\(3-\dfrac{16}{11}=\dfrac{33}{11}-\dfrac{16}{11}\\ =\dfrac{33-16}{11}=\dfrac{17}{11}\)

.

\(\dfrac{9}{8}-1=\dfrac{9}{8}-\dfrac{8}{8}=\dfrac{9-8}{8}\\ =\dfrac{1}{8}\)

\(3-\dfrac{16}{11}=\dfrac{33-16}{11}=\dfrac{17}{11}\)

\(\dfrac{9}{8}-1=\dfrac{9-8}{8}=\dfrac{1}{8}\)

\(\dfrac{19}{5}-\dfrac{3}{15}\)

\(=\dfrac{19}{5}-\dfrac{1}{5}\)

\(=\dfrac{19-1}{5}\)

\(=\dfrac{18}{5}\)

19/5 - 3/15

= 57/15 - 3/15

= 54/15

= 18/5

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 \(x\in\left\{0;1;2\right\}\)

Vì chữ số 1 thuộc lớp nghìn của số tạo thành nên để số tạo thành là lẻ thì chữ số hàng đơn vị là 5

Với 3 số 7 ; 1 ; 2 thuộc lớp nghìn , ta có :

3 cách chọn chữ số hàng trăm nghìn

2 cách chọn chữ số hàng chục nghìn

1 cách chọn chữ số hàng nghìn

Với 2 số còn lại ( số 0 và số 6 ) , ta có :

2 cách chọn chữ số hàng trăm

1 cách chọn chữ số hàng nghìn

Như vậy , từ các tấm thẻ số : 7 , 1 , 2 , 0 , 5 , 6 lập được số các số lẻ có sáu chữ số mà lớp nghìn của chữ soó đó gồm các chữ số 7 , 1 , 2 là :

\(3\times2\times1\times3\times2=36\) ( số )

Đáp số : 36 số

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.

\(\dfrac{4}{18}-\dfrac{1}{9}\)

\(=\dfrac{4}{18}-\dfrac{2}{18}\)

\(=\dfrac{4-2}{18}\)

\(=\dfrac{2}{18}\)

\(=\dfrac{1}{9}\)

\(\dfrac{4}{18}-\dfrac{1}{9}\)

\(=\dfrac{2}{9}-\dfrac{1}{9}\)

\(=\dfrac{2-1}{9}\)

\(=\dfrac{1}{9}\)

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 \(x=\dfrac{-3}{20}\)

Sossssssss

Mình sửa đề nhé: (x^3+y^3):(x+y)

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

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

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