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Bài 1:
a) \(-\frac{4}{5}-\frac{8}{25}\left(\frac{-5}{2}-0,125\right)\\ =-\frac{4}{5}-\frac{8}{25}\left(\frac{-5}{2}-\frac{1}{8}\right)\\ =-\frac{4}{5}-\frac{8}{25}\left(\frac{-20}{8}-\frac{1}{8}\right)\\ =-\frac{4}{5}-\frac{8}{25}\cdot\frac{-21}{8}\\ =-\frac{4}{5}-\frac{-21}{25}\\ =\frac{-4}{5}+\frac{21}{25}\\ =\frac{-20}{25}+\frac{21}{25}=\frac{1}{25}\)
c) \(5\frac{1}{2}-4\frac{2}{3}:\frac{16}{9}-3\frac{1}{3}:\frac{16}{9}\\ =5\frac{1}{2}-\left(4\frac{2}{3}:\frac{16}{9}+3\frac{1}{3}:\frac{16}{9}\right)\\ =5\frac{1}{2}-\left(4\frac{2}{3}+3\frac{1}{3}\right):\frac{16}{9}\\ =5\frac{1}{2}-8\cdot\frac{9}{16}\\ =\frac{11}{2}-\frac{9}{2}=\frac{2}{2}=1\)
Bài 2:
a) \(\left(20\%x+\frac{2}{5}x-2\right):\frac{1}{3}=-2013\\ \left(\frac{1}{5}x+\frac{2}{5}x-2\right)\cdot3=-2013\\ \left[x\left(\frac{1}{5}+\frac{2}{5}\right)-2\right]=\left(-2013\right):3\\ x\cdot\frac{3}{5}-2=-671\\ x\cdot\frac{3}{5}=-671+2\\ x\cdot\frac{3}{5}=-669\\ x=\left(-669\right):\frac{3}{5}\\ x=\left(-669\right)\cdot\frac{5}{3}\\ x=-1115\)Vậy x = -1115
b) \(\left(4,5-2\left|x\right|\right)\cdot1\frac{4}{7}=\frac{11}{14}\\ \left(\frac{9}{2}-2\left|x\right|\right)\cdot\frac{11}{7}=\frac{11}{14}\\ \frac{9}{2}-2\left|x\right|=\frac{11}{14}:\frac{11}{7}\\ \frac{9}{2}-2\left|x\right|=\frac{11}{14}\cdot\frac{7}{11}\\ \frac{9}{2}-2\left|x\right|=\frac{1}{2}\\ 2\left|x\right|=\frac{9}{2}-\frac{1}{2}\\ 2\left|x\right|=4\\ \left|x\right|=4:2\\ \left|x\right|=2\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)Vậy x ∈ {2 ; -2}
a) \(\frac{2}{3}+\frac{1}{3}\cdot\left(-\frac{2}{5}\right)\\ =\frac{2}{3}+\frac{-2}{15}\\ =\frac{10}{15}+\frac{-2}{15}\\ =\frac{8}{15}\)
b) \(0,75\cdot1\frac{7}{9}-1\frac{2}{5}:\frac{-21}{20}\\ =\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}\cdot\frac{-20}{21}\\ =\frac{4}{3}-\frac{-4}{3}\\ =\frac{4}{3}+\frac{4}{3}\\ =\frac{4}{3}\cdot2\\ =\frac{8}{3}\)
c) \(\frac{-2}{17}+\frac{15}{23}+\frac{15}{-17}-\frac{-4}{19}+\frac{8}{23}\\ =\frac{-2}{17}+\frac{15}{23}+\frac{-15}{17}+\frac{4}{19}+\frac{8}{23}\\ =\left(\frac{-2}{17}+\frac{-15}{17}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\\ =\left(-1\right)+1+\frac{4}{19}\\ =0+\frac{4}{19}\\ =\frac{4}{19}\)
d) \(2019^0\cdot\left(6-2\frac{4}{5}\right)\cdot3\frac{1}{8}-1\frac{3}{5}:25\%\\ =1\cdot\left(\frac{30}{5}-\frac{14}{5}\right)\cdot\frac{25}{8}-\frac{8}{5}:\frac{1}{4}\\ =1\cdot\frac{16}{5}\cdot\frac{25}{8}-\frac{8}{5}\cdot4\\ =\frac{16}{5}\cdot\frac{25}{8}-\frac{32}{5}\\ =\frac{50}{5}-\frac{32}{5}\\ =\frac{18}{5}\)
e) \(\left(\frac{7}{8}-\frac{1}{2}\right)\cdot2\frac{2}{3}-\frac{3}{7}\cdot\left(2,5^2\right)\\ =\left(\frac{7}{8}-\frac{4}{8}\right)\cdot\frac{8}{3}-\frac{3}{7}\cdot6,25\\ =\frac{3}{8}\cdot\frac{8}{3}-\frac{3}{7}\cdot\frac{25}{4}\\ =1-\frac{75}{28}\\ =\frac{28}{28}-\frac{75}{28}\\ =\frac{-47}{28}\)
a, \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-2}{5}\right)\)
= \(\frac{2}{3}+\frac{-2}{15}=\frac{8}{15}\)
b, \(0,75.1\frac{7}{9}-1\frac{2}{5}:\frac{-21}{20}\)
= \(\frac{3}{4}.\frac{16}{9}-\frac{7}{5}.\frac{-20}{21}\)
= \(\frac{4}{3}-\left(\frac{-4}{3}\right)=\frac{8}{3}\)
c, \(\frac{-2}{17}+\frac{15}{23}+\frac{15}{-17}+\frac{4}{19}+\frac{8}{23}\)
= \(\left(\frac{-2}{17}+\frac{-15}{17}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\)
= \(\left(-1\right)+1+\frac{4}{19}=0+\frac{4}{19}=\frac{4}{19}\)
d, \(\left(6-2\frac{4}{5}\right).3\frac{1}{8}-1\frac{3}{5}:25\%\)
=> \(\left(6-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}:25\%\)
= \(\frac{16}{5}.\frac{25}{8}-\frac{8}{5}.25:100\)
= 10 - 0,4 = 9,6
e, \(\left(\frac{7}{8}-\frac{1}{2}\right).2\frac{2}{3}-\frac{3}{7}.\left(2,5^2\right)\)
=> \(\frac{3}{8}.\frac{8}{3}-\frac{3}{7}.6,25\)
= \(1-\frac{75}{28}=\frac{-47}{28}\)
Bài 1:
\(a,22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
=\(\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)
=\(\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)
=\(\frac{70}{4}+\frac{2}{4}-\frac{5}{4}\)
=\(\frac{67}{4}\)
\(b,1,4.\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{12}{15}+\frac{10}{15}\right):\frac{11}{5}\)
=\(\frac{3}{7}-\frac{22}{15}.\frac{5}{11}\)
=\(\frac{3}{7}-\frac{2}{3}\)
=\(-\frac{5}{21}\)
\(c,125\%.\left(-\frac{1}{2}\right)^2:\left(1\frac{5}{6}-1,6\right)+2016^0\)
=\(\frac{5}{4}.\frac{1}{4}:\left(\frac{11}{6}-\frac{8}{5}\right)+1\)
=\(\frac{5}{16}:\frac{7}{30}+1\)
=\(\frac{131}{56}\)
\(d,1,4.\frac{15}{49}-\left(20\%+\frac{2}{3}\right):2\frac{1}{5}\)
=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{1}{5}+\frac{2}{3}\right):\frac{11}{5}\)
=\(\frac{3}{7}-\frac{13}{15}:\frac{11}{5}\)
=\(\frac{3}{7}-\frac{13}{33}\)
=\(\frac{8}{231}\)
Bài đ làm giống hệt như bài c
Bài 2 :
\(a,\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)
=>\(\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}=1\\x=\frac{1}{4}:\frac{3}{4}=\frac{1}{3}\end{matrix}\right.\)
Vậy x ∈{1;\(\frac{1}{3}\)}
\(b,\frac{5}{3}.x-\frac{2}{5}.x=\frac{19}{10}\)
=>\(\frac{19}{15}.x=\frac{19}{10}\)
=>\(x=\frac{19}{10}:\frac{19}{15}=\frac{3}{2}\)
Vậy x ∈ {\(\frac{3}{2}\)}
c,\(\left|2.x-\frac{1}{3}\right|=\frac{2}{9}\)
=>\(\left[{}\begin{matrix}2.x-\frac{1}{3}=\frac{2}{9}\\2.x-\frac{1}{3}=-\frac{2}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2.x=\frac{2}{9}+\frac{1}{3}=\frac{5}{9}\\2.x=-\frac{2}{9}+\frac{1}{3}=\frac{1}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{5}{9}:2=\frac{5}{18}\\x=\frac{1}{9}:2=\frac{1}{18}\end{matrix}\right.\)
Vậy x∈{\(\frac{5}{18};\frac{1}{18}\)}
\(d,x-30\%.x=-1\frac{1}{5}\)
=\(70\%x=-\frac{6}{5}\)
=\(\frac{7}{10}.x=-\frac{6}{5}\)
=>\(x=-\frac{6}{5}:\frac{7}{10}=-\frac{12}{7}\)
Vậy x∈{\(-\frac{12}{7}\)}
Bài 2
a/
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{3}{4}\\\frac{3}{4}.x=\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}\\x=\frac{1}{4}:\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=\frac{1}{3}\)
b/ Đặt x làm thừa số chung rồi tính như bình thường
c/ Tương tự câu a
d/ Tương tự câu b
Bài 1:
d)
= \(\frac{-5}{9}\left(\frac{6}{13}+\frac{7}{13}\right)+\frac{5}{23}.\frac{7}{9}\)
= \(\frac{-5}{9}.1+\frac{35}{207}\)
= \(\frac{-80}{207}\)
Bài 2:
a) 20%x + 0,4x = 4,5
x( 20% + 0,4 ) = 4,5
x. 0,6 = 4,5
x = 4,5 : 0,6
x = 7,5