Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(b,=-\dfrac{40}{30}-\dfrac{12}{30}-\dfrac{45}{30}=-\dfrac{97}{30}\\ c,=\left(\dfrac{4}{5}+\dfrac{7}{10}\right)+\dfrac{2}{7}=\dfrac{3}{2}+\dfrac{2}{7}=\dfrac{25}{14}\\ d,=\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{1}{2}+\dfrac{3}{8}\\ =\left(\dfrac{2}{3}+\dfrac{1}{2}\right)+\left(\dfrac{7}{4}+\dfrac{3}{8}\right)=\dfrac{7}{6}+\dfrac{17}{8}=\dfrac{79}{24}\)
c: \(\dfrac{4}{5}-\dfrac{-2}{7}-\dfrac{-7}{10}\)
\(=\dfrac{56}{70}+\dfrac{20}{70}+\dfrac{49}{70}\)
\(=\dfrac{125}{70}=\dfrac{25}{14}\)
\(\left(-\dfrac{1}{3}\right)\cdot\dfrac{1}{3}+\left(\dfrac{15}{2}\right)^7:\left(\dfrac{15}{2}\right)^5-\left(\left(-2\right)^2\right)^3\)
\(=\dfrac{-1}{9}+\dfrac{225}{4}-64\)
\(=-\dfrac{283}{36}\)
a: =>|1/3x|=3:2,7=10/9
=>1/3=10/9 hoặc 1/3x=-10/9
=>x=10/3 hoặc x=-10/3
b: =>2|2x-1|=19-7=12
=>|2x-1|=6
=>2x-1=6 hoặc 2x-1=-6
=>2x=7 hoặc 2x=-5
=>x=7/2 hoặc x=-5/2
c: |x|>2
=>x>2 hoặc x<-2
\(a,=\dfrac{13}{50}\cdot\dfrac{50}{13}\cdot\left(-\dfrac{31}{2}\right)\cdot\dfrac{169}{2}=-\dfrac{5239}{2}\\ b,=\dfrac{-\dfrac{49}{100}\cdot\left(-125\right)}{-\dfrac{343}{27}\cdot\dfrac{81}{16}\cdot\left(-1\right)}=\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{245}{4}\cdot\dfrac{16}{1029}=\dfrac{20}{21}\)
a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}=\dfrac{13}{50}.-75:\dfrac{13}{50}.\dfrac{169}{2}=-\dfrac{75.169}{2}=-\dfrac{12675}{2}\)
b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}=\dfrac{0,49.\left(-125\right)}{-\dfrac{343}{27}.\dfrac{81}{16}.\left(-1\right)}=-\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{20}{21}\)
\(a.\)
\(\left[6.\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)
\(=\left[6.\dfrac{1}{9}+1+1\right]:\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right):\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right).\left(-\dfrac{3}{4}\right)\)
\(=-2\)
\(b.\)
\(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}\)
\(=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}\)
\(=\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}\)
\(=\dfrac{72}{5}\)
\(\dfrac{\left(\dfrac{2}{3}\right)^3\cdot\left(-\dfrac{3}{4}\right)^2\cdot\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2\cdot\left(-\dfrac{5}{12}\right)^3}\)\(=\dfrac{\dfrac{2^3\cdot\left(-3\right)^2\cdot\left(-1\right)}{3^3\cdot4^2}}{\dfrac{2^2\cdot\left(-5\right)^3}{5^2\cdot12^3}}\)
\(=\dfrac{\dfrac{2^3\cdot3^2\cdot\left(-1\right)}{3^3\cdot\left(2^2\right)^2}}{\dfrac{2^2\cdot\left(-5\right)^3}{5^2\cdot\left(2^2\cdot3\right)^3}}=\dfrac{\dfrac{2^3\cdot3^2\cdot\left(-1\right)}{3^3\cdot2^4}}{\dfrac{2^2\cdot5^2\cdot\left(-5\right)}{5^2\cdot2^6\cdot3^3}}=\dfrac{\dfrac{-1}{3\cdot2}}{\dfrac{-5}{2^4\cdot3^3}}\)
\(=\dfrac{-\dfrac{1}{6}}{27\cdot16}=-\dfrac{1}{6}:432=-\dfrac{1}{6}\cdot\dfrac{1}{432}=-\dfrac{1}{2592}\)
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^2}\)
\(=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\left(\dfrac{2}{5}.-\dfrac{5}{12}\right)^2}\)
\(=\dfrac{\dfrac{1}{3}.\dfrac{1}{2}.\left(-1\right)}{\left(-\dfrac{1}{6}\right)^2}=\dfrac{-\dfrac{1}{6}}{\dfrac{1}{36}}=-6\)
\(\dfrac{2^3.5.7(5^2.7^3)}{(2.5.7^2)^2}\)
= \(\dfrac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
= 2.5
= 10