\(F=\dfrac{1}{2\times\left(1+2\right)}+\dfrac{1}{2\times\left(1+2+3\right)}+....+\dfrac{1}{2\times\left(1+2+3+...+9\right)}\)
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MH
26 tháng 1 2022
\(A=\dfrac{-19}{9}.\dfrac{1}{2}-\dfrac{4}{11}.\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)=-\dfrac{23}{18}\)
\(B=\left(-\dfrac{15}{6}\right):\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}.\dfrac{-11}{2}=\dfrac{25}{4}\)
\(C=\dfrac{3}{4}.\left(-8\right)-\dfrac{1}{3}.\dfrac{-7}{2}-\dfrac{5}{18}=-\dfrac{46}{9}\)
26 tháng 1 2022
\(A=\dfrac{-19}{18}+\dfrac{4}{9}-\dfrac{2}{3}=\dfrac{-19}{18}+\dfrac{8}{18}-\dfrac{12}{18}=\dfrac{-23}{18}\)
\(B=\dfrac{-5}{2}\cdot\dfrac{-2}{1}-\dfrac{7}{12}+\dfrac{11}{6}=\dfrac{5\cdot12-7+22}{12}=\dfrac{75}{12}=\dfrac{25}{4}\)
24 tháng 8 2023
a) \(\dfrac{2}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{2}{3}\)
b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)
c) \(\left(\dfrac{1}{3}-\dfrac{2}{15}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
31 tháng 7 2022
1: \(A=\dfrac{-25}{27}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}=\dfrac{-2}{3}-\dfrac{2}{3}=\dfrac{-4}{3}\)
2: \(B=\dfrac{10.3-\left(9.5-4.5\right)\cdot2}{1.2-1.5}=\dfrac{10.3-10}{-0.3}=-1\)
c: \(=\dfrac{3}{49}\left(\dfrac{19}{2}-\dfrac{5}{2}\right)-\left(\dfrac{1}{20}-\dfrac{5}{20}\right)^2\cdot\left(\dfrac{-7}{14}-\dfrac{193}{14}\right)\)
\(=\dfrac{3}{49}\cdot7-\dfrac{1}{25}\cdot\dfrac{-200}{14}\)
\(=\dfrac{3}{7}+\dfrac{8}{14}=1\)
29 tháng 11 2022
1: =>|1/4x^2+1/45|=1/20
=>1/4x^2+1/45=1/20 hoặc 1/4x^2+1/45=-1/20
=>1/4x^2=1/36
=>x^2=1/36:1/4=1/9
=>x=1/3 hoặc x=-1/3
2: =(x^2-3)(x^2-2x)
=x(x-2)(x^2-3)
5 tháng 8 2018
Ta có:
\(B=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}\\ =\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}=\dfrac{72}{5}\)
Vậy B = \(\dfrac{72}{5}\)
23 tháng 12 2021
\(D=\left(-8\right).\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=\left(-8\right)\dfrac{1}{2}:\dfrac{13}{12}=\left(-4\right).\dfrac{12}{13}=-\dfrac{48}{13}\)
VN
24 tháng 7 2018
a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)
=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)
=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)
=> \(x=\dfrac{-1}{11}\)
HS
24 tháng 7 2018
Đây toán 8 mà? :v
a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)
\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)
\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)
\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)
\(\Leftrightarrow\left(11+1\right)x=0\)
\(\Leftrightarrow11x+1=0;x=0\)
\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)
Vậy....
PN
9 tháng 7 2017
a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
PN
9 tháng 7 2017
d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
