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HN
9 tháng 4 2017
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(-\dfrac{5}{6}x=\dfrac{5}{12}\)
\(x=-\dfrac{1}{2}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\)
\(3x-3.7=-\dfrac{19}{2}\)
\(3x=-5.8\)
\(x=-\dfrac{29}{15}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)
\(2+\dfrac{3}{4}x=\dfrac{21}{8}\)
\(\dfrac{3}{4}x=\dfrac{5}{8}\)
\(x=\dfrac{5}{6}\)
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=-\dfrac{3}{20}\)
R
30 tháng 3 2018
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
25 tháng 4 2017
Bài 1: a) Ta có : \(\dfrac{-3}{x}=\dfrac{x}{-27}\Leftrightarrow\left(-3\right).\left(-27\right)=x.x\Leftrightarrow81=x^2\Leftrightarrow9^2=x^2\Leftrightarrow x=9\)
b) Do \(\dfrac{2}{3}\) của x là -150 nên x là: (-150) : \(\dfrac{2}{3}\) = -225
c) \(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{2}-\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{18}\)
\(\Leftrightarrow x+2=18\)
\(\Leftrightarrow x=16\)
Bài 2:
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right)\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\)
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right).0\)
\(A=0\)
16 tháng 5 2018
3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)
=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)
=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)
=> \(x\) = \(\dfrac{2}{5}\)
16 tháng 5 2018
4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)
=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)
=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)
=> \(3x=\dfrac{1}{9}\)
=> \(x=\dfrac{1}{9}:3\)
=> \(x=\dfrac{1}{27}\)
Q
11 tháng 4 2017
a) \(\dfrac{1}{2}.\left(\dfrac{2}{9}+\dfrac{3}{7}-\dfrac{5}{27}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{41}{63}-\dfrac{5}{27}\right)\)
\(=\dfrac{1}{2}.\dfrac{88}{189}\)
\(=\dfrac{44}{189}\)
b) \(\left(\dfrac{-5}{28}+1,75+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\dfrac{-69}{20}\)
\(=\dfrac{-12}{23}\)
c) \(\dfrac{1}{3}.\dfrac{5}{7}-\dfrac{7}{27}.\dfrac{36}{14}\)
\(=\dfrac{5}{21}-\dfrac{7}{27}.\dfrac{36}{14}\)
\(=\dfrac{5}{21}-\dfrac{2}{3}\)
\(=\dfrac{-3}{7}\)
d) \(70,5-528:\dfrac{15}{2}\)
\(=70,5-\dfrac{352}{5}\)
\(=\dfrac{1}{10}\)
11 tháng 4 2017
em không trả lời được câu hỏi của chị nhưng chị có thể giúp em đăng bài toán lên bằng cách nào không
6 tháng 2 2018
bài này áp dụng quy tắc nhân chéo nha :vv
a) \(\dfrac{x}{5}=\dfrac{2}{5}\Leftrightarrow5x=2.5=10\Leftrightarrow x=\dfrac{10}{5}=2\)
b) \(\dfrac{3}{8}=\dfrac{6}{x}\Leftrightarrow3x=6.8=48\Leftrightarrow x=\dfrac{48}{3}=16\)
c)\(\dfrac{1}{9}=\dfrac{x}{27}\Leftrightarrow9x=27\Leftrightarrow x=\dfrac{27}{9}=3\)
d)\(\dfrac{4}{x}=\dfrac{8}{6}\Leftrightarrow8x=4.6=24\Leftrightarrow x=\dfrac{24}{8}=3\)
e) \(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Leftrightarrow3x+4x=-6+20\\ \Leftrightarrow7x=14\Leftrightarrow x=\dfrac{14}{7}=2\)
g) \(\dfrac{x}{-2}=\dfrac{-8}{x}\Leftrightarrow x^2=\left(-2\right)\left(-8\right)=16\\ \Rightarrow x=\pm4\)
MV
2 tháng 5 2017
2)
\(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}\\ D=\dfrac{3+1}{3}+\dfrac{3^2+1}{3^2}+\dfrac{3^3+1}{3^3}+...+\dfrac{3^{98}+1}{3^{98}}\\ D=\dfrac{3}{3}+\dfrac{1}{3}+\dfrac{3^2}{3^2}+\dfrac{1}{3^2}+\dfrac{3^3}{3^3}+\dfrac{1}{3^3}+...+\dfrac{3^{98}}{3^{98}}+\dfrac{1}{3^{98}}\\ D=1+\dfrac{1}{3}+1+\dfrac{1}{3^2}+1+\dfrac{1}{3^3}+...+1+\dfrac{1}{3^{98}}\\ D=\left(1+1+1+...+1\right)+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\\ D=98+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\)
Gọi \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\) là \(C\)
\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\\ 3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\\ 3C-C=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\\ 2C=1-\dfrac{1}{3^{98}}\\ C=\left(1-\dfrac{1}{3^{98}}\right):2\\ C=1:2-\dfrac{1}{3^{98}}:2\\ C=\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}\)
\(D=98+C=98+\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}=98\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}< 100\)
Vậy \(D< 100\)
\(\dfrac{x}{27}=\dfrac{2}{9}-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{x}{27}=\dfrac{2}{9}-\dfrac{3}{9}\)
\(\Leftrightarrow\dfrac{x}{27}=-\dfrac{1}{9}\)
\(\Leftrightarrow9x=-27\)
\(\Leftrightarrow x=-\dfrac{27}{9}\)
\(\Leftrightarrow x=-3\)
\(\dfrac{x}{27}=\dfrac{2}{9}-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{x}{27}=\dfrac{2}{9}-\dfrac{3}{9}\)
\(\Leftrightarrow\dfrac{x}{27}=-\dfrac{1}{9}\)
Suy ra :
\(9x=-27\)
\(\Rightarrow x=-3\)