75% + 1,1 : \(\left(\dfrac{2}{5}-1\dfrac{1}{2}\right)-\left(\dfrac{1}{3}\right)^2\)
\(75\%-\left(\dfrac{5}{2}+\dfrac{5}{3}\right)+\left(-\dfrac{1}{2}\right)^3\)
\(4\dfrac{1}{3}.\dfrac{1}{6}-\dfrac{1}{2}.4\dfrac{1}{3}+\left|\dfrac{-13}{3}.\dfrac{7}{15}\right|\)
= 3/4 - 25/6 - 1/8 = -85/24
= 13/18 - 13/6 + 91/45 = 26/45
Tính:
\(a,\left(\dfrac{5}{12}-\dfrac{3}{4}\right).\dfrac{-6}{5}+\dfrac{-7}{6}\)
\(b,\left(5\dfrac{1}{2}-5\right)^2+\dfrac{-1}{6}:\dfrac{5}{3}\)
\(c,75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)
Thực hiện phép tính:
\(75\%-\left(\dfrac{5}{2}+\dfrac{5}{3}\right)+\left(\dfrac{-1}{2}\right)^2\)
\(75\%-\left(\dfrac{5}{2}+\dfrac{5}{3}\right)+\left(\dfrac{-1}{2}\right)^2\)
\(=\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{5}{3}+\dfrac{1}{4}\)
\(=\dfrac{9}{12}-\dfrac{30}{12}-\dfrac{20}{12}+\dfrac{3}{12}\)
\(=\dfrac{-38}{12}=\dfrac{-19}{6}\)
\(\dfrac{1}{5}\).\(\left(x+\dfrac{1}{5}\right)\)\(+\)\(\dfrac{2}{5}\)\(\left(x+\dfrac{5}{3}\right)\)\(=\)\(\dfrac{98}{75}\)
help me please :D
\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ =>\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ =>\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{2}{3}-\dfrac{1}{25}=\dfrac{3}{5}\\ =>x=1\)
\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ \Rightarrow\left(\dfrac{1}{5}x+\dfrac{2}{5}x\right)+\left(\dfrac{1}{25}+\dfrac{2}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x+\dfrac{53}{75}=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{53}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{45}{75}=\dfrac{3}{5}\\ \Rightarrow x=\dfrac{3}{5}:\dfrac{3}{5}\\ \Rightarrow x=1\)
\(\dfrac{1}{5}\)x(X+\(\dfrac{1}{5}\))+\(\dfrac{2}{5}\)x(X+\(\dfrac{5}{3}\)) = \(\dfrac{98}{75}\)
=> \(\dfrac{1}{5}\)X+\(\dfrac{2}{5}\)+\(\dfrac{6}{15}\)X+\(\dfrac{31}{15}\) = \(\dfrac{98}{75}\)
=> (\(\dfrac{1}{5}\)X+\(\dfrac{6}{15}\)X)+(\(\dfrac{2}{5}\)+\(\dfrac{31}{15}\)) =\(\dfrac{98}{75}\)
=> X x(\(\dfrac{1}{5}\)+\(\dfrac{6}{15}\))+(\(\dfrac{6}{15}\)+\(\dfrac{31}{15}\)) =\(\dfrac{98}{75}\)
=> X x(\(\dfrac{3}{15}\)+\(\dfrac{6}{15}\))+\(\dfrac{37}{15}\) = \(\dfrac{98}{75}\)
=>X x\(\dfrac{9}{15}\)+\(\dfrac{37}{15}\) =\(\dfrac{98}{75}\)
=>X x\(\dfrac{9}{15}\) =\(\dfrac{98}{75}\)-\(\dfrac{37}{15}\)
=>X =\(\dfrac{-29}{25}\):\(\dfrac{9}{15}\)
=>X =\(\dfrac{-29}{15}\)
\(\left(0,25+\dfrac{3}{5}-75\%+1\dfrac{2}{5}\right): \left(-2\dfrac{1}{2}\right)\)
= \(\left(\dfrac{25}{100}+\dfrac{3}{5}-\dfrac{75}{100}+\dfrac{7}{5}\right):\dfrac{-5}{2}\)
= \(\left(\dfrac{25}{100}+\dfrac{3}{5}_{ }+\dfrac{-75}{100}+\dfrac{7}{5}\right).\dfrac{-2}{5}\)
= \(\left(\dfrac{1}{4}+\dfrac{3}{5}+\dfrac{-3}{4}+\dfrac{7}{5}\right).\dfrac{-2}{5}\)
= \(\left[\left(\dfrac{1}{4}+\dfrac{-3}{4}\right)+\left(\dfrac{3}{5}+\dfrac{7}{5}\right)\right].\dfrac{-2}{5}\)
= \(\left(\dfrac{-2}{4}+\dfrac{10}{5}\right).\dfrac{-2}{5}\)
= \(\left(\dfrac{-10}{20}+\dfrac{40}{20}\right).\dfrac{-2}{5}\)
= \(\dfrac{30}{20}.\dfrac{2}{5}\)
= \(\dfrac{30.2}{20.5}\)
= \(\dfrac{6.1}{10.1}\)
= \(\dfrac{6}{10}=\dfrac{3}{5}\)
= (25100+35−75100+75):−52(25100+35−75100+75):−52
= (25100+35+−75100+75).−25(25100+35+−75100+75).−25
= (14+35+−34+75).−25(14+35+−34+75).−25
= [(14+−34)+(35+75)].−25[(14+−34)+(35+75)].−25
= (−24+105).−25(−24+105).−25
= (−1020+4020).−25(−1020+4020).−25
= 3020.253020.25
= 30.220.530.220.5
= 6.110.16.110.1
= 610=35
- kết quả là \(\dfrac{-3}{5}\)nha tại mình quên dấu (-)
Tìm x, biết:
a) x+\(\dfrac{1}{6}\)=\(\dfrac{-3}{8}\) b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
c) \(\dfrac{1}{2}x\)+\(\dfrac{1}{8}x=\dfrac{3}{4}\) d) 75%-\(1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)
\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)
\(\Leftrightarrow x=-\dfrac{13}{24}\)
\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{6}{5}\)
\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)
\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)
\(=\dfrac{1}{5}\)
a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)
\(x=\dfrac{-3}{8}-\dfrac{1}{6}\)
\(x=\dfrac{-13}{24}\)
vậy x =....
b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=\dfrac{17}{12}\)
\(x=\dfrac{3}{4}-\dfrac{17}{12}\)
\(x=\dfrac{-2}{3}\)
vậy x =....
Cho mk hỏi :
tìm x,biết
a,\(-3\dfrac{1}{2}\)x -0,75-1,25x=\(\left(\dfrac{-1}{2}\right)^2:\dfrac{-3}{4}+\dfrac{1}{6}\)
b, \(\dfrac{-2}{3}-\left(\dfrac{x}{2}-75\%\right)=\left(\dfrac{3}{-4}-\dfrac{9}{8}\right)^2:\dfrac{-3}{32}-1\dfrac{1}{3}\)
\(\text{Thực hiện phép tính một cách hợp lý nhất (nếu có thể):}\)
\(1\)) \(\left(\dfrac{-8}{13}:\dfrac{3}{7}+\dfrac{-5}{13}:\dfrac{3}{7}\right).\dfrac{\left(-4\right)^3.\left|-3\right|}{7}\)
\(2\)) \(75\)\(\text{%}\)\(-\left(\dfrac{5}{2}+\dfrac{5}{3}\right)+\left(-\dfrac{1}{2}\right)^2\)
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 !
tính
a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)
b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)
c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)
e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)
h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{5}{8}\)
\(=\dfrac{3}{8}+\dfrac{5}{8}\)
\(=1\)