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\(\left(85\dfrac{7}{30}-83\dfrac{5}{18}\right):2\dfrac{2}{3}=0,01x:4\)
\(\Leftrightarrow[\left(85-83\right)+\left(\dfrac{7}{30}-\dfrac{5}{18}\right)]:2\dfrac{2}{3}=0,01x:4\)
\(\Leftrightarrow1\dfrac{43}{45}:2\dfrac{2}{3}=0,01x:4\)
\(\Leftrightarrow\dfrac{88}{45}:\dfrac{8}{3}=0,01x:4\)
\(\Leftrightarrow\dfrac{88}{45}.\dfrac{3}{8}=0,01x:4\)
\(\Leftrightarrow\dfrac{11}{15}=0,01x:4\)
\(\Leftrightarrow0,01x:4=\dfrac{11}{15}\)
\(\Leftrightarrow x=\dfrac{11}{15}.4:0,01\)
\(\Leftrightarrow x=\dfrac{880}{3}\)
Vậy x = \(\dfrac{880}{3}\)
Ta có:
\(\left(85\dfrac{7}{30}-83\dfrac{5}{18}\right):2\dfrac{2}{3}=0.01x:4\\ \left(\left(85-83\right)+\left(\dfrac{7}{30}-\dfrac{5}{18}\right)\right):\dfrac{8}{3}=0.01x:4\\ \dfrac{88}{45}:\dfrac{8}{3}=0.01x:4\\ \dfrac{11}{15}=0.01x:4\\ \dfrac{44}{15}=0.01x\\ x=\dfrac{44}{15}:0.01\\ x=\dfrac{880}{3}\)
Vậy \(x=\dfrac{880}{3}\)
\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)
\(=\dfrac{-1}{8}\)
\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{71}{35}\)
\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)
\(=\left(-3\right)+\left(\dfrac{-5}{2}\right)+\left(\dfrac{-7}{5}\right)+\dfrac{5}{7}\)
\(=\dfrac{-433}{70}\)
a: \(=\dfrac{3}{8}\left(72+\dfrac{1}{5}-51-\dfrac{1}{5}\right)=\dfrac{3}{8}\cdot21=\dfrac{63}{8}\)
b: \(=25\cdot\dfrac{-1}{125}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{2}-\dfrac{1}{2}=-1\)
c: \(=4\left(35+\dfrac{1}{6}\right)\cdot\dfrac{-1}{5}-\left(45+\dfrac{1}{6}\right)\cdot\dfrac{-1}{5}\)
\(=\dfrac{-1}{5}\left(140+\dfrac{2}{3}-45-\dfrac{1}{6}\right)=-\dfrac{191}{10}\)
\(\dfrac{\text{45^{10^{ }}}.5^{10}}{75^{10}}=\dfrac{9^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}=\dfrac{9^{10}}{3^{10}}=3^{10}\)
\(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}=\dfrac{2^5.\left(0,4\right)^5}{\left(0,4\right)^6}=\dfrac{2^5}{0,4}=\dfrac{32}{0,4}=80\)
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: \(=\dfrac{-5}{4}\left(35+\dfrac{1}{6}-45-\dfrac{1}{6}\right)=\dfrac{-5}{4}\cdot\left(-10\right)=\dfrac{50}{4}=\dfrac{25}{2}\)
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. \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)+\left(\dfrac{-5}{41}-\dfrac{36}{41}\right)+0,5\)
\(=1+\left(-1\right)+0,5\)
\(=0,5\)
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. \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2\)
\(=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)
Mình chỉ cho đáp án thôi,sai thì châm chước cho mìn nha!
a)-91 phần200
b)-25phần 4
c)5 phần 2
d)2
e)0
a, ( 0,36-2,18) : ( 3,8 + 0,2)
= -1,82 : 4
=-0,455 hay -91/200
b, 3/8*19/1/3-3/8*33/1/3
=3/8*(19/1/3-33/1/3)
=3/8*(-14)
=-21/4
\(\dfrac{-2}{45}:\dfrac{8}{3}=0,01x:4\)
\(0,01x:4=\dfrac{-2}{45}\times\dfrac{3}{8}\)
\(0,01x:4=\dfrac{-6}{360}\)
\(0,01x:4=\dfrac{-1}{60}\)
\(0,01x=\dfrac{-1}{60}\times\dfrac{1}{4}\)
\(0,01x=\dfrac{-1}{240}\)
\(x=\dfrac{-1}{240}:0,01\)
\(x=\dfrac{-1}{240}\times100\)
\(x=\dfrac{-100}{240}=-2,4\)
\(\dfrac{-2}{45}:\dfrac{8}{3}=0,01x:4\)
\(\dfrac{-2}{45}\times\dfrac{3}{8}=0,01x:4\)
\(\dfrac{-6}{360}=0,01x:4\)
\(\dfrac{-1}{60}=0,01x:4\)
\(0,01x=\dfrac{-1}{60}\times\dfrac{4}{1}\)
\(0,01x=\dfrac{-4}{60}\)
\(0,01x=\dfrac{-1}{15}\)
\(x=\dfrac{-1}{15}:0,01\)
\(x=\dfrac{-1}{15}\times100\)
\(x=\dfrac{-100}{15}=\dfrac{-20}{3}\)
Vậy \(x=\dfrac{-20}{3}\).