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a) \(\dfrac{3}{4}=\dfrac{15}{20}\)
b) \(\dfrac{4}{5}=\dfrac{12}{15}\)
c) \(\dfrac{-4}{9}=\dfrac{-16}{36}\)
d) \(\dfrac{7}{-13}=\dfrac{21}{-39}\)
a ) \(\dfrac{3}{4}\)= \(\dfrac{15}{20}\)
b )\(\dfrac{4}{5}\)= \(\dfrac{12}{15}\)
c) \(\dfrac{-4}{9}\)=\(\dfrac{-16}{36}\)
d) \(\dfrac{7}{-13}\)=\(\dfrac{21}{-39}\)
a)\(\dfrac{1}{4}=\dfrac{2}{8}=\dfrac{4}{16}=\dfrac{8}{32}=......\)
b)\(\dfrac{-3}{4}=\dfrac{-6}{8}=\dfrac{-9}{12}=\dfrac{-12}{16}.......\)
\(1=\dfrac{2}{2}=\dfrac{-4}{-4}\)\(=\dfrac{6}{6}=\dfrac{-8}{-8}=\dfrac{10}{10}\)
a) \(\dfrac{1}{6};\dfrac{1}{3};\dfrac{1}{2};...\)
\(\Rightarrow\dfrac{1}{6};\dfrac{2}{6};\dfrac{3}{6};...\)
Dãy có quy luật tăng dần lên 1 đơn vị ở tử số
\(\Rightarrow\) Số tiếp theo của dãy là: \(\dfrac{4}{6}\)
b) \(\dfrac{1}{8};\dfrac{5}{24};\dfrac{7}{24};...\)
\(\Rightarrow\dfrac{3}{24};\dfrac{5}{24};\dfrac{7}{24};...\)
Dãy có quy luật tăng dần lên 2 đơn vị ở tử số
\(\Rightarrow\) Số tiếp theo của dãy là: \(\dfrac{9}{24}\)
c) \(\dfrac{1}{5};\dfrac{1}{4};\dfrac{1}{3};...\)
\(\dfrac{4}{20};\dfrac{5}{20};\dfrac{6}{20};...\)
Dãy có quy luật tăng dần lên 1 đơn vị ở tử số
\(\Rightarrow\) Số tiếp theo của dãy là: \(\dfrac{7}{20}\)
d) \(\dfrac{4}{15};\dfrac{3}{10};\dfrac{1}{3};...\)
\(\Rightarrow\dfrac{8}{30};\dfrac{9}{30};\dfrac{11}{30};...\)
Dãy có quy luật tăng dần lên 1 đơn vị ở tử số
\(\Rightarrow\) Số tiếp theo của dãy là: \(\dfrac{12}{30}\)
a) \(\dfrac{37}{40}-0,64\\ =\dfrac{37}{40}-\dfrac{16}{25}\\ =\dfrac{185}{200}-\dfrac{128}{200}\\ =\dfrac{57}{200}\)
b) \(130\dfrac{25}{28}-120\dfrac{12}{35}\\ =\dfrac{3665}{28}-\dfrac{4212}{35}\\ =\dfrac{18325}{140}-\dfrac{16848}{140}\\ =\dfrac{211}{20}\)
a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)
=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)
=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)
=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)
b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)
=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)
c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)
=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)
= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)
d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)
=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)
=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)
e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)
=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)
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. \(\dfrac{6}{12}\)
b.\(\dfrac{-5}{-7}\)
c.\(\dfrac{-7}{8}\)
d.\(\dfrac{3}{-6}\)
a)6/12
b)-5/-7
c)-7/8
d)3/-6