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dễ mà bạn đây là bài cơ bản lớp 6 dấy
câu a nhé bạn bạn nếu ko làm kiểu khó thì đổi về phaan số bình thường nà sau đó tính trong ngoặc trước rồi tính xoong bỏ dấu ngoặc nhưng ko đổi dấu né thế lad đc tương tự như các câu dưới
a)\(8\frac{2}{3}:2\frac{1}{6}-2\frac{27}{51}=\frac{26}{3}.\frac{6}{13}-\frac{43}{17}=4-\frac{43}{17}=\frac{25}{17}\)
b)\(\frac{27}{20}.\frac{15}{4}+\frac{19}{8}=\frac{119}{16}\)
c)\(\left(\frac{1}{12}+\frac{5}{6}\right)+\left(\frac{13}{35}+\frac{23}{35}\right)=\frac{11}{12}+\frac{36}{35}=\frac{817}{420}\)
d)\(\frac{24}{37}.\left(\frac{13}{18}+\frac{2}{9}+\frac{1}{18}\right)=\frac{24}{37}.1=\frac{24}{37}\)
2) Tinh nhanh:
a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)
= \(\dfrac{5}{598}\)
b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)
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 , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)
=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)
\(A=\frac{7}{12}+\frac{5}{12}:6-\frac{1}{36}\)
\(=\frac{7}{12}+\frac{5}{72}-\frac{11}{36}\)
\(=\frac{42}{72}+\frac{5}{72}-\frac{22}{72}\)
\(=\frac{25}{72}\)
Trả lời
\(A=\frac{7}{12}+\frac{5}{12}\div6-\frac{11}{36}\)
\(A=\frac{7}{12}+\frac{5}{12}\times\frac{1}{6}-\frac{11}{36}\)
\(A=\frac{7}{12}+\frac{5}{72}-\frac{11}{36}\)
\(A=\frac{25}{72}\)
\(B=12\frac{1}{3}-\frac{5}{7}\div\left(24-23\frac{5}{7}\right)\)
\(B=\frac{37}{3}-\frac{5}{7}\div\left(24-\frac{166}{7}\right)\)
\(B=\frac{37}{3}-\frac{5}{7}\div\frac{2}{7}\)
\(B=\frac{37}{3}-\frac{5}{7}\times\frac{7}{2}\)
\(B=\frac{37}{3}-\frac{5}{2}\)
\(B=\frac{59}{6}\)
Bài làm:
Ta có:
\(A=\frac{7}{12}+\frac{5}{12}\div6-\frac{11}{36}\)
\(A=\frac{7}{12}+\frac{5}{72}-\frac{11}{36}\)
\(A=\frac{42+5-22}{72}=\frac{25}{72}\)
và
\(B=12\frac{1}{3}-\frac{5}{7}\div\left(24-23\frac{5}{7}\right)\)
\(B=\frac{37}{3}-\frac{5}{7}\div\left(24-23-\frac{5}{7}\right)\)
\(B=\frac{37}{3}-\frac{5}{7}\div\left(1-\frac{5}{7}\right)\)
\(B=\frac{37}{3}-\frac{5}{7}+1\)
\(B=\frac{265}{21}\)
1 a, Ta có: \(36^5\): \(18^5\)= \(\left(36:18\right)^5\)= \(2^5\)= \(32\)
\(1b.\)\(24\)\(5^3\)+ \(5^2\). \(5^3\)= \(5^2\). \(\left(5^3+24\right)\)= \(25.149\)= \(3725\)
@@ dùng máy tính mà tính
Anh làm mẫu 1 phần
\(\frac{\frac{2}{2017}+\frac{2}{2018}}{\frac{5}{2017}+\frac{5}{2018}}=\frac{2.\left(\frac{1}{2017}+\frac{1}{2018}\right)}{5.\left(\frac{1}{2017}+\frac{1}{2018}\right)}=\frac{2}{5}\)