tính hợp lí (nếu có thể):\(\left(\dfrac{1}{3}\right)^3.\dfrac{9}{4}-\left(\dfrac{-1}{2}\right)^2\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1, \(\dfrac{-3}{5}:\dfrac{7}{5}-\dfrac{3}{5}:\dfrac{7}{5}+2\dfrac{3}{5}\)
\(=\left(\dfrac{-3}{5}-\dfrac{3}{5}\right):\dfrac{7}{5}+\dfrac{13}{5}\)
\(=\dfrac{-6}{5}:\dfrac{7}{5}+\dfrac{13}{5}\)
\(=\dfrac{-6}{7}+\dfrac{13}{5}\\ =\dfrac{61}{35}\)
2, \(0,75-\left(2\dfrac{1}{3}+0,75\right)+3^2\cdot\left(-\dfrac{1}{9}\right)\)
\(=0,75-\dfrac{37}{12}+9\cdot\left(-\dfrac{1}{9}\right)\)
\(=-\dfrac{7}{3}+\left(-1\right)\\ -\dfrac{10}{3}\)
A= \(\dfrac{11}{9}\).\(\dfrac{-3}{2}\)- \(\dfrac{11}{9}\).\(\dfrac{15}{2}\)+(2021)0
= \(\dfrac{11}{9}\)(\(\dfrac{-3}{2}\)-\(\dfrac{15}{2}\)) + 1
= \(\dfrac{11}{9}\)(-9) + 1
= -11 + 1
= -10
Lần sau bn ko đc phép chuyển \(\left(-2021\right)^0\) sang \(\left(2021\right)^0\) đou nha
\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)
\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)
Giải:
1) (-8/13:3/7+-5/13:3/7).(-4)3.|-3|/7
=[7/3.(-8/13+-5/13)].-192/7
=[7/3.(-1)].-192/7
=-7/3.-192/7
=64
2) 75%-(5/2+5/3)+(-1/2)2
=3/4-25/6+1/4
=(3/4+1/4)-25/6
=1-25/6
=-19/6
Chúc bạn học tốt!
1) \(\left(\dfrac{-8}{13}:\dfrac{3}{7}+\dfrac{-5}{13}:\dfrac{3}{7}\right).\dfrac{\left(-4\right).|-3|}{7}\)
= \(\left[\left(\dfrac{-8}{13}+\dfrac{-5}{13}\right):\dfrac{3}{7}\right].\dfrac{-64.3}{7}\)
= \(\left[-1:\dfrac{3}{7}\right].\dfrac{-192}{7}\)
= \(\dfrac{-7}{3}.\dfrac{-192}{7}\)
= \(64\)
2) \(75\%-\left(\dfrac{5}{2}+\dfrac{5}{3}\right)+\left(-\dfrac{1}{2}\right)^2\)
= \(\dfrac{3}{4}-\dfrac{25}{6}+\dfrac{1}{4}\)
= \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)-\dfrac{25}{6}\)
= \(1-\dfrac{25}{6}\)
= \(\dfrac{-19}{6}\)
Chúc bạn học tốt !
\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)
Lời giải :
a ) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=2,5\)
b ) \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)
\(=\dfrac{3}{7}\left(19-33\right)\)
\(=\dfrac{3}{7}\left(-14\right)\)
\(=-6\)
c ) \(9\left(-\dfrac{1}{3}\right)^3+\dfrac{1}{3}\)
\(=9\left(-\dfrac{1}{27}\right)+\dfrac{1}{3}\)
\(=-\dfrac{1}{3}+\dfrac{1}{3}\)
\(=0\)
d ) \(15\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right)\div\left(-\dfrac{5}{7}\right)\)
\(=-10\left(-\dfrac{7}{5}\right)\)
\(=14\)
(\(\dfrac{1}{3}\))3.\(\dfrac{9}{4}\) - (\(-\dfrac{1}{2}\))2
= \(\dfrac{1}{3}\).(\(\dfrac{1}{2}\))2 - (\(\dfrac{1}{2}\))2
= (\(\dfrac{1}{2}\))2.(\(\dfrac{1}{3}\) - 1)
= \(\dfrac{1}{4}\).\(\dfrac{-2}{3}\)
= - \(\dfrac{1}{6}\)
- 1/6