Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
\(a)\dfrac{-5}{21}-\dfrac{1}{3}+3\dfrac{1}{2}.\left(\dfrac{-2}{3}\right)^3\)
\(=\dfrac{-5}{21}+\dfrac{-7}{21}+\dfrac{7}{2}.\dfrac{-8}{27}\)
\(=-\dfrac{4}{7}+\dfrac{-28}{27}\)
\(=\dfrac{-108}{189}+\dfrac{-196}{189}\)
\(=-\dfrac{304}{189}\)
\(b)-2\dfrac{1}{3}+\left(\dfrac{3}{8}-\dfrac{3}{4}\right)^3:\dfrac{5}{9}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\left(\dfrac{3}{8}-\dfrac{6}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\left(-\dfrac{3}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\dfrac{-27}{512}.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\dfrac{-243}{2560}-\dfrac{1}{2}\)
\(=\dfrac{-17920}{7680}+\dfrac{-729}{7680}+\dfrac{-3840}{7680}\)
\(=\dfrac{-22489}{7680}\)
Bạn tính hai vế à.!? Hay tính vế thứ nhất rồi với vế thứ 2.!???
a. \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)+\left(\dfrac{-5}{41}-\dfrac{36}{41}\right)+0,5\)
\(=1+\left(-1\right)+0,5\)
\(=0,5\)
b. \(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=-12:\left(\dfrac{-1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)
\(=-1728\)
c. \(\dfrac{7}{23}.\left[\left(-\dfrac{8}{6}\right)-\dfrac{45}{18}\right]\)
\(=\dfrac{7}{23}.\dfrac{-23}{6}\)
\(=\dfrac{-7}{6}\)
d. \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}.\dfrac{7}{5}\)
\(=\left(23\dfrac{1}{4}-13\dfrac{1}{4}\right).\dfrac{7}{5}\)
\(=10.\dfrac{7}{5}\)
\(=14\)
e. \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2\)
\(=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)
1, \(x\left(x+\dfrac{2}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)
2, a, \(\left|x+\dfrac{4}{6}\right|\ge0\)
Để \(\left|x+\dfrac{4}{6}\right|\) đạt GTNN thì \(\left|x+\dfrac{4}{6}\right|=0\)
\(\Leftrightarrow x+\dfrac{4}{6}=0\Rightarrow x=\dfrac{-2}{3}\)
Vậy, ...
b, \(\left|x-\dfrac{1}{3}\right|\ge0\)
Để \(\left|x-\dfrac{1}{3}\right|\) đạt GTLN thì \(\left|x-\dfrac{1}{3}\right|=0\)
\(\Leftrightarrow x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)
Vậy, ...
1)
a)
\(x\cdot\left(x+\dfrac{2}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2)
a)
\(\left|x+\dfrac{4}{6}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x+\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{-4}{6}\Leftrightarrow x=\dfrac{-2}{3}\)
Vậy \(Min_{\left|x+\dfrac{4}{6}\right|}=0\text{ khi }x=\dfrac{-2}{3}\)
b)
\(\left|x-\dfrac{1}{3}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(Min_{\left|x-\dfrac{1}{3}\right|}=0\text{ khi }x=\dfrac{1}{3}\)
a: \(=5-2\cdot\dfrac{1}{4}=5-\dfrac{1}{2}=\dfrac{9}{2}\)
b: \(=\left(\dfrac{7}{2}\right)^3+\dfrac{1}{2}=\dfrac{343}{8}+\dfrac{1}{2}=\dfrac{347}{8}\)
c: \(=\left(5+\dfrac{5}{27}-\dfrac{5}{27}\right)+\left(\dfrac{7}{23}+\dfrac{16}{23}\right)-\dfrac{1}{2}=5+1-\dfrac{1}{2}=5+\dfrac{1}{2}=5.5\)
e: \(=\dfrac{-5}{4}\left(35+\dfrac{1}{6}-45-\dfrac{1}{6}\right)=\dfrac{-5}{4}\cdot\left(-10\right)=\dfrac{50}{4}=\dfrac{25}{2}\)
Bài 1:
a) Ta có:
\(-3\dfrac{1}{5}=-2,8\)
\(\dfrac{37}{10}=3,7\)
Vì \(-3,25< -2,8< 3,7\)
Nên \(-3,25< -3\dfrac{1}{5}< \dfrac{37}{10}\)
b) Ta có:
\(\dfrac{-567}{568}>-1\)
\(\dfrac{-568}{567}< -1\)
\(\Leftrightarrow\dfrac{-567}{568}>-1>\dfrac{-568}{567}\)
\(\Leftrightarrow\dfrac{-567}{568}>\dfrac{-568}{567}\)
c) \(-0,625=-\dfrac{5}{8}\)
Vì \(8>7\)
\(\Leftrightarrow\dfrac{1}{8}< \dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{-5}{8}>\dfrac{-5}{7}\)
\(\Leftrightarrow-0,625>\dfrac{-5}{7}\)
Vậy ...
Giải:
a) \(3,6-\dfrac{-5}{6}.\left(-2,4\right).\dfrac{3}{5}\)
\(=\dfrac{18}{5}-\dfrac{-5}{6}.\left(-\dfrac{12}{5}\right).\dfrac{3}{5}\)
\(=\dfrac{18}{5}-\dfrac{6}{5}\)
\(=\dfrac{12}{5}\)
Vậy ...
b) \(\dfrac{1}{4}-0,5-6\dfrac{1}{2}+\dfrac{5}{8}\)
\(=\dfrac{1}{4}-\dfrac{1}{2}-6\dfrac{1}{2}+\dfrac{5}{8}\)
\(=-\dfrac{49}{8}\)
Vậy ...
c) \(1,1-\left(-1,2\right)-\left|-1,3\right|-2\dfrac{3}{4}\)
\(=1,1+1,2-1,3-2,75\)
\(=-\dfrac{7}{4}=-1,75\)
Vậy ...