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20 tháng 12 2018
a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{2}{63}\)
20 tháng 12 2018
b) \(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}\)
Vậy.........
8 tháng 7 2019
\(b,\left(\sqrt{1\frac{9}{16}-\sqrt{\frac{9}{16}}}\right):5\)
\(=\left(\sqrt{\frac{25}{16}-\frac{3}{4}}\right):5\)
\(=\sqrt{\frac{13}{16}}:5\)
\(=\frac{\sqrt{13}}{4}:5\)
\(=\frac{\sqrt{13}}{20}\)
22 tháng 12 2016
\(\frac{5}{9}+\left[\frac{4.2}{6}-\frac{3.3}{6}\right]-\frac{\sqrt{4^2}}{9}=\frac{5}{9}+\frac{\left(-1\right)}{6}-\frac{4}{9}=\left(\frac{5}{9}-\frac{4}{9}\right)-\frac{1}{6}=\frac{1}{9}-\frac{1}{6}\)\(\frac{2}{18}-\frac{3}{18}=-\frac{1}{18}\)
22 tháng 12 2016
\(\frac{5}{9}+\left(\frac{8}{6}-\frac{9}{6}\right)-4\times\frac{1}{9}=\frac{-1}{18}\)
27 tháng 8 2021
\(\sqrt{\frac{1}{9}+\frac{1}{16}}\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}\)
1 tháng 2 2020
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
18 tháng 10 2018
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
18 tháng 10 2018
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
NN
3
3 tháng 6 2018
\(\left|x+\frac{1}{2}\right|-\frac{2}{3}=\sqrt{\frac{16}{9}}\)
\(\left|x+\frac{1}{2}\right|-\frac{2}{3}=\frac{4}{3}\)
\(\left|x+\frac{1}{2}\right|=\frac{4}{3}+\frac{2}{3}\)
\(\left|x+\frac{1}{2}\right|=2\)
TH1: \(x+\frac{1}{2}=2\)
\(x=2-\frac{1}{2}\)
\(x=\frac{3}{2}\)
TH2: \(x+\frac{1}{2}=-2\)
\(x=-2-\frac{1}{2}\)
\(x=\frac{-5}{2}\)
KL: x =3/2 hoặc x= -5/2
3 tháng 6 2018
\(\left|x+\frac{1}{2}\right|-\frac{2}{3}=\sqrt{\frac{16}{9}}\)
\(\Leftrightarrow\left|x+\frac{1}{2}\right|-\frac{2}{3}=\frac{4}{3}\)
\(\Leftrightarrow\left|x+\frac{1}{2}\right|=\frac{8}{3}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{8}{3}\\x+\frac{1}{2}=-\frac{8}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=-\frac{19}{6}\end{cases}}\)
vậy.....
\(\left|x+\frac{1}{2}\right|-\frac{2}{3}=\sqrt{\frac{16}{9}}\)
\(\Leftrightarrow\left|x+\frac{1}{2}\right|-\frac{2}{3}=\frac{4}{3}\)
\(\Leftrightarrow\left|x+\frac{1}{2}\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=2\\x+\frac{1}{2}=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{5}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{5}{2}\right\}\)
\(|x+\frac{1}{2}|-\frac{2}{3}=\sqrt{\frac{16}{9}}\)
<=> \(|x+\frac{1}{2}|-\frac{2}{3}=\frac{4}{3}\)
<=> \(|x+\frac{1}{2}|=\frac{4}{3}+\frac{2}{3}\)
<=> \(|x+\frac{1}{2}|=\frac{6}{3}=2\)
TH1: x + 1/2 = 2
x = 2 - 1/2 = 3/2
TH2: x + 1/2 = -2
x = -2 -1/2 = -5/2
Vậy:...