}{8}+\dfrac{2}{5}-\dfrac{3}{10}=\dfrac{25}{40}+\dfrac{16}{40}-\dfrac{12}{40}=\dfrac{29}{40}\)

h) \(\dfrac{3}{4}-\left(-\dfrac{5}{3}\right)+\left(\dfrac{1}{12}+\dfrac{2}{9}\right)=\dfrac{3}{4}+\dfrac{5}{3}+\left(\dfrac{3}{36}+\dfrac{8}{36}\right)=\dfrac{27}{36}+\dfrac{60}{36}+\dfrac{11}{36}=\dfrac{98}{36}=\dfrac{49}{18}\)

a) \(\dfrac{-3}{21}+\dfrac{-2}{7}+\dfrac{1}{3}=\dfrac{-3}{21}-\dfrac{6}{21}+\dfrac{7}{21}\\ =\dfrac{-3-6+7}{21}=-\dfrac{2}{21}\)

b) \(\dfrac{-13}{15}+\dfrac{5}{-18}+\dfrac{4}{9}=\dfrac{-78}{90}-\dfrac{25}{90}+\dfrac{40}{90}\\ =\dfrac{-78-25+40}{90}=\dfrac{-63}{90}=-\dfrac{7}{10}\)

c) \(\dfrac{-2}{5}-\left(\dfrac{-3}{11}\right)=-\dfrac{2}{5}+\dfrac{3}{11}\\ =-\dfrac{22}{55}+\dfrac{15}{55}=-\dfrac{7}{55}\)

a: Trên tia Ox, ta có: OA<OB

nên A nằm giữa O và B

=>OA+AB=OB

=>AB+4=6

=>AB=2(cm)

b: C là trung điểm của OA

=>\(CO=CA=\dfrac{OA}{2}=2\left(cm\right)\)

Vì AO và AB là hai tia đối nhau

nên AC và AB là hai tia đối nhau

=>A nằm giữa hai điểm B và C

Ta có: A nằm giữa B và C

mà AB=AC(=2cm)

nên A là trung điểm của BC

d: Các góc đỉnh D trong hình vẽ là: \(\widehat{ODC};\widehat{ODA};\widehat{ODB};\widehat{CDA};\widehat{CDB};\widehat{ADB}\)