Bài 1: Tính ( hợp lý nếu có thể )

\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)

\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)

\(=-1+1+\dfrac{2}{-5}\)

\(=0+\dfrac{2}{-5}\)

\(=\dfrac{2}{-5}\)

\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)

\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)

\(=0+\dfrac{2}{3}\)

\(=\dfrac{2}{3}\)

\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)

\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)

\(=-1+1\)

\(=0\)

\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)

\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)

\(=\dfrac{1}{6}+\dfrac{-1}{6}\)

\(=0\)

Bài 2: Tìm x,biết:

a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)

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

\(x=\dfrac{2}{15}\)

Vậy \(x=\dfrac{2}{15}\)

b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)

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

\(x=\dfrac{1}{3}+\dfrac{2}{3}\)

\(x=\dfrac{3}{3}=1\)

Vậy \(x=1\)

c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!

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

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

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

Vậy \(x=\dfrac{1}{44}\)

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

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

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

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

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

Vậy \(x=-\dfrac{3}{20}\)

a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)

b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)

c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)

d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)

lưu ý mk ko chép đầu bài

mình cần gấp lắm đến chiều mai là phải nộp rùi

giúp mình nha thanks cá bạn trước ko có tâm trạng mà cười nữa

phamthiminhtrang

\(a,x=\frac{3}{6}-\frac{8}{16}\)

\(\Rightarrow x=0\)

\(b,\frac{12}{16}:x=\frac{32}{64}\)

\(x=\frac{12}{16}:\frac{32}{64}\)

\(x=\frac{12}{16}\cdot\frac{64}{32}\)

\(x=\frac{3}{8}\)

\(a,\)\(x\)\(=\frac{3}{6}-\frac{8}{16}=\frac{1}{2}-\frac{1}{2}=0\)

\(b,\)\(\frac{12}{16}\)\(:\)\(x\)\(=\frac{32}{64}\)

\(=>\)        \(x\)\(=\)\(\frac{12}{16}:\frac{32}{64}\)

                        \(x\) \(=\)\(\frac{12}{16}.\frac{64}{32}\)

                        \(x\)\(=\)\(\frac{3}{4}.2\)

                        \(x\)\(=\)\(\frac{6}{4}=\frac{3}{2}\)

\(2-\left|\frac{3}{4}-x\right|=\frac{7}{12}\)

\(\left|\frac{3}{4}-x\right|=\frac{17}{12}\)

\(\Rightarrow\frac{3}{4}-x=\orbr{\begin{cases}\frac{17}{12}\\-\frac{17}{12}\end{cases}\Rightarrow x=\orbr{\begin{cases}-\frac{2}{3}\\\frac{13}{6}\end{cases}}}\)

\(\frac{1}{2}x+\frac{1}{8}x=\frac{3}{4}\)

\(x\left(\frac{1}{2}+\frac{1}{8}\right)=\frac{3}{4}\)

\(x.\frac{5}{8}=\frac{3}{4}\)

\(\Rightarrow x=\frac{3}{4}:\frac{5}{8}\)

\(\Rightarrow x=\frac{6}{5}\)

Bài giải:

a) Ta có: \(-8⋮x\) và \(12⋮x\Rightarrow x\inƯC\left(-8;12\right)\)

\(Ư\left(-8\right)=\left\{-1;-2;-4;-8;1;2;4;8\right\}\)

\(Ư\left(12\right)=\left\{-1;-2;-3;-4;-6;-12;1;2;3;4;6;12\right\}\)

\(\RightarrowƯC\left(-8;12\right)=\left\{-1;-2;-4;1;2;4\right\}\)

Vậy: \(x\in\left\{-1;-2;-4;1;2;4\right\}\)

b)

toán này ko phải toán lớp 6 rùi

dễ quá 

a,8/3x + 26/3= 10/3

8/3x = 10/3- 26/3 = -16/3

=>x = -16/3 : 8/3 = -2

b, (2/3-1/2)x = 5/12

1/6x = 5/12

=>x = 5/2

xong rùi đó 

nhớ tk nha