\(-\left(\dfrac{3}{7}+\dfrac{3}{8}\right)-\left(-\dfrac{3}{8}+\dfrac{4}{7}\right)\\ =-\dfrac{3}{7}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{7}\\ =\left(-\dfrac{3}{7}-\dfrac{4}{7}\right)+\left(\dfrac{3}{8}-\dfrac{3}{8}\right)\\ =\dfrac{-7}{7}+0\\ =-1\)

$-(\frac37+\frac38)-(-\frac38+\frac47)$

$=-\frac37-\frac38+\frac38-\frac47$

$=-\frac37-\frac47=-\frac77=-1$

\(B=\dfrac{2a+3}{a-2}=\dfrac{2\left(a-2\right)+7}{a-2}\\ =2+\dfrac{7}{a-2}\) (a nguyên, a khác 2)

Để B đạt gt nguyên thì: \(\dfrac{7}{a-2}\) cũng phải đạt gt nguyên

\(\Rightarrow7⋮\left(a-2\right)\)

\(\Rightarrow a-2\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\\ \Rightarrow a\in\left\{3;1;9;-5\right\}\left(TMDK\right)\)

a) \(5\dfrac{1}{3}:\left(\dfrac{5}{4}-x\right)=0,8\\ \Rightarrow\dfrac{5}{4}-x=5\dfrac{1}{3}:0,8\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{16}{3}:\dfrac{4}{5}\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{16}{3}\times\dfrac{5}{4}\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{20}{3}\\ \Rightarrow x=\dfrac{5}{4}-\dfrac{20}{3}\\ \Rightarrow x=-\dfrac{65}{12}\)

b) \(\dfrac{3}{10}x-2\dfrac{1}{3}=\dfrac{-28}{5}:\dfrac{2}{15}\\ \Rightarrow\dfrac{3}{10}x-\dfrac{7}{3}=\dfrac{-28}{5}\times\dfrac{15}{2}\\ \Rightarrow\dfrac{3}{10}x-\dfrac{7}{3}=-42\\ \Rightarrow\dfrac{3}{10}x=-42+\dfrac{7}{3}\\ \Rightarrow\dfrac{3}{10}x=\dfrac{-119}{3}\\ \Rightarrow x=\dfrac{-119}{3}:\dfrac{3}{10}\\ \Rightarrow x=-\dfrac{1190}{9}\)

a) 

\(5\dfrac{1}{3}:\left(\dfrac{5}{4}-x\right)=0,8\\ \Rightarrow\dfrac{16}{3}:\left(\dfrac{5}{4}-x\right)=\dfrac{4}{5}\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{16}{3}:\dfrac{4}{5}\\ \Rightarrow\dfrac{5}{4}-x=\dfrac{20}{3}\\ \Rightarrow x=\dfrac{5}{4}-\dfrac{20}{3}\\ \Rightarrow x=\dfrac{-65}{12}\)

b) 

\(\dfrac{3}{10}x-2\dfrac{1}{3}=\dfrac{-28}{5}:\dfrac{2}{15}\\ \Rightarrow\dfrac{3}{10}x-\dfrac{7}{3}=\dfrac{-28}{5}\cdot\dfrac{15}{2}\\ \Rightarrow\dfrac{3}{10}x-\dfrac{7}{3}=-42\\ \Rightarrow\dfrac{3}{10}x=-42+\dfrac{7}{3}\\ \Rightarrow\dfrac{3}{10}x=-\dfrac{119}{3}\\ \Rightarrow x=\dfrac{-119}{3}:\dfrac{3}{10}\\ =-\dfrac{1190}{9}\)

AA + BB = CBC

11 x (A + B) = CBC

⇒ B = C x 2

Ta xét các trường hợp:

CBC = 121₁ = 242₂ = 363₃ = 484₄

- Nếu CBC = 121 thì B = 2;  A = 9 và C = 1 (chọn)

Với các trường hợp khác, do đều được tạo bởi các số quá lớn dẫn đến A là số có 2 chữ số nên chỉ có 1 trường hợp nêu trên.

\(\dfrac{-9}{11}< \dfrac{7}{a}< \dfrac{-9}{13}\\ \Rightarrow\dfrac{63}{-77}< \dfrac{63}{9a}< \dfrac{63}{-91}\\ \Rightarrow-77>9a>-91\)

Với \(a\inℤ\Rightarrow9a⋮9\)

Do đó \(9a\in\left\{-81;-90\right\}\)

\(\Rightarrow a\in\left\{-9;-10\right\}\)

Vậy số hữu tỉ thỏa mãn là: \(\dfrac{7}{-9};\dfrac{7}{-10}\)

\(D=1-\dfrac{2}{5\cdot10}-\dfrac{2}{10\cdot15}-\dfrac{2}{15\cdot20}-...-\dfrac{2}{2020\cdot2025}\)

\(D=1-\left(\dfrac{2}{5\cdot10}+\dfrac{2}{10\cdot15}+\dfrac{2}{15\cdot20}+...+\dfrac{2}{2020\cdot2025}\right)\)

Đặt \(A=\dfrac{2}{5\cdot10}+\dfrac{2}{10\cdot15}+\dfrac{2}{15\cdot20}+...+\dfrac{2}{2020\cdot2025}\)

\(A=\dfrac{2}{5}\cdot\left(\dfrac{1}{5}-\dfrac{1}{10}\right)+\dfrac{2}{5}\cdot\left(\dfrac{1}{10}-\dfrac{1}{15}\right)+\dfrac{2}{5}\cdot\left(\dfrac{1}{15}-\dfrac{1}{20}\right)+...+\dfrac{2}{5}\cdot\left(\dfrac{1}{2020}-\dfrac{1}{2025}\right)\)

\(A=\dfrac{2}{5}\cdot\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{20}+...+\dfrac{1}{2020}-\dfrac{1}{2025}\right)\)

\(A=\dfrac{2}{5}\cdot\left(\dfrac{1}{5}-\dfrac{1}{2025}\right)\)

\(A=\dfrac{2}{5}\cdot\dfrac{404}{2025}\)

\(A=\dfrac{808}{10125}\)

Thay vào D được:

\(D=1-\dfrac{808}{10125}\)

\(D=\dfrac{9317}{10125}\)

Vậy \(D=\dfrac{9317}{10125}\)

Gọi phân số cần tìm có dạng là \(\dfrac{x}{15}\)  \(\left(x\inℤ\right)\)

Theo đề bài ta có:

\(-\dfrac{3}{5}< \dfrac{x}{15}< -\dfrac{1}{6}\)

\(\Rightarrow-\dfrac{18}{30}< \dfrac{2x}{30}< -\dfrac{5}{30}\)

\(\Rightarrow-18< 2x< -5\)

\(\Rightarrow-9< x< -\dfrac{5}{2}\)

\(\Rightarrow x\in\left\{-8,-7,-6,-5,-4,-3\right\}\)

Suy ra các phân số cần tìm là \(-\dfrac{8}{15};-\dfrac{7}{15};-\dfrac{6}{15};-\dfrac{5}{15};-\dfrac{4}{15};-\dfrac{3}{15}\)

Vậy có 6 phân số thỏa mãn đề bài

phân số thoả mãn đề bài có dạng: \(\dfrac{x}{15}\); \(x\) \(\in\) Z

Theo bài ra ta có: 

                \(\dfrac{-3}{5}\) <  \(\dfrac{x}{15}\) <  \(\dfrac{-1}{6}\)

        \(\dfrac{-3\times6}{5\times6}\) <  \(\dfrac{x\times2}{15\times2}\) < \(\dfrac{-1\times5}{6\times5}\)

              \(\dfrac{-18}{30}\) <  \(\dfrac{x\times2}{30}\) < \(\dfrac{-5}{30}\) 

   30 x  \(\dfrac{-18}{30}\) <  \(\dfrac{x\times2}{30}\) < \(\dfrac{-5}{30}\) x 30

             - 18 <  \(x\times\) 2 < - 5 

            - 18 < \(x\) \(\times\) 2 < - 5

            -18 : 2 < \(x\) < - 5 : 2

           - 9 <  \(x\) < - 2\(\dfrac{1}{2}\)

     Vì \(x\in\) Z nên \(x\) \(\in\) {- 8; - 7; - 6; -5; - 4; - 3}

Vậy có 6 phân số thoả mãn yêu cầu đề bài. 

h) Mình sửa đề vế phải là -1/27 nhé 

\(\left(2x+1\right)^3=-\dfrac{1}{27}=\left(-\dfrac{1}{3}\right)^3\\ \Rightarrow2x+1=-\dfrac{1}{3}\\ \Rightarrow2x=-\dfrac{1}{3}-1=-\dfrac{4}{3}\\ \Rightarrow x=-\dfrac{4}{3}:2=-\dfrac{2}{3}\)

i) \(\left|2x-\dfrac{1}{3}\right|-\dfrac{2}{5}=0\Rightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{2}{5}\\ \Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{2}{5}\\2x-\dfrac{1}{3}=-\dfrac{2}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{11}{15}\\2x=-\dfrac{1}{15}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{30}\\x=-\dfrac{1}{30}\end{matrix}\right.\)

d) \(\left|2x-1\right|=\left|x-2\right|\Rightarrow\left[{}\begin{matrix}2x-1=x-2\\2x-1=-\left(x-2\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=1-2\\2x-1=-x+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\2x+x=1+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\3x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

e) \(\left(x-\dfrac{2}{3}\right)+1=2x-\dfrac{5}{4}\Rightarrow x-\dfrac{2}{3}+1=2x-\dfrac{5}{4}\\ \Rightarrow2x-x=\dfrac{5}{4}-\dfrac{2}{3}+1\\ \Rightarrow x=\dfrac{19}{12}\)

f) \(\left|2x-1\right|-2=3\Rightarrow\left|2x-1\right|=5\\ \Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Ý bạn là sao nhỉ?

Sao cho số thực a,b,c thỏa mãn a=b=c rồi lại chứng minh a=b=c